NoteNextra-origin/content/Math4201/Math4201_L10.md

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:

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

Proof

1. By definition of continuous functions, \forall V open in Y, f^{-1}(V) is open in X.

2. 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).

3. Similar to 2 but use forall.

Properties of continuous functions

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

1. f^{-1}(V) is open in X for any open set V\subset Y.
2. f is continuous at any point x\in X.
3. f^{-1}(C) is closed in X for any closed set C\subset Y.
4. Assume \mathcal{B} is a basis for Y, then f^{-1}(\mathcal{B}) is open in X for any B\in \mathcal{B}.
5. 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.