# $f:I\rightarrow X$ where $X$ is hausdorff show that $X$ is metrizable.

compactnessconnectednessgeneral-topologymetric-spacessecond-countable

This question comes from section 44 problem 4 of Munkres.

Let $$X$$ be a Hausdorff space. Let $$I=[0,1]$$. Show that if there is a continuous surjective map $$f : I \rightarrow X$$, then $$X$$ is compact, connected, weakly locally connected, and metrizable. [Hint: Show $$f$$ is a perfect map.]

I have already shown everything except that it is metrizable. I know that since $$X$$ is compact it is also paracompact, so I was going to use the Smirnov metrization theorem which states that if $$X$$ is paracompact and locally metrizable that $$X$$ is metrizable.

I am struggling with how to set up the neighborhood $$U$$ of $$x\in X$$, such that $$U$$ is metrizable. Any hints would be great in this direction.

Alternatively I thought about proving that $$X$$ was second-countable but I was also struggling with that. So any hints this direction would also be welcome.

Situation. $$X$$ is Hausdorff, $$(Y,d)$$ is a compact metric space, and $$f:Y \to X$$ is a continuous surjection.

Then $$X$$ is compact.

Definition. Let $$n\geq 1$$ be an integer and $$x,x'\in X$$ a point. Write $$\tilde{F}_n := \{ (y_1,y_2)\in Y\times Y | d(f^{-1}(f(y_1)),y_2) \geq 1/n\}.$$ Then $$\tilde{F}_n\subset Y\times Y$$ is a closed symmetric subset. Write $$\Delta \subset X\times X$$ for the diagonal.

Claim 1.

• $$\tilde{F}_n \cap (f\times f)^{-1}(\Delta) = \emptyset$$.
• $$\bigcap_{n\geq 1}(Y\times Y \setminus \tilde{F}_n) = (f\times f)^{-1}(\Delta)$$.
• $$\bigcap_{n\geq 1}(X\times X \setminus (f\times f)(\tilde{F}_n)) = \Delta$$.

Proof. If $$f(y_1) = f(y_2)$$, then $$y_2\in f^{-1}(f(y_1))$$. Hence $$d(f^{-1}(f(y_1)),y_2) = 0$$. This implies the first equality. The second equality follows from the compactness of $$f^{-1}(f(y))$$ (for any $$y\in Y$$). The third equality follows from the second equality. $$\square$$

Definition. For any $$\tilde{A}\subset X\times X$$, we write $$\tilde{A}^{\mathrm{op}} := \{(a,b)|(b,a)\in \tilde{A}\}$$, and $$\tilde{A}^{\mathrm{sym}} := \tilde{A}\cap \tilde{A}^{\mathrm{op}}$$. Write $$\tilde{W}_n:= (X\times X \setminus (f\times f)(F_n))^{\mathrm{sym}}$$. Then $$\tilde{W}_n$$ is open and symmetric. Write $$\mathcal{U}_n := \{U \ | \ U\text{ is open}, U\times U \subset \tilde{W}_n\}$$ for the open covering of $$X$$. Write \begin{align*} U_n(x) &:= \{ x'\in X | \forall y'\in f^{-1}(x'), \exists y\in f^{-1}(x), d(y,y') < 1/n\}, \\ W_n(x) &:= \{x'\in X | (x,x')\in \tilde{W}_n\}. \end{align*}

Claim 2. For any $$x\in X$$ and any open subset $$x\in U\subset X$$, there exists $$n\geq 1$$ such that $$U_n(x)\subset U$$.

Proof. Take an integer $$n \gg 1$$ such that $$d(f^{-1}(x),f^{-1}(X\setminus U)) > 1/n$$. Then $$U_n(x) \subset U$$. $$\square$$

Claim 3. $$W_n(x) \subset U_n(x)$$.

Proof. Assume that $$(x,x')\in W_n$$. Then $$(f\times f)^{-1}(x,x') \cap F_n = \emptyset$$. Hence for any $$y'\in f^{-1}(x')$$, $$d(f^{-1}(x),y') < 1/n$$, i.e., $$\exists y\in f^{-1}(x), d(y,y') < 1/n$$. This implites that $$x'\in U_n(x)$$. $$\square$$

Claim 4. If $$U\times U \subset W_n$$, and $$x\in U$$, then $$U\subset W_n(x)$$.

Proof. If $$x'\in U$$, then $$(x,x')\in U\times U \subset W_n$$. Thus $$x'\in W_n(x)$$. $$\square$$

Definition. Since $$X$$ is compact (hence in particular, paracompact), we can take a finite star (open) refinement $$\mathcal{V}_n$$ of the open covering $$\{U\cap V | U\in \mathcal{U}_n, V\in \mathcal{V}_{n-1}\}$$, inductively. Write $$\mathcal{V}_n(x) := \bigcup \{V\in \mathcal{V}_n | x\in V\} \subset X.$$ Then it follows immediately that $$\mathcal{V}_n(x) \subset \mathcal{V}_{n-1}(x)$$. Finally, we write $$\mathcal{V} := \bigcup_{n\geq 1}\mathcal{V}_n$$. Then $$\mathcal{V}$$ is a countable set of open subsets of $$X$$.

Claim 5. $$\mathcal{V}$$ is a basis of $$X$$. In particular, $$X$$ is second countable.

Proof. Let $$x\in X$$ be a point and $$x\in U\subset X$$ an open neighborhood of $$x$$. By Claim 2 and Claim 3, there exists an integer $$n\geq 1$$ such that $$W_n(x) \subset U$$. Since $$\mathcal{V}_n$$ is a star refinement of $$\mathcal{U}_n$$, $$\mathcal{V}_n(x) \times \mathcal{V}_n(x) \subset W_n$$. Hence, by Claim 4, $$\mathcal{V}_n(x) \subset W_n(x) \subset U$$. This implies that $$\exists V\in \mathcal{V}_n$$ such that $$x\in V \subset U$$. $$\square$$

Conclusion. $$X$$ is metrizable.

Proof. This follows directly from Uryshon metrization theorem. $$\square$$