Math4201 Topology I (Lecture 10)

Continuity

Continuous functions

Let $X,Y$ be topological spaces and $f:X\to Y$ . For any $x\in X$ and any open neighborhood $V$ of $f(x)$ in $Y$ , $f^{-1}(V)$ contains an open neighborhood of $x$ in $X$ .

Lemma for continuous functions

Let $f:X\to Y$ be a function, then:

$A\subseteq Y$ : $f^{-1}(A^c) = (f^{-1}(A))^c$ .
$\{A_\alpha\}_{\alpha\in I}\subseteq Y$ : $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$ .
$\{A_\alpha\}_{\alpha\in I}\subseteq Y$ : $f^{-1}(\bigcap_{\alpha\in I} A_\alpha) = \bigcap_{\alpha\in I} f^{-1}(A_\alpha)$ .

Proof

By definition of continuous functions, $\forall V$ open in $Y$ , $f^{-1}(V)$ is open in $X$ .
It is sufficient to shoa that $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$ if and only if $x\in \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$ .

This condition holds if and only if $\exists \alpha\in I$ such that $f(x)\in A_\alpha$ .

Which is equivalent to $\exists \alpha\in I$ such that $x\in f^{-1}(A_\alpha)$ .

So $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$

In particular, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$ .

Similar to 2 but use forall.

Properties of continuous functions

A function $f:X\to Y$ is continuous if and only if:

$f^{-1}(V)$ is open in $X$ for any open set $V\subset Y$ .
$f$ is continuous at any point $x\in X$ .
$f^{-1}(C)$ is closed in $X$ for any closed set $C\subset Y$ .
Assume $\mathcal{B}$ is a basis for $Y$ , then $f^{-1}(\mathcal{B})$ is open in $X$ for any $B\in \mathcal{B}$ .
For any $A\subseteq X$ , $f(\overline{A})\subseteq \overline{f(A)}$ .

Proof

Showing $1\iff 3$ :

Use the lemma for continuous functions (1)

Showing $1\iff 4$ :

$1 \implies 4$ :

Because any $B\in \mathcal{B}$ is open in $Y$ , so $f^{-1}(B)$ is open in $X$ .

$4 \implies 1$ :

Let $V\subset Y$ be an open set. Then there are basis elements $\{B_\alpha\}_{\alpha\in I}$ such that $V=\bigcup_{\alpha\in I} B_\alpha$ .

So $f^{-1}(V) = f^{-1}(\bigcup_{\alpha\in I} B_\alpha) = \bigcup_{\alpha\in I} f^{-1}(B_\alpha)$ (by lemma (2)) is a union of open sets, so $f^{-1}(V)$ is open in $X$ .

Showing $1\implies 5$ :

Take $A\subseteq X$ and $x\in \overline{A}$ . It suffices to show $f(x)$ is an element of the closure of $f(A)$ . This is equivalent to say that any open neighborhood $V$ of $f(x)$ intersects $f(A)$ has a non-trivial intersection with $f(A)$ .

For any such $V$ , 1 implies that $f^{-1}(V)$ is open in $X$ . Moreover, $x\in f^{-1}(V)$ because $f(x)\in V$ .

This means that $f^{-1}(V)$ is an open neighborhood of $x$ . Since $x\in \overline{A}$ , we have $f^{-1}(V)\cap A\neq \emptyset$ and contains a point $x'\in X$ .

So $x'\in f^{-1}(V)\cap A$ , this implies that $f(x')\in V$ and $f(x')\in f(A)$ , so $f(x')\in V\cap f(A)$ .

Note

This verifies our claim. Proof of $5\implies 1$ is similar and left as an exercise.

Example of property 5

Let $X=(0,1)\cup (1,2)$ and $Y=\mathbb{R}$ equipped with the subspace topology induced by the standard topology on $\mathbb{R}$ .

Let $f:X\to Y$ be the inclusion map, $f(x)=x$ for all $x\in X$ . This is continuous.

Let $A=(0,1)\cup (1,2)$ . Then $\overline{A}=A$ . So $f(\overline{A})=f(A)=(0,1)\cup (1,2)$ .

However, $\overline{f(A)}=\overline{(0,1)\cup (1,2)}=[0,2]$ .

So $f(\overline{A})\subsetneq \overline{f(A)}$ .

Definition of homeomorphism

A homeomorphism $f:X\to Y$ is a continuous map of topological spaces that is a bijection and $f^{-1}:Y\to X$ is also continuous.

Example of homeomorphism

Let $X=\mathbb{R}$ and $Y=\mathbb{R}+$ with standard topology.

$f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ is continuous and bijective.

$f^{-1}:\mathbb{R}^+\to \mathbb{R}$ be defined by $f^{-1}(y)=\ln(y)$ is continuous and homeomorphism.

Epsilon delta definition of continuity

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function where we use the standard topology on $\mathbb{R}$ .

Then property 4 implies that for any open interval $(a,b)\in \mathbb{R}$ , $f^{-1}((a,b))$ is open in $\mathbb{R}$ .

Now take an arbitrary $x\in \mathbb{R}$ and $\epsilon > 0$ . In particular $f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$ is an open set containing $x$ .

In particular, there is an open interval (by the standard topology on $\mathbb{R}$ ) $(c,d)$ such that $x\in (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$ .

Let $\delta = \min\{x-c, d-x\}$ . Then $(x-\delta, x+\delta)\subseteq (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$ .

This says that if $|y-x| < \delta$ , then $|f(y)-f(x)| < \epsilon$ .