NoteNextra-origin/content/Math4201/Math4201_L15.md

Math4201 Topology I (Lecture 15)

Continue on convergence of sequences

Closure and convergence

Proposition in metric space, every convergent sequence converges to a point in the closure

If X is a metric space, A\subset X, and x\in \overline{A}, then there is a sequence \{x_n\}_{n=1}^\infty\subseteq A such that x_n\to x.

Proof

Let U_n=B_{\frac{1}{n}}(x) be a sequence of open neighborhoods of x.

Since x\in \overline{A} and U_n is an open neighborhood of x, then U_n\cap A\neq \emptyset. Take x_n\in U_n\cap A, we claim that \{x_n\}_{n=1}^\infty\subseteq A that x_n\to x.

Take an open neighborhood U of x. Then by the definition of metric topology, there is r>0 such that B_r(x)\subseteq U.

let N be such that \frac{1}{N}<r. Then for an n\geq N, we have \frac{1}{n}\leq \frac{1}{N}<r. This implies that B_{\frac{1}{n}}(x)\subseteq B_r(x)\subseteq U.

So x_n\in B_{\frac{1}{n}}(x)\subseteq U for all n\geq N.

Corollary \mathbb{R}^\omega with the box topology is not metrizable

\mathbb{R}^\omega is \text{Map}(\mathbb{N},\mathbb{R})=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots.

Note that \mathbb{R}^\omega is Hausdorff. So not all Hausdorff spaces are metrizable.

Proof

Otherwise, the last proposition holds for \mathbb{R}^\omega with the box topology.

Last time we showed that this is not the case.

Definition of first countability axiom

A topological space (X,\mathcal{T}) satisfies the first countability axiom if for any point x\in X, there is a sequence of open neighborhoods of x, \{V_n\}_{n=1}^\infty such that any open neighborhood U of x contains one of V_n.

Note that if we take U_n=V_1\cap V_2\cap \cdots \cap V_n, then any open neighborhood U that contains V_N, then it also contains U_n for all n\geq N.

As the previous prof, for metric space, it is natural to try V_n=B_{\frac{1}{n}}(x) for some x\in X.

Proposition on first countability axiom

Rewrite the Proposition of metric space, every convergent sequence converges to a point in the closure

We can have the following:

If (X,\mathcal{T}) satisfies the first countability axiom, then every convergent sequence converges to a point in the closure.

We can easily prove this by takeing the sequence of open neighborhoods \{V_n\}_{n=1}^\infty instead of U_n=B_{\frac{1}{n}}(x).

Proposition of continuous functions

Let f:X\to Y be a map between two topological spaces.

1. If f is continuous, then for any convergent sequence \{x_n\}_{n=1}^\infty in X converging to x, the sequence \{f(x_n)\}_{n=1}^\infty converges to f(x).
2. If X is equipped with the metric topology and for any convergent sequence \{x_n\}_{n=1}^\infty\to x in X, the sequence \{f(x_n)\}_{n=1}^\infty\to f(x) in Y, then f is continuous.

Exercise

Find an example of a function f:X\to Y which is not continuous but for any convergent sequence in X, \{x_n\}_{n=1}^\infty\to x, the sequence \{f(x_n)\}_{n=1}^\infty\to f(x).

Solution

Consider X=\mathbb{R} with complement finite topology and Y=\mathbb{R} with the standard topology.

Take identity function f(x)=x.

This function is not continuous by trivially taking (0,1)\subseteq \mathbb{R} and the complement of (0,1) is not a finite set, so the function is not continuous.

However, for every convergent sequence in X, \{x_n\}_{n=1}^\infty\to x, the sequence \{f(x_n)\}_{n=1}^\infty\to f(x) trivially.

Proof

Part 1:

Let f:X\to Y be a continuous map and

\{x_n\}_{n=1}^\infty\subseteq X

converges to x.

Want to show that \{f(x_n)\}_{n=1}^\infty converges to f(x).

i.e. for any open neighborhood U of f(x), we want to show f(x_n) is eventually in U. Take f^{-1}(U). This is an open neighborhood of x since x_n\to x.

There is N such that \forall n\geq N, we have x_n\in f^{-1}(U), f(x_n)\in U.

This implies that \{f(x_n)\}_{n=1}^\infty converges to f(x).

Part 2:

Let f:X\to Y be a map between two topological spaces with X being metric such that for any convergent sequence in X, \{x_n\}_{n=1}^\infty\to x in X, we have f(x_n)\to f(x) in Y.

Want to show that f is continuous.

Recall that it suffice to show that for any A\subseteq X, f(\overline{A})\subseteq \overline{f(A)}.

Take y\in f(\overline{A}). Then y=f(x) with x\in \overline{A}. By the previous proposition, there is a sequence \{x_n\}_{n=1}^\infty\subseteq A (since X is a metric space) such that x_n\to x.

By our assumption,

\{f(x_n)\}_{n=1}^\infty\to f(x) \tag{*}

Note that \{f(x_n)\}_{n=1}^\infty is a sequence in f(A) and (*) implies that y=f(x)\in f(A) (by the first part of the proposition). This gives us the claim.

Note

The second part of the proposition is also true when X is not a metric space but satisfies the first countability axiom.

Equivalent formulation of continuity

If (X,d) and (Y,d') are metric spaces and f:X\to Y is a map, then f is continuous if and only if for any x_0\in X and any \epsilon > 0, there exists \delta > 0 such that if \forall n\in X, d(x_n,x_0)<\delta, then d'(f(x_n),f(x_0))<\epsilon.

Proof similar to the X=Y=\mathbb{R} case.