# Continuous Maps and Homeomorphisms: Continuity Recast through Preimages of Open Sets
> Continuity is defined by requiring preimages of open sets to be open; we prove this agrees with ε-δ continuity on metric spaces, and compute in full why R is homeomorphic to (0,1) but not to [0,1].
> https://rikai.mugen-giken.com/en/mathematics/topology/continuous-maps
## 0. Key points
- Continuity can be defined by a single requirement: that the **preimage** of every open set of the codomain be open in the domain. No distance, no $\varepsilon$, no $\delta$ is needed.
- For maps between metric spaces this definition agrees **exactly** with continuity in the $\varepsilon$-$\delta$ sense (). The order of the quantifiers in the $\varepsilon$-$\delta$ definition translates directly into the language of open sets.
- We use preimages rather than images because taking preimages commutes with unions, intersections and complements. A continuous map need not send open sets to open sets ().
- A homeomorphism is a continuous bijection **whose inverse is also continuous**. This last condition is independent of the others: drop it and the notion collapses ().
- Two homeomorphic spaces share every property that can be stated using the topology alone. Conversely, one proves that two spaces are *not* homeomorphic by exhibiting a topological property possessed by only one of them.
- We have $\mathbb{R} \cong (0,1)$ but $\mathbb{R} \not\cong [0,1]$. The reason, however, is **not** boundedness: boundedness is not a topological property ( and the Appendix).
## 1. Motivation: does continuity survive the loss of distance?
The definition of continuity one meets first in calculus goes back to Cauchy and Weierstrass and reads as follows. A function $f : \mathbb{R} \to \mathbb{R}$ is continuous at a point $a$ if
$$
\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x \in \mathbb{R},\quad |x - a| < \delta \implies |f(x) - f(a)| < \varepsilon .
$$
One may read this as a contract: if you want the error away from $f(a)$ to stay below $\varepsilon$, then keep the distance from $a$ below $\delta$.
But the definition depends entirely on the absolute value $|x - a|$ — that is, on a **distance**. Does this mean that continuity cannot be discussed where no distance is given in advance? In fact, situations with no distance, or with no canonical choice of one, occur throughout mathematics. The Zariski topology on the zero set of a family of polynomials, the quotient of a group by a subgroup, the topology of pointwise convergence on a set of functions — none of these is easy, or even possible, to define by a metric. Yet we speak freely of "continuous homomorphisms" and "continuous actions".
So let us ask the question again: which part of the $\varepsilon$-$\delta$ definition is genuinely needed? In the display above, the condition $|f(x) - f(a)| < \varepsilon$ says that $f(x)$ lies in the open ball $B(f(a), \varepsilon)$ centred at $f(a)$, and $|x - a| < \delta$ says that $x$ lies in the open ball $B(a,\delta)$. The definition therefore becomes
$$
\forall \varepsilon > 0,\ \exists \delta > 0,\quad f\bigl(B(a,\delta)\bigr) \subset B\bigl(f(a), \varepsilon\bigr) .
$$
Here the specific shape "ball of radius $\varepsilon$" is no longer used. What is used is only the relation **"given an open set containing $f(a)$, return an open set containing $a$"**. Discard the numerical datum of a radius and keep only the family of open sets, and this condition can still be written down. That is the definition of continuity on a topological space.
We can go one step further. The global condition "continuous at every point" collapses into a strikingly compact form that requires no point-by-point inspection: the preimage of every open set is open. Why preimages and not images, and why this one line suffices — the article begins by checking these two matters carefully.
Once continuous maps are available, we can define what it means for two topological spaces to be isomorphic. Isomorphisms of sets are bijections; isomorphisms of groups are bijective homomorphisms. Isomorphisms of topological spaces are **homeomorphisms**: continuous bijections whose inverses are continuous as well. The familiar slogan that a coffee cup and a doughnut are the same object says precisely that this notion of isomorphism cannot tell them apart. Turning the slogan into a theorem is the goal of this article.
## 2. Preliminaries: notation and the algebra of preimages
We use the definition of a topological space given in [Topological spaces: definitions and basic notions](/en/mathematics/topology/topological-spaces) (). Let us fix the notation.
A pair $(X, \mathcal{O}_X)$ consisting of a set $X$ and a family $\mathcal{O}_X \subset 2^{X}$ of subsets is a **topological space** if
- (O1) $\emptyset \in \mathcal{O}_X$ and $X \in \mathcal{O}_X$;
- (O2) the union of any number (possibly infinitely many) of members of $\mathcal{O}_X$ again belongs to $\mathcal{O}_X$;
- (O3) the intersection of finitely many members of $\mathcal{O}_X$ again belongs to $\mathcal{O}_X$.
The members of $\mathcal{O}_X$ are called **open sets**. A subset whose complement is open is called **closed**. The **closure** $\overline{A}$ of $A \subset X$ is the intersection of all closed sets containing $A$; it is the smallest closed set containing $A$. When no confusion can arise we suppress $\mathcal{O}_X$ and speak simply of "the topological space $X$".
A subset $A \subset X$ carries the **subspace topology** $\mathcal{O}_A = \{\, U \cap A : U \in \mathcal{O}_X \,\}$. Throughout, subsets of $\mathbb{R}$ (intervals, in particular) always carry the subspace topology inherited from the usual topology of $\mathbb{R}$.
For a map $f : X \to Y$ and $B \subset Y$ we define the **preimage**
$$
f^{-1}(B) := \{\, x \in X : f(x) \in B \,\} .
$$
Here $f^{-1}$ is a piece of notation; it does not presuppose that $f$ is invertible. The reason preimages are so convenient is contained in the following three identities. Let $\{B_\lambda\}_{\lambda \in \Lambda}$ be a family of subsets of $Y$ and $B \subset Y$. Then
$$
f^{-1}\Bigl(\bigcup_{\lambda} B_\lambda\Bigr) = \bigcup_{\lambda} f^{-1}(B_\lambda), \qquad
f^{-1}\Bigl(\bigcap_{\lambda} B_\lambda\Bigr) = \bigcap_{\lambda} f^{-1}(B_\lambda), \qquad
f^{-1}(Y \setminus B) = X \setminus f^{-1}(B) .
$$
Each is proved by unwinding the definitions. For the first, say: $x \in f^{-1}(\bigcup_\lambda B_\lambda)$ $\iff$ $f(x) \in \bigcup_\lambda B_\lambda$ $\iff$ $f(x) \in B_\lambda$ for some $\lambda$ $\iff$ $x \in f^{-1}(B_\lambda)$ for some $\lambda$ $\iff$ $x \in \bigcup_\lambda f^{-1}(B_\lambda)$. For the third: $x \in f^{-1}(Y \setminus B)$ $\iff$ $f(x) \notin B$ $\iff$ $x \notin f^{-1}(B)$.
**Images**, by contrast, have no such property. Take $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^{2}$, and $A = [-1, 0]$, $B = [0,1]$. Then $A \cap B = \{0\}$, so $f(A \cap B) = \{0\}$; but $f(A) = f(B) = [0,1]$, so $f(A) \cap f(B) = [0,1]$, and the two do not agree. In general one has only $f(A \cap B) \subset f(A) \cap f(B)$, and for complements not even an inclusion holds in either direction. Since the axioms (O2) and (O3) are conditions about unions and intersections, preimages — which commute with both — are the better fit.
Finally we recall the notion of a **base** (). A family $\mathcal{B} \subset \mathcal{O}_Y$ is a **base** of the topology $\mathcal{O}_Y$ if every open set $V \in \mathcal{O}_Y$ can be written as a union of members of $\mathcal{B}$. For instance the family of all open intervals $\{\, (a,b) : a < b \,\}$ is a base for the usual topology of $\mathbb{R}$, and in a metric space $(Y, d)$ the family of all open balls $\{\, B_d(y, r) : y \in Y,\ r > 0 \,\}$ is a base.
## 3. Continuous maps: definition and equivalent conditions
### 3.1. The definition
Let $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ be topological spaces. A map $f : X \to Y$ is **continuous** if
$$
\forall V \in \mathcal{O}_Y,\quad f^{-1}(V) \in \mathcal{O}_X ,
$$
that is, if the preimage of every open subset of $Y$ is open in $X$.
Note that both topologies, that of $X$ and that of $Y$, occur in the definition. Continuity is not a property of the map $f$ alone but of the **triple** consisting of $f$ together with the topologies chosen on the domain and codomain. The same map of sets may be continuous for one choice and not for another ().
Let $f : X \to Y$ be a map and $x \in X$. We say $f$ is **continuous at the point $x$** if for every open set $V \subset Y$ containing $f(x)$ there exists an open set $U \subset X$ containing $x$ with $f(U) \subset V$.
### 3.2. Equivalent conditions
The next theorem is the foundation of the article. Thanks to it we may always pick whichever condition is most convenient. When open sets are awkward to handle — when we want to estimate a closure, say — condition (iv) serves; for arguments carried out point by point, condition (iii) does.
Let $X$ and $Y$ be topological spaces and $f : X \to Y$ a map. The following four conditions are equivalent.
- **(i)** For every open set $V \subset Y$, the set $f^{-1}(V)$ is open in $X$ (that is, $f$ is continuous).
- **(ii)** For every closed set $F \subset Y$, the set $f^{-1}(F)$ is closed in $X$.
- **(iii)** $f$ is continuous at every point of $X$ in the sense of .
- **(iv)** For every subset $A \subset X$ we have $f(\overline{A}) \subset \overline{f(A)}$.
(i) $\Rightarrow$ (iii). Let $x \in X$ and let $V \subset Y$ be an open set containing $f(x)$. Put $U := f^{-1}(V)$; by (i) this is open in $X$. Since $f(x) \in V$ we have $x \in U$, and moreover $f(U) = f(f^{-1}(V)) \subset V$ (if $u \in f^{-1}(V)$ then $f(u) \in V$ by definition). Hence the condition of holds at $x$. As $x$ was arbitrary, (iii) follows.
(iii) $\Rightarrow$ (i). Let $V \subset Y$ be open. If $f^{-1}(V) = \emptyset$, it is open by (O1). Otherwise, for each $x \in f^{-1}(V)$ we have $f(x) \in V$, and $V$ is an open set containing $f(x)$; so by (iii) there is an open set $U_x \subset X$ containing $x$ with $f(U_x) \subset V$. The inclusion $f(U_x) \subset V$ is equivalent to $U_x \subset f^{-1}(V)$. Consequently
$$
f^{-1}(V) = \bigcup_{x \in f^{-1}(V)} \{x\} \subset \bigcup_{x \in f^{-1}(V)} U_x \subset f^{-1}(V) ,
$$
so $f^{-1}(V) = \bigcup_{x} U_x$. This is a union of open sets, hence open by (O2).
(i) $\Leftrightarrow$ (ii). Let $F \subset Y$ be closed, so that $Y \setminus F$ is open; by the complement rule for preimages from §2,
$$
f^{-1}(Y \setminus F) = X \setminus f^{-1}(F) .
$$
Assuming (i), the left-hand side is open, so $f^{-1}(F)$ is closed and (ii) follows. Conversely, assume (ii) and let $V \subset Y$ be open. Then $Y \setminus V$ is closed, so $f^{-1}(Y \setminus V) = X \setminus f^{-1}(V)$ is closed, whence $f^{-1}(V)$ is open and (i) follows. We used here that the correspondence "$F$ closed $\iff$ $Y \setminus F$ open" is a bijection between the closed sets and the open sets.
(ii) $\Rightarrow$ (iv). Let $A \subset X$ be arbitrary and put $C := \overline{f(A)}$. Since $C$ is closed in $Y$, (ii) gives that $f^{-1}(C)$ is closed in $X$. From $f(A) \subset C$ we get $A \subset f^{-1}(f(A)) \subset f^{-1}(C)$. Now $\overline{A}$ is the smallest closed set containing $A$, and $f^{-1}(C)$ is a closed set containing $A$, so $\overline{A} \subset f^{-1}(C)$, that is, $f(\overline{A}) \subset C = \overline{f(A)}$.
(iv) $\Rightarrow$ (ii). Let $F \subset Y$ be closed and put $A := f^{-1}(F)$. Then $f(A) = f(f^{-1}(F)) \subset F$, and since $F$ is closed, $\overline{f(A)} \subset \overline{F} = F$. By (iv),
$$
f(\overline{A}) \subset \overline{f(A)} \subset F ,
$$
and passing to preimages on both sides gives $\overline{A} \subset f^{-1}(F) = A$. As always $A \subset \overline{A}$, so $\overline{A} = A$: the set $A = f^{-1}(F)$ is closed.
We have proved (i) $\Leftrightarrow$ (iii), (i) $\Leftrightarrow$ (ii) and (ii) $\Leftrightarrow$ (iv), so all four conditions are equivalent.
Condition (iv) is the topological version of the intuition that $f$ does not destroy limits. It says that if $x$ is an adherent point of $A$ (a point of $\overline{A}$), then $f(x)$ is an adherent point of $f(A)$. Translated into the language of sequences it corresponds to "$x_n \to x$ implies $f(x_n) \to f(x)$"; but in a general topological space sequences alone do not characterise continuity, so the closure formulation is the correct generalisation.
### 3.3. Testing on a base, and composition
Inspecting all open sets is laborious. In practice it suffices to inspect a base.
Let $X$ and $Y$ be topological spaces and let $\mathcal{B}$ be a base for the topology of $Y$. A map $f : X \to Y$ is continuous if and only if $f^{-1}(B)$ is open in $X$ for every $B \in \mathcal{B}$.
Necessity is immediate: $\mathcal{B} \subset \mathcal{O}_Y$, so if $f$ is continuous we simply apply to $B \in \mathcal{B}$.
For sufficiency, let $V \subset Y$ be any open set. Since $\mathcal{B}$ is a base, there are an index set $\Lambda$ and members $B_\lambda \in \mathcal{B}$ $(\lambda \in \Lambda)$ with $V = \bigcup_{\lambda \in \Lambda} B_\lambda$. By the commutation of preimages with unions from §2,
$$
f^{-1}(V) = f^{-1}\Bigl(\bigcup_{\lambda} B_\lambda\Bigr) = \bigcup_{\lambda} f^{-1}(B_\lambda) .
$$
By hypothesis each $f^{-1}(B_\lambda)$ is open, so by (O2) their union is open. Hence $f$ is continuous.
Let $X$, $Y$, $Z$ be topological spaces.
1. The identity map $\mathrm{id}_X : X \to X$ is continuous.
2. If $f : X \to Y$ and $g : Y \to Z$ are both continuous, then the composite $g \circ f : X \to Z$ is continuous.
(1) For every open set $U \subset X$ we have $\mathrm{id}_X^{-1}(U) = U$, which is open.
(2) First we verify the set identity $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$: indeed $x \in (g\circ f)^{-1}(W)$ $\iff$ $g(f(x)) \in W$ $\iff$ $f(x) \in g^{-1}(W)$ $\iff$ $x \in f^{-1}(g^{-1}(W))$. Now let $W \subset Z$ be open. By continuity of $g$ () the set $g^{-1}(W)$ is open in $Y$, and then by continuity of $f$ the set $f^{-1}(g^{-1}(W))$ is open in $X$. By the identity above this equals $(g\circ f)^{-1}(W)$, so $g \circ f$ is continuous.
is exactly the statement that topological spaces as objects and continuous maps as morphisms form a category $\mathbf{Top}$. What its isomorphisms are is the subject of §5.
Let $Y$ be a topological space, give $A \subset Y$ the subspace topology, and let $\iota : A \to Y$ be the inclusion map. The following two facts will be used repeatedly in the examples below.
**(1) The inclusion is continuous.** If $V \subset Y$ is open then $\iota^{-1}(V) = V \cap A$, which is open in $A$ by the definition of the subspace topology.
**(2) The codomain may be replaced by a subspace.** Let $X$ be a topological space and $f : X \to A$ a map. Then $f$ is continuous if and only if $\iota \circ f : X \to Y$ is continuous. Indeed, the open sets of $A$ are those of the form $V \cap A$ with $V$ open in $Y$, and the values of $f$ always lie in $A$, so
$$
f^{-1}(V \cap A) = \{\, x \in X : f(x) \in V \ \text{and}\ f(x) \in A \,\} = \{\, x \in X : f(x) \in V \,\} = (\iota \circ f)^{-1}(V) .
$$
Hence the condition "the $f$-preimage of every open set of $A$ is open" and the condition "the $(\iota \circ f)$-preimage of every open set of $Y$ is open" are, literally, conditions about the same family of sets.
In short: "continuous maps into $\mathbb{R}$ whose values lie in $(0,1)$" and "continuous maps into $(0,1)$" are the same thing.
### 3.4. Examples
Give a set $X$ the **discrete topology** $\mathcal{O}_X = 2^{X}$ (every subset is open). Then for any topological space $Y$ and any map $f : X \to Y$, the set $f^{-1}(V)$ is a subset of $X$ and hence automatically open. So $f$ is continuous.
Give a set $Y$ the **indiscrete topology** $\mathcal{O}_Y = \{\emptyset, Y\}$. Then for any map $f : X \to Y$ we have $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(Y) = X$, both open in $X$ by (O1). So $f$ is continuous.
A constant map $f \equiv y_0$ is always continuous. If $V \subset Y$ is open then $f^{-1}(V) = X$ when $y_0 \in V$ and $f^{-1}(V) = \emptyset$ when $y_0 \notin V$, and both are open by (O1).
On the other hand, put the usual topology on $X = \mathbb{R}$ and the discrete topology on $Y = \mathbb{R}$. Then the identity map $\mathrm{id} : X \to Y$ is not continuous: the set $\{0\}$ is open in $Y$, but $\mathrm{id}^{-1}(\{0\}) = \{0\}$ is not open in the usual topology of $\mathbb{R}$ (any open interval $(-\delta, \delta)$ containing $0$ also contains points other than $0$, so it is not contained in $\{0\}$). The same map of sets loses continuity once the topology is changed.
The map $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^{2}$, is continuous (this is checked in ). But the image of the open set $(-1,1)$ is
$$
f\bigl((-1,1)\bigr) = \{\, x^{2} : |x| < 1 \,\} = [0, 1) .
$$
The point $0$ belongs to this image, yet for every $\delta > 0$ we have $-\delta/2 \in (-\delta, \delta)$ and $-\delta/2 \notin [0,1)$, so no neighbourhood of $0$ is contained in $[0,1)$. Hence $[0,1)$ is not open and $f$ is not an open map.
Next consider $g : \mathbb{R} \to \mathbb{R}$, $g(x) = 1/(1 + x^{2})$. The denominator never vanishes, since $1 + x^{2} \ge 1 > 0$, so this is a continuous rational function. The set $\mathbb{R}$ is closed in $\mathbb{R}$; but $x^{2}$ ranges over all of $[0,\infty)$, so $1 + x^{2}$ ranges over $[1,\infty)$ and therefore
$$
g(\mathbb{R}) = \Bigl\{\, \tfrac{1}{1+x^{2}} : x \in \mathbb{R} \,\Bigr\} = (0, 1] .
$$
Since $1/n \in (0,1]$ while $1/n \to 0 \notin (0,1]$, the set $(0,1]$ is not closed. Hence $g$ is not a closed map either.
These two examples show that continuity is not a condition about images.
Let us prove that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^{2}$, is continuous without using $\varepsilon$ or $\delta$, by appealing to . Take the family of all open intervals as a base for the usual topology of $\mathbb{R}$ and compute $f^{-1}((a,b)) = \{\, x \in \mathbb{R} : a < x^{2} < b \,\}$ for $a < b$, distinguishing cases.
**Case $b \le 0$.** Since $x^{2} \ge 0 \ge b$, no $x$ satisfies $x^{2} < b$. Hence $f^{-1}((a,b)) = \emptyset$, which is open by (O1).
**Case $a < 0 < b$.** The inequality $a < x^{2}$ holds automatically because $x^{2} \ge 0 > a$, so the condition reduces to $x^{2} < b$. This is equivalent to $|x| < \sqrt{b}$, so $f^{-1}((a,b)) = (-\sqrt{b}, \sqrt{b})$, which is open.
**Case $0 \le a < b$.** The map $t \mapsto \sqrt{t}$ is strictly increasing on $[0,\infty)$, so $a < x^{2} < b$ is equivalent to $\sqrt{a} < |x| < \sqrt{b}$. Hence
$$
f^{-1}\bigl((a,b)\bigr) = (-\sqrt{b}, -\sqrt{a}) \cup (\sqrt{a}, \sqrt{b}) ,
$$
a union of two open intervals and therefore open by (O2). When $a = 0$ this becomes $(-\sqrt{b},0) \cup (0,\sqrt{b})$.
In every case the preimage of a basic open set is open, so $f$ is continuous by . Observe that we never had to write down a single $\varepsilon$ or $\delta$.
## 4. Agreement with the ε-δ definition
We now prove that generalises the notion of continuity from calculus. First we recall the topology of a metric space.
In a metric space $(X, d)$, for $a \in X$ and $r > 0$ we define the **open ball** $B_d(a, r) := \{\, x \in X : d(a,x) < r \,\}$, and we call
$$
\mathcal{O}_d := \{\, U \subset X : \forall a \in U,\ \exists r > 0,\ B_d(a,r) \subset U \,\}
$$
the **metric topology** determined by $d$. That $\mathcal{O}_d$ satisfies the axioms of a topology (), and that the family of all open balls is a base for $\mathcal{O}_d$, were verified in the previous article. Let us at least repeat the proof that open balls are open. Let $x \in B_d(a,r)$ and put $s := r - d(a,x) > 0$. For $y \in B_d(x,s)$ the triangle inequality gives
$$
d(a,y) \le d(a,x) + d(x,y) < d(a,x) + s = r ,
$$
so $B_d(x,s) \subset B_d(a,r)$. Hence $B_d(a,r) \in \mathcal{O}_d$.
Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, each with its metric topology. For a map $f : X \to Y$ the following two conditions are equivalent.
- **(i)** $f$ is continuous in the sense of ; that is, $f^{-1}(V)$ is open in $X$ for every open set $V \subset Y$.
- **(ii)** For every $a \in X$ and every $\varepsilon > 0$ there exists $\delta > 0$ such that $d_Y(f(x), f(a)) < \varepsilon$ holds for all $x \in X$ with $d_X(x, a) < \delta$.
(i) $\Rightarrow$ (ii). Let $a \in X$ and $\varepsilon > 0$ be arbitrary and put $V := B_{d_Y}(f(a), \varepsilon)$. As seen above, $V$ is open in $Y$, so $f^{-1}(V)$ is open in $X$ by (i). Since $d_Y(f(a), f(a)) = 0 < \varepsilon$ we have $f(a) \in V$, hence $a \in f^{-1}(V)$. Applying the definition of the metric topology $\mathcal{O}_{d_X}$ to $U = f^{-1}(V)$ and $a$, we obtain $\delta > 0$ with $B_{d_X}(a, \delta) \subset f^{-1}(V)$. This says precisely that $d_X(x,a) < \delta$ implies $f(x) \in V = B_{d_Y}(f(a), \varepsilon)$, that is, $d_Y(f(x), f(a)) < \varepsilon$.
(ii) $\Rightarrow$ (i). Let $V \subset Y$ be open and let $a \in f^{-1}(V)$ be arbitrary. From $f(a) \in V$ and the definition of the metric topology there is $\varepsilon > 0$ with $B_{d_Y}(f(a), \varepsilon) \subset V$. Applying (ii) to this $\varepsilon$ and to $a$ yields $\delta > 0$ such that $d_X(x,a) < \delta$ implies $d_Y(f(x), f(a)) < \varepsilon$, that is, $f(x) \in B_{d_Y}(f(a),\varepsilon) \subset V$. In other words $B_{d_X}(a, \delta) \subset f^{-1}(V)$. Since $a \in f^{-1}(V)$ was arbitrary, the definition of the metric topology shows that $f^{-1}(V)$ is open in $X$. Hence $f$ is continuous.
Reading the proof again makes it clear where each part of the $\varepsilon$-$\delta$ definition went.
| $\varepsilon$-$\delta$ language | Topological language |
|---|---|
| for every $\varepsilon > 0$ | for every open set $V$ containing $f(a)$ |
| there exists $\delta > 0$ | there exists an open set $U$ containing $a$ |
| $d_X(x,a) < \delta$ | $x \in U$ |
| $d_Y(f(x), f(a)) < \varepsilon$ | $f(x) \in V$ |
The order of quantifiers $\forall \varepsilon\, \exists \delta\, \forall x$ carries over verbatim to $\forall V\, \exists U\, \forall x \in U$. What the topology discards is only the numerical value $\varepsilon$; the logical structure — a target accuracy is prescribed first, and a tolerance is chosen afterwards — is preserved completely.
The condition that $\delta$ may be chosen independently of $a$ (uniform continuity) cannot be expressed by this translation. The set $U$ is chosen for each $a$ separately, and topology has no way of saying "a $U$ of the same size". See [Continuous functions and uniform continuity](/en/mathematics/real-analysis/uniform-continuity) and the Appendix of this article for details.
## 5. Homeomorphisms: what it means to be topologically the same
### 5.1. Definition and criteria
Let $f : X \to Y$ be a map between topological spaces. We call $f$ an **open map** if $f(U)$ is open in $Y$ for every open set $U \subset X$, and a **closed map** if $f(F)$ is closed in $Y$ for every closed set $F \subset X$.
As shows, being continuous, being an open map and being a closed map are mutually independent conditions. A homeomorphism requires continuity and (as a consequence) openness at the same time.
Let $X$ and $Y$ be topological spaces. A map $f : X \to Y$ is a **homeomorphism** if
1. $f$ is a bijection,
2. $f$ is continuous, and
3. the inverse map $f^{-1} : Y \to X$ is continuous as well.
If a homeomorphism $f : X \to Y$ exists we say that $X$ and $Y$ are **homeomorphic** and write $X \cong Y$.
That condition 3 cannot be dropped is verified in . For groups and vector spaces the inverse of a bijective homomorphism is automatically a homomorphism, so no analogue of condition 3 was needed there. That this fails for topological spaces is the first surprise.
Let $X$ and $Y$ be topological spaces and $f : X \to Y$ a continuous bijection. The following three conditions are equivalent.
- **(i)** $f$ is a homeomorphism.
- **(ii)** $f$ is an open map.
- **(iii)** $f$ is a closed map.
Since $f$ is a bijection, the inverse map $g := f^{-1} : Y \to X$ is defined. We first prove the key identity
$$
g^{-1}(A) = f(A) \qquad (A \subset X \text{ arbitrary}) .
$$
If $y \in g^{-1}(A)$ then $g(y) \in A$ and $y = f(g(y)) \in f(A)$. Conversely, if $y \in f(A)$ then $y = f(a)$ for some $a \in A$, and $g(y) = g(f(a)) = a \in A$, so $y \in g^{-1}(A)$. This proves the identity.
(i) $\Leftrightarrow$ (ii). By , continuity of $g$ means that $g^{-1}(U)$ is open in $Y$ for every open set $U \subset X$. Taking $A = U$ in the identity above gives $g^{-1}(U) = f(U)$, so this condition says that $f(U)$ is open for every open $U$, that is, that $f$ is an open map. Since $f$ is by hypothesis a continuous bijection, continuity of $g$ is equivalent to $f$ being a homeomorphism.
(ii) $\Leftrightarrow$ (iii). Since $f$ is a bijection, $f(X \setminus A) = Y \setminus f(A)$ holds for every $A \subset X$. Indeed, for $y \in Y$ bijectivity of $f$ gives a unique $x_0 \in X$ with $y = f(x_0)$, and $y \in f(X\setminus A) \iff x_0 \notin A \iff y \notin f(A)$ (the second equivalence again uses uniqueness of $x_0$). Now assume (ii) and let $F \subset X$ be closed. Then $X \setminus F$ is open, so $f(X\setminus F) = Y \setminus f(F)$ is open, hence $f(F)$ is closed and (iii) follows. Conversely, if $U \subset X$ is open then $X\setminus U$ is closed, so $f(X \setminus U) = Y \setminus f(U)$ is closed and therefore $f(U)$ is open.
On any collection of topological spaces, the relation "$X \cong Y$" is reflexive, symmetric and transitive.
Reflexivity. The map $\mathrm{id}_X$ is a bijection, it is continuous by part (1) of , and its inverse is again $\mathrm{id}_X$, hence continuous. So $\mathrm{id}_X$ is a homeomorphism and $X \cong X$.
Symmetry. Let $f : X \to Y$ be a homeomorphism. Then $g := f^{-1}$ is a bijection, continuous by condition 3 of , and $g^{-1} = f$ is continuous by condition 2. So $g$ is a homeomorphism and $Y \cong X$.
Transitivity. Let $f : X \to Y$ and $h : Y \to Z$ be homeomorphisms. Then $h \circ f$ is a composite of bijections, hence a bijection, and continuous by part (2) of . Moreover $(h\circ f)^{-1} = f^{-1} \circ h^{-1}$, and both $h^{-1}$ and $f^{-1}$ are continuous, so part (2) of again shows that $(h\circ f)^{-1}$ is continuous. Hence $h \circ f$ is a homeomorphism and $X \cong Z$.
For equivalence relations in general see [Relations and equivalence relations: what does "the same" mean?](/mathematics/foundations/equivalence-relations). By , the enterprise of classifying topological spaces up to homeomorphism is meaningful. A large part of topology is devoted to manufacturing tools that distinguish these equivalence classes.
A property $P$ of topological spaces is a **topological property** (a topologically invariant property) if $X \cong Y$ and $X$ has $P$ together imply that $Y$ has $P$.
Any property defined using only the family of open sets is, by definition, a topological property. A homeomorphism $f$ induces via $U \mapsto f(U)$ a bijection between $\mathcal{O}_X$ and $\mathcal{O}_Y$ (by part (ii) of and its converse direction), so a statement phrased in the language of open sets holds in $X$ if and only if it holds in $Y$. Compactness ([Compactness](/mathematics/topology/compactness)), connectedness ([Connectedness](/mathematics/topology/connectedness)) and the Hausdorff property ([Separation axioms and metrisability](/mathematics/topology/separation-axioms)) are all topological properties.
### 5.2. Examples
Define $h : \mathbb{R} \to (0,1)$ by
$$
h(x) := \frac{1}{2} + \frac{x}{2(1 + |x|)} .
$$
First we check the range. From $|x|/(1+|x|) < 1$ we get $|x/(1+|x|)| < 1$, hence $h(x) \in (0,1)$.
Next we compute the inverse explicitly. Put $t = h(x)$ and $u := 2t - 1 = x/(1+|x|)$. If $x \ge 0$ then $u = x/(1+x) \ge 0$ and $|u| = u$, so from $u(1+x) = x$ we get $x = u/(1-u) = u/(1-|u|)$. If $x < 0$ then $|x| = -x$, so $u = x/(1-x) < 0$ and $|u| = -u$, and from $u(1-x) = x$ we get $x = u/(1+u) = u/(1-|u|)$. In both cases we obtain the same formula
$$
x = \frac{u}{1 - |u|}, \qquad u = 2t-1 .
$$
If $t \in (0,1)$ then $|2t-1| < 1$, so the denominator is positive and the formula makes sense.
Conversely, given $u \in (-1,1)$, put $x := u/(1-|u|)$. Then $|x| = |u|/(1-|u|)$, hence
$$
1 + |x| = \frac{(1-|u|) + |u|}{1-|u|} = \frac{1}{1-|u|}, \qquad
\frac{x}{1+|x|} = \frac{u}{1-|u|}\cdot (1-|u|) = u .
$$
Thus $u \mapsto u/(1-|u|)$ is a bijection $(-1,1) \to \mathbb{R}$ with inverse $x \mapsto x/(1+|x|)$. Consequently $h$ is a bijection and
$$
h^{-1}(t) = \frac{2t-1}{1 - |2t-1|} \qquad (0 < t < 1) .
$$
Now for continuity. The map $x \mapsto |x|$ is continuous, since $\bigl||x| - |y|\bigr| \le |x-y|$ (take $\delta = \varepsilon$), and $1 + |x| \ge 1 > 0$, so the quotient $x \mapsto x/(1+|x|)$ is continuous; composing with a continuous affine map shows that $h : \mathbb{R} \to \mathbb{R}$ is continuous ( and ). Since the values lie in $(0,1)$, part (2) of shows that $h : \mathbb{R} \to (0,1)$ is continuous. Similarly $t \mapsto 2t-1$ is continuous, and $u \mapsto u/(1-|u|)$ is continuous on $(-1,1)$ because the denominator is positive there; combined with continuity of the inclusion $(0,1) \hookrightarrow \mathbb{R}$ (part (1) of ) this gives that $h^{-1} : (0,1) \to \mathbb{R}$ is continuous.
Hence $h$ is a homeomorphism and $\mathbb{R} \cong (0,1)$. For $a < b$ the affine map $s \mapsto a + (b-a)s$ gives $(0,1) \cong (a,b)$, so by transitivity () every bounded open interval is homeomorphic to $\mathbb{R}$. One could reach the same conclusion using $\tan : (-\pi/2, \pi/2) \to \mathbb{R}$, but the rational formula above is more elementary in that it assumes nothing about trigonometric functions.
Give $S^{1} := \{\, (x,y) \in \mathbb{R}^{2} : x^{2}+y^{2} = 1 \,\}$ the subspace topology from $\mathbb{R}^{2}$ and consider
$$
f : [0, 2\pi) \to S^{1}, \qquad f(t) = (\cos t, \sin t)
$$
(where $[0,2\pi)$ carries the subspace topology from $\mathbb{R}$).
**$f$ is a bijection.** From $\cos^{2}t + \sin^{2}t = 1$ the values lie in $S^{1}$. For surjectivity, let $(x,y) \in S^{1}$; then $|x| \le 1$, and since $\cos$ is a strictly decreasing bijection from $[0,\pi]$ onto $[-1,1]$, the number $t_0 := \arccos x \in [0,\pi]$ is defined. We have $\sin t_0 \ge 0$ and $\sin^{2} t_0 = 1 - x^{2} = y^{2}$, so if $y \ge 0$ then $t := t_0$ works, and if $y < 0$ then $t := 2\pi - t_0 \in (\pi, 2\pi)$ works (because $\cos(2\pi - t_0) = \cos t_0 = x$ and $\sin(2\pi-t_0) = -\sin t_0 = y$). Injectivity follows from the addition formulas: if $f(s) = f(t)$ with $0 \le s \le t < 2\pi$, then
$$
\cos(t-s) = \cos t \cos s + \sin t \sin s = \cos^{2}s + \sin^{2}s = 1 ,
$$
and in the range $0 \le t - s < 2\pi$ the equation $\cos\theta = 1$ holds only for $\theta = 0$, so $s = t$.
**$f$ is continuous.** Since $\cos$ and $\sin$ are continuous on $\mathbb{R}$, the map $t \mapsto (\cos t, \sin t)$ is continuous as a map $\mathbb{R} \to \mathbb{R}^{2}$ (this can be shown directly by an $\varepsilon$-$\delta$ estimate componentwise); as the values lie in $S^{1}$, shows it is continuous as a map into $S^{1}$, and so is its restriction to $[0,2\pi)$.
**The inverse is not continuous.** By it suffices to show that $f$ is not an open map. Put $U := [0, \pi)$. Since $U = (-\pi, \pi) \cap [0,2\pi)$, the set $U$ is open in $[0,2\pi)$. Its image is
$$
f(U) = \{\, (\cos t, \sin t) : 0 \le t < \pi \,\} ,
$$
and for $0 \le t < \pi$ we have $\sin t \ge 0$, so $f(U)$ contains no point with negative second coordinate. On the other hand, any open subset of $S^{1}$ containing $p := (1,0) = f(0) \in f(U)$ contains, by the definition of the subspace topology, a set $B_{\mathbb{R}^{2}}(p, \varepsilon) \cap S^{1}$ for some $\varepsilon > 0$. Taking $\eta > 0$ small, the point
$$
q_\eta := (\cos(2\pi - \eta), \sin(2\pi-\eta)) = (\cos \eta, -\sin \eta) \in S^{1}
$$
converges to $p$ as $\eta \to 0^{+}$, so for $\eta$ small enough $q_\eta \in B_{\mathbb{R}^{2}}(p,\varepsilon) \cap S^{1}$. But the second coordinate of $q_\eta$ is $-\sin\eta < 0$, so $q_\eta \notin f(U)$. Hence no open set containing $p$ is contained in $f(U)$, and $f(U)$ is not open in $S^{1}$.
Therefore $f$ is not an open map, and by it is not a homeomorphism. Intuitively, $f$ glues the two ends of an interval together to form a circle. The gluing can be done continuously, but the operation of cutting the circle open (the inverse map) is discontinuous at the cut.
Checking by hand whether a continuous bijection is a homeomorphism is laborious. The theorem "a continuous bijection from a compact space to a Hausdorff space is a homeomorphism" (), which we prove in a later article, removes the need for it. The reason we did not get a homeomorphism in is that the domain $[0,2\pi)$ is not compact. See [Compactness](/mathematics/topology/compactness) and [Separation axioms and metrisability](/mathematics/topology/separation-axioms) for details.
## 6. How to prove that two spaces are not homeomorphic
To prove that two spaces *are* homeomorphic it is enough to construct a single map: an existence proof. To prove that they are *not*, one must establish the universal statement that no map whatsoever is a homeomorphism. An exhaustive search is out of the question, so we use instead. Find one topological property $P$ that $X$ has and $Y$ lacks; then the assumption $X \cong Y$ would force $Y$ to have $P$ as well, a contradiction.
A["Direction 1: construct an explicit bijection h"]
Q --> B["Direction 2: look for a topological property P"]
A --> A1["Check that h is continuous"]
A1 --> A2["Check that the inverse of h is continuous"]
A2 --> A3["Conclusion: X and Y are homeomorphic"]
B --> B1["X has P and Y does not have P"]
B1 --> B2["Conclusion: X and Y are not homeomorphic"]`} />
**(1) $\mathbb{R} \not\cong [0,1]$.** Suppose a homeomorphism $h : [0,1] \to \mathbb{R}$ existed. In particular $h$ is a continuous real-valued function on the closed interval $[0,1]$, so by the extreme value theorem of Weierstrass () it attains a maximum: there is $c \in [0,1]$ with $h(s) \le h(c) =: M$ for all $s \in [0,1]$. Hence $h([0,1]) \subset (-\infty, M]$. But $h$ is surjective, so $h([0,1]) = \mathbb{R}$, while $M + 1 \in \mathbb{R}$ and $M + 1 > M$ give $M+1 \notin (-\infty, M]$ — a contradiction. Therefore $\mathbb{R} \not\cong [0,1]$.
**(2) $[0,1] \not\cong [0,1)$.** Suppose a homeomorphism $h : [0,1] \to [0,1)$ existed. As in (1), $h$ attains a maximum $M = h(c)$ on $[0,1]$, and by surjectivity $h([0,1]) = [0,1)$, so $M$ would have to be a greatest element of the set $[0,1)$. But $[0,1)$ has no greatest element: for any $t \in [0,1)$ put $t' := (t+1)/2$; then $t < t' < 1$, so $t' \in [0,1)$ and $t' > t$, and $t$ cannot be greatest. This contradiction gives $[0,1] \not\cong [0,1)$.
Another powerful invariant is connectedness. A topological space $X$ is **connected** if there is no decomposition $X = V_1 \cup V_2$ with $V_1 \cap V_2 = \emptyset$, with $V_1$ and $V_2$ both open in $X$, and with $V_1 \ne \emptyset$ and $V_2 \ne \emptyset$. That continuous maps preserve this property follows at once from alone.
Let $X$ and $Y$ be topological spaces and $f : X \to Y$ a continuous surjection. If $X$ is connected, then so is $Y$.
Suppose $Y$ were not connected, and take a decomposition $Y = V_1 \cup V_2$ with $V_1 \cap V_2 = \emptyset$ and with $V_1, V_2$ nonempty open subsets of $Y$.
By continuity of $f$ (), both $f^{-1}(V_1)$ and $f^{-1}(V_2)$ are open in $X$. By the properties of preimages from §2,
$$
f^{-1}(V_1) \cup f^{-1}(V_2) = f^{-1}(V_1 \cup V_2) = f^{-1}(Y) = X, \qquad
f^{-1}(V_1) \cap f^{-1}(V_2) = f^{-1}(V_1 \cap V_2) = f^{-1}(\emptyset) = \emptyset .
$$
Moreover, since $V_i \ne \emptyset$ we may pick $y_i \in V_i$, and by surjectivity of $f$ there is $x_i \in X$ with $f(x_i) = y_i$. Then $x_i \in f^{-1}(V_i)$, so $f^{-1}(V_i) \ne \emptyset$.
Hence $X = f^{-1}(V_1) \cup f^{-1}(V_2)$ is a decomposition contradicting the connectedness of $X$. Therefore $Y$ is connected.
Connectedness is a topological property: if $X \cong Y$ and $X$ is connected, then $Y$ is connected.
Let $f : X \to Y$ be a homeomorphism. By , $f$ is continuous and bijective, in particular a continuous surjection. So applies: if $X$ is connected, so is $Y$.
is a theorem of wide application. Combined with the fact that intervals of $\mathbb{R}$ are connected (proved in [Connectedness](/mathematics/topology/connectedness) as ), it gives that the image of a continuous function on an interval is connected, hence an interval — and the intermediate value theorem follows as a corollary. That a basic theorem of analysis drops out of the general theory of topology is, I think, one of the rewards of this abstraction. In we use this invariant to distinguish $\mathbb{R}$ from $\mathbb{R}^{2}$.
## 7. Exercises
Let $X$ and $Y$ be topological spaces and let $A, B$ be closed subsets of $X$ with $X = A \cup B$. Let $f : X \to Y$ be a map. Show that if the restrictions $f|_A : A \to Y$ and $f|_B : B \to Y$ are both continuous (with $A$ and $B$ carrying the subspace topology), then $f$ is continuous (the pasting lemma).
First we record an auxiliary fact: if $A$ is closed in $X$ and $C \subset A$ is closed in the subspace $A$, then $C$ is closed in $X$. Indeed, by the definition of the subspace topology the open sets of $A$ have the form $U \cap A$ with $U$ open in $X$, so the closed sets of $A$ have the form $A \setminus (U \cap A) = A \cap (X \setminus U)$, that is, they are intersections of $A$ with closed subsets of $X$. Since $A$ is itself closed in $X$, such a set is the intersection of two closed subsets of $X$ and hence closed in $X$.
Now let $F \subset Y$ be an arbitrary closed set. If $x \in f^{-1}(F)$ then $x \in X = A \cup B$, so $x \in A$ or $x \in B$; in the first case $x \in (f|_A)^{-1}(F)$, in the second $x \in (f|_B)^{-1}(F)$. Conversely, if $x \in (f|_A)^{-1}(F)$ then $x \in A \subset X$ and $f(x) = f|_A(x) \in F$, so $x \in f^{-1}(F)$; the same argument applies to $(f|_B)^{-1}(F)$. Hence
$$
f^{-1}(F) = (f|_A)^{-1}(F) \cup (f|_B)^{-1}(F) .
$$
Since $f|_A$ is continuous, part (ii) of shows that $(f|_A)^{-1}(F)$ is closed in $A$, hence closed in $X$ by the auxiliary fact. Likewise $(f|_B)^{-1}(F)$ is closed in $X$. The union of two closed sets is closed (the dual of (O3), which says that the intersection of two open sets is open), so $f^{-1}(F)$ is closed in $X$. As $F$ was arbitrary, the implication (ii) $\Rightarrow$ (i) of shows that $f$ is continuous.
The hypothesis that both $A$ and $B$ are closed cannot be dropped. Take $X = [0,2]$, $A = [0,1]$ (closed in $X$) and $B = (1,2]$ (not closed in $X$), and define $f : X \to \mathbb{R}$ to be $0$ on $A$ and $1$ on $B$. The restrictions $f|_A$ and $f|_B$ are constant maps, hence continuous by . But $f$ itself is not continuous: the set $V := (-1/2, 1/2)$ is open in $\mathbb{R}$, while $f^{-1}(V) = [0,1]$ is not open in $X = [0,2]$ (any open subset of $X$ containing $1$ contains $(1-\delta, 1+\delta) \cap [0,2]$ for some $\delta > 0$, and this set necessarily contains points greater than $1$).
Write $\mathbb{R}_\ell$ (the lower limit topology, or Sorgenfrey line) for the space $\mathbb{R}$ equipped with the topology having the family of half-open intervals $\mathcal{B} = \{\, [a,b) : a < b \,\}$ as a base. Show that, with $\mathbb{R}$ carrying its usual topology, the identity map $\mathrm{id} : \mathbb{R}_\ell \to \mathbb{R}$ is continuous, while $\mathrm{id} : \mathbb{R} \to \mathbb{R}_\ell$ is not.
**First part.** We use . The open intervals form a base for the usual topology of $\mathbb{R}$, so it suffices to show that $\mathrm{id}^{-1}((a,b)) = (a,b)$ is open in $\mathbb{R}_\ell$ for $a < b$. Here
$$
(a,b) = \bigcup_{a < c < b} [c, b) .
$$
Indeed, each $[c,b)$ on the right is contained in $(a,b)$ because $c > a$; conversely, if $x \in (a,b)$ we may take $c = x$, and then $x \in [x, b)$ with $a < x < b$. The right-hand side is a union of members of $\mathcal{B}$ and hence open in $\mathbb{R}_\ell$. So $\mathrm{id} : \mathbb{R}_\ell \to \mathbb{R}$ is continuous.
**Second part.** The set $[0,1) \in \mathcal{B}$ is open in $\mathbb{R}_\ell$, but $\mathrm{id}^{-1}([0,1)) = [0,1)$ is not open in the usual topology. Indeed, for $0 \in [0,1)$ and any $\delta > 0$ we have $-\delta/2 \in (-\delta, \delta)$ and $-\delta/2 \notin [0,1)$, so no open interval containing $0$ is contained in $[0,1)$. Hence $\mathrm{id} : \mathbb{R} \to \mathbb{R}_\ell$ is not continuous.
**How to read the conclusion.** The map $\mathrm{id} : \mathbb{R}_\ell \to \mathbb{R}$ is a continuous bijection, but its inverse $\mathrm{id} : \mathbb{R} \to \mathbb{R}_\ell$ is not continuous, so condition 3 of fails and it is not a homeomorphism. Alongside , this is further evidence that condition 3 is independent.
Show that the space $S^{1} \setminus \{N\}$, obtained from $S^{1} = \{\, (x,y) \in \mathbb{R}^{2} : x^{2}+y^{2}=1 \,\}$ by removing the north pole $N = (0,1)$, is homeomorphic to $\mathbb{R}$, using the stereographic projection $\pi(x,y) = x/(1-y)$.
For $(x,y) \in S^{1}\setminus\{N\}$ we have $y \ne 1$, so $\pi(x,y) = x/(1-y)$ is defined. As a candidate for the inverse take
$$
\sigma(t) := \left( \frac{2t}{t^{2}+1},\ \frac{t^{2}-1}{t^{2}+1} \right) \qquad (t \in \mathbb{R}) .
$$
**The values of $\sigma$ lie in $S^{1}\setminus\{N\}$.** Since $(2t)^{2} + (t^{2}-1)^{2} = 4t^{2} + t^{4} - 2t^{2} + 1 = t^{4}+2t^{2}+1 = (t^{2}+1)^{2}$, the sum of the squares of the two coordinates of $\sigma(t)$ equals $1$, so $\sigma(t) \in S^{1}$. Moreover the second coordinate equals $1$ only if $t^{2}-1 = t^{2}+1$, that is $-1 = 1$, which never happens. Hence $\sigma(t) \ne N$.
**$\pi \circ \sigma = \mathrm{id}_{\mathbb{R}}$.** Since $1 - \dfrac{t^{2}-1}{t^{2}+1} = \dfrac{(t^{2}+1)-(t^{2}-1)}{t^{2}+1} = \dfrac{2}{t^{2}+1}$, we get
$$
\pi(\sigma(t)) = \frac{2t/(t^{2}+1)}{2/(t^{2}+1)} = t .
$$
**$\sigma \circ \pi = \mathrm{id}$.** For $(x,y) \in S^{1}\setminus\{N\}$ put $t := x/(1-y)$. Using $x^{2} = 1-y^{2}$,
$$
t^{2}+1 = \frac{x^{2} + (1-y)^{2}}{(1-y)^{2}} = \frac{(1-y^{2}) + 1 - 2y + y^{2}}{(1-y)^{2}} = \frac{2(1-y)}{(1-y)^{2}} = \frac{2}{1-y} ,
$$
and hence $t^{2}-1 = \dfrac{2}{1-y} - 2 = \dfrac{2y}{1-y}$. Substituting these,
$$
\frac{2t}{t^{2}+1} = \frac{2x/(1-y)}{2/(1-y)} = x, \qquad
\frac{t^{2}-1}{t^{2}+1} = \frac{2y/(1-y)}{2/(1-y)} = y ,
$$
so $\sigma(\pi(x,y)) = (x,y)$. Therefore $\pi$ is a bijection with $\pi^{-1} = \sigma$.
**Continuity.** The map $\pi$ is a rational expression whose denominator does not vanish on the open subset $\{\, (x,y) : y \ne 1 \,\}$ of $\mathbb{R}^{2}$, hence continuous there, and so is its restriction to $S^{1}\setminus\{N\}$ (composition with the inclusion; and ). The map $\sigma$ is continuous as a map $\mathbb{R} \to \mathbb{R}^{2}$ because $t^{2}+1 \ge 1 > 0$, and since its values lie in $S^{1}\setminus\{N\}$, part (2) of shows that it is continuous as a map $\mathbb{R} \to S^{1}\setminus\{N\}$. Hence $\pi$ is a homeomorphism and $S^{1}\setminus\{N\} \cong \mathbb{R}$.
That removing a single point turns the circle into a line is exactly the "cutting open" of , made legitimate by deleting a point of the domain beforehand.
Show that $\mathbb{R}$ and $\mathbb{R}^{2}$ are not homeomorphic. You may use, as results from [Connectedness](/mathematics/topology/connectedness), that every nonempty interval of $\mathbb{R}$ is connected and that $\mathbb{R}^{2}\setminus\{p\}$ is connected for every $p \in \mathbb{R}^{2}$.
**Auxiliary fact (restrictions are homeomorphisms).** If $h : X \to Y$ is a homeomorphism and $A \subset X$, then $h|_A : A \to h(A)$ is a homeomorphism. Indeed, $h$ is injective, so $h|_A$ is a bijection onto $h(A)$. For continuity: the open sets of $h(A)$ have the form $W \cap h(A)$ with $W$ open in $Y$, and
$$
(h|_A)^{-1}(W \cap h(A)) = A \cap h^{-1}(W) ,
$$
which is open in $A$ because $h^{-1}(W)$ is open in $X$ by continuity of $h$. Applying the same argument to $h^{-1} : Y \to X$ and $h(A) \subset Y$ gives continuity of $(h|_A)^{-1}$.
**Main argument.** Suppose $\mathbb{R} \cong \mathbb{R}^{2}$ and take a homeomorphism $h : \mathbb{R} \to \mathbb{R}^{2}$. Put $p := h(0)$. Since $h$ is a bijection, $h(\mathbb{R}\setminus\{0\}) = \mathbb{R}^{2}\setminus\{p\}$, and by the auxiliary fact
$$
h|_{\mathbb{R}\setminus\{0\}} : \mathbb{R}\setminus\{0\} \longrightarrow \mathbb{R}^{2}\setminus\{p\}
$$
is a homeomorphism.
But $\mathbb{R}\setminus\{0\} = (-\infty, 0) \cup (0, \infty)$, where $(-\infty,0)$ and $(0,\infty)$ are open in $\mathbb{R}$ and hence open in the subspace $\mathbb{R}\setminus\{0\}$; they are disjoint, and neither is empty (they contain $-1$ and $1$ respectively). So $\mathbb{R}\setminus\{0\}$ is not connected.
On the other hand, $\mathbb{R}^{2}\setminus\{p\}$ is connected by hypothesis. By connectedness is a topological property, so it is impossible for one of two homeomorphic spaces to be connected and the other not. This is a contradiction, and therefore $\mathbb{R} \not\cong \mathbb{R}^{2}$.
**Remark.** The same argument shows $\mathbb{R} \not\cong \mathbb{R}^{n}$ for $n \ge 2$. It cannot be used to show $\mathbb{R}^{2} \not\cong \mathbb{R}^{3}$, however, since removing a point does not destroy connectedness in either case. The general statement $\mathbb{R}^{m}\not\cong\mathbb{R}^{n}$ for $m \ne n$ (invariance of dimension) requires stronger topological invariants, such as homology groups.
## References
- J. R. Munkres, *Topology*, 2nd edition, Prentice Hall, 2000 — Chapter 2, §18 "Continuous Functions", collects the equivalent conditions, the pasting lemma and homeomorphisms treated here.
- S. Willard, *General Topology*, Addison-Wesley, 1970 (reprinted by Dover, 2004) — a systematic treatment of the characterisations of continuous maps and homeomorphisms.
- Kazuo Matsuzaka, *Shugo, Iso Nyumon* (Introduction to Sets and Topology), Iwanami Shoten, 1968 (in Japanese) — explains the passage from metric spaces to topological spaces in graded steps, for beginners.
- Fuichi Uchida, *Shugo to Iso* (Sets and Topology), Shokabo, 1986 (in Japanese) — the chapters on continuous maps and homeomorphisms contain many examples along the same lines as this article.
- Teiji Takagi, *Kaiseki Gairon* (Introduction to Analysis), revised 3rd ed., Iwanami Shoten, 1961 (in Japanese) — contains the proof of the Weierstrass theorem that a continuous function on a closed interval attains a maximum, which we used in .
## Appendix: what topology cannot see — boundedness, completeness, uniform continuity
**Topology forgets how close things are.** In we proved $\mathbb{R} \cong (0,1)$. From this single fact it follows immediately that several important metric notions are not topological properties.
**Boundedness is not a topological property.** The interval $(0,1)$ is bounded with respect to the usual metric (its diameter is $1$), whereas $\mathbb{R}$ is not bounded; yet the two are homeomorphic. So the argument "bounded, or unbounded, therefore not homeomorphic" is invalid. Boundedness depends on the metric $d$, and a different metric inducing the same topology can make a space bounded or unbounded at will. Indeed, $d'(x,y) := \min\{1, |x-y|\}$ is a metric on $\mathbb{R}$ that determines the same family of open sets as the usual metric, and $(\mathbb{R}, d')$ is bounded.
**Completeness is not a topological property either.** The space $\mathbb{R}$ is complete for the usual metric, while $(0,1)$ is not: the sequence $x_n = 1/(n+1)$ consists of points of $(0,1)$ and satisfies $|x_n - x_m| \to 0$, so it is a Cauchy sequence, but its limit $0$ does not belong to $(0,1)$. Nevertheless $\mathbb{R} \cong (0,1)$. For completeness see [Completeness of the real numbers and Cauchy sequences](/en/mathematics/real-analysis/completeness-and-cauchy).
**Uniform continuity is not a topological property.** Consider the inverse $h^{-1} : (0,1) \to \mathbb{R}$, $h^{-1}(t) = (2t-1)/(1-|2t-1|)$, of the homeomorphism of . For $t \in [1/2, 1)$ we have $2t-1 \ge 0$, that is $|2t-1| = 2t-1$, so
$$
h^{-1}(t) = \frac{2t-1}{1-(2t-1)} = \frac{2t-1}{2(1-t)} .
$$
Put $t_n := 1 - 1/n$ and $s_n := 1 - 1/(2n)$ for $n \ge 2$. Since $1 - 1/n \ge 1/2$ and $1 - 1/(2n) \ge 3/4$, both lie in $[1/2,1)$, and
$$
h^{-1}(t_n) = \frac{2(1-1/n)-1}{2/n} = \frac{1 - 2/n}{2/n} = \frac{n}{2} - 1, \qquad
h^{-1}(s_n) = \frac{1 - 1/n}{2 \cdot 1/(2n)} = \frac{1-1/n}{1/n} = n - 1 .
$$
We have $|t_n - s_n| = 1/(2n) \to 0$, whereas
$$
\bigl| h^{-1}(t_n) - h^{-1}(s_n) \bigr| = \Bigl| \frac{n}{2} - 1 - (n-1) \Bigr| = \frac{n}{2} \longrightarrow \infty .
$$
So for $\varepsilon = 1$, no matter which $\delta > 0$ is chosen, any $n$ with $1/(2n) < \delta$ gives $|t_n - s_n| < \delta$ and $|h^{-1}(t_n) - h^{-1}(s_n)| \ge 1$. Hence $h^{-1}$ is not uniformly continuous, while the identity map on $\mathbb{R}$ is. Uniform continuity is not preserved under homeomorphism.
**To summarise**, what a topology remembers is only the qualitative information of which sets lie near a point; the quantitative information of how large the distances are has been discarded. To handle uniform continuity, completeness or boundedness one must pass to a category of maps preserving a finer structure than the topology — a metric, or a uniformity. Conversely, a theorem proved using topological properties alone is robust: it survives any change of metric. Keeping this dividing line in mind makes it visible exactly where distance enters the arguments of [Continuous functions and uniform continuity](/en/mathematics/real-analysis/uniform-continuity).