# compact metric spaces

Let $$(X,d_X)$$ and $$(Y, d_Y)$$ be compact metric spaces.

1. Show that for a nonempty set $$A\subseteq X,$$ the function $$f : X\to \mathbb{R}, f(x) = \inf \{d_X(x,a) : a\in A\}$$ is continuous.
2. Show that if a map $$f : X\to Y$$ is both continuous and bijective, then its inverse $$f^{-1} : Y\to X$$ is also continuous.
3. Let $$C$$ be the Cantor set with metric from $$(\mathbb{R},|\cdot|)$$ and $$P = \prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$$ with metric from $$(\ell_1, \lVert \cdot \rVert_1).$$ Show that there exists a continuous bijection $$f : C\to P.$$

For $$1),$$ let $$x\in X$$ and let $$\epsilon > 0.$$ Choose $$\delta = \epsilon.$$ Then if $$y\in X$$ is such that $$d(x,y) < \epsilon, we$$ have \begin{align}d(f(x), f(y)) &= \inf\{d(x,a) | a \in A \} - \inf\{d(y,a) | a\in A\}\\ &\leq \inf \{d(x,a) | a\in A\} - (\inf \{d(x,a) - d(x,y) | a \in A\})\\ &= \inf \{d(x,a) | a\in A\} - (\inf \{d(x,a) | a \in A\} - d(x,y)\}\\ &= d(x,y) < \epsilon.\end{align}

Is this part correct?

For $$2),$$ suppose $$f:X\to Y$$ is continuous and bijective. Let $$y\in Y$$ and $$\epsilon > 0.$$ We want to show that there exists $$\delta > 0$$ so that if $$d_Y(x,y) < \delta$$ then $$d_X(f^{-1}(x), f^{-1}(y)) < \epsilon$$. Let $$(y_n)$$ be a sequence in $$Y$$ such that $$y_n\to y.$$ Then $$(f^{-1}(y_n))\subseteq X$$ and since $$X$$ is compact, it has a convergent subsequence, say $$(f^{-1}(y_{n_k}))_k.$$ Since $$f$$ is bijective, we can write $$y_n = f(x_n)$$ for each $$n\in\mathbb{N},$$ where $$x_n \in X.$$ Let $$x = \lim_{k\to\infty} f^{-1}(y_{n_k}) = \lim_{k\to \infty} x_{n_k}.$$ Since $$f$$ is continuous, $$f(x) = \lim_{k\to\infty} f(x_{n_k}).$$ But here I'm stuck as I'm not sure how to show $$f^{-1}$$ is continuous from here.

As for the third part, every element of the Cantor set can be uniquely written in the form $$\sum_{k=1}^\infty \frac{a_k}{3^k}, a_k \in \{0,2\}\forall k.$$ Then it suffices to show that the function $$f : C\to P, f(\sum_{k=1}^\infty \frac{a_k}{3^k}) = \prod_{k=1}^\infty \{\frac{a_k/2}{2^k}\}$$ is continuous. I think one could show it's sequentially continous, but I'm not really sure about the details.

• For 3, consider choosing $\delta = min_k \frac{2^k}{3^k} \leq \epsilon$ Feb 1 at 16:22

I think (1) is correct, but I wouldn't bother with the infima: let $$a,b\in A$$. Then

$$|d(x,a)-d(y,b)|\leq |d(x,a)-d(a,y)|+|d(a,y)-d(y,b)|\leq d(x,y)+d(a,b)$$ where we used the triangle inequality. Take infimum on both sides as $$a\in A$$, while $$b$$ is (for now) fixed, so $$|f(x)-d(y,b)|\leq d(x,y)+0=d(x,y)$$. Now this is true for all $$b\in A$$, so take infimum over $$b$$ and conclude that $$|f(x)-f(y)|\leq d(x,y)$$, I think this is more neat.

For (2) this is a standard topological fact: If $$f:X\to Y$$ is a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism. Indeed, all we need to do is show that $$f^{-1}$$ is continuous: if $$E\subset X$$ is a closed subset, we want to show that the inverse image of $$E$$ through $$f^{-1}$$ is closed in $$Y$$, which is the same as $$f(E)$$ is closed in $$Y$$. Now $$X$$ is compact and $$E$$ is a closed subset, so $$E$$ is also compact. Now $$f$$ is continuous, so $$f(E)$$ is a compact subset of $$Y$$. Now $$Y$$ is Hausdorff, so all compact sets are closed, thus $$f(E)$$ is closed in $$Y$$. Sometimes it helps viewing things in a more abstract frame; for example the metrics here are not essentially needed, so the extra information causes extra confusion.

For (3) you have a correct guess: Indeed, every element of the Cantor set $$C$$ can be written uniquely as $$\sum_{k=1}^\infty\frac{a_k}{3^k}$$, where $$a_k$$ is either $$0$$ or $$2$$. But it is important to note that the converse is also true: Any sum $$\sum_{k=1}^\infty\frac{a_k}{3^k}$$ where $$a_k\in\{0,2\}$$ is an element of the cantor set. Now define $$f:C\to P$$ as $$f\big(\sum_{k=1}^\infty\frac{a_k}{3^k}\big)=(\frac{a_k}{2^{k+1}})_{k=1}^\infty$$

• $$f$$ is well-defined: since $$a_k$$ is either $$0$$ or $$2$$, we have that $$a_k/2^{k+1}$$ is either $$0$$ or $$1/2^k$$, so each $$a_k/2^{k+1}$$ does indeed belong to $$\{0,\frac{1}{2^k}\}$$, so $$f$$ does indeed take values in $$\prod_{k=1}^\infty\{0,\frac{1}{2^k}\}$$.

• $$f$$ is one to one: this is obvious.

• $$f$$ is surjective: indeed, let $$x=(x_k)_{k=1}^\infty\in P$$, so each $$x_k$$ is either $$0$$ or $$1/2^k$$. Set $$a_k=2^{k+1}\cdot x_k$$. Then $$a_k$$ is either $$0$$ or $$2$$, so if $$t=\sum_{k=1}^\infty\frac{a_k}{3^k}\in C$$ then $$f(t)=x$$.

• $$f$$ is continuous: Let $$\varepsilon>0$$ and fix a point $$x\in C$$, say $$x=\sum_{k=1}^\infty\frac{a_k}{3^k}$$, where $$a_k\in\{0,2\}$$. Let $$y\in C$$ be another element, say $$y=\sum_{k=1}^\infty\frac{b_k}{3^k}$$ where $$b_k\in\{0,2\}$$. Note that $$d_P(f(x),f(y))=\|\{\frac{a_k}{2^{k+1}}\}_{k=1}^\infty-\{\frac{b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\|\{\frac{a_k-b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\sum_{k=1}^\infty\frac{|a_k-b_k|}{2^{k+1}}$$

Claim: Let $$m\geq1$$ be an integer. Then if $$|x-y|<\frac{1}{3^m}$$, then $$a_k=b_k$$ for all $$k=1,\dots,m$$.

Proof of the claim: I will only describe the idea here. Note that the construction of the Cantor set is done in steps: in each step we divide the segments in 3 pieces and throw out the middle part. In the expression $$x=\sum_{k=0}^\infty\frac{a_k}{3^k}$$ with $$a_k=0$$ or $$2$$, we have the following interpretation: in the first step of the construction, we divide $$[0,1]$$ in three segments and we only keep $$[0,1/3]$$ and $$[2/3,1]$$. If $$a_1=0$$, then $$x$$ lies in $$[0,1/3]$$. If $$a_1=2$$, then $$x$$ lies in $$[2/3,1]$$. So if $$|x-y|<1/3$$ where $$y\in C,y=\sum_{k=1}^\infty\frac{b_k}{3^k}$$ with $$b_k=0$$ or $$2$$, then $$b_1=a_1$$: otherwise we would have $$x\in[0,1/3]$$ and $$y\in[2/3,1]$$ or $$x\in [2/3,1]$$ and $$y\in[0,1/3]$$, which contradicts the fact that $$|x-y|<1/3$$. Let's assume that $$a_1=0$$, so $$x\in[0,1/3]$$. Now again, in the second step of the construction, we cut each of $$[0,1/3]$$ and $$[2/3,1]$$ in three pieces each and throw out the middle third of each. Let's look at $$a_2$$. If $$a_2=0$$, then $$x$$ lies in the left segment of $$[0,1/3]$$ that is left after the second step, i.e. $$x\in[0,1/3^2]$$. If $$a_2=2$$, then $$x$$ lies in the right segment of $$[0,1/3]$$ that is left after the second step, i.e. $$x\in[2/3^2,1/3]$$. Now if $$|x-y|<1/3^2$$ then we also have $$|x-y|<1/3$$, so $$y$$ lies in $$[0,1/3]$$ as we explained for the first step, i.e. $$b_1=0$$. Now if $$b_2\neq a_2$$, we have that $$x,y$$ are in different thirds of $$[0,1/3]$$ after the application of the second step of the construction, so $$|x-y|\geq 1/3^2$$, a contradiction. I believe that you get the idea.

Now let $$m$$ be so large that $$\sum_{k={m+1}}^\infty\frac{1}{2^k}<\varepsilon$$. Taking $$\delta=\frac{1}{3^m}$$; from the claim we have that if $$y\in C$$ satisfies $$|x-y|<\delta$$, then $$a_k=b_k$$ for all $$k=1,\dots,m$$, so $$d_P(f(x),f(y))=\sum_{k=m+1}^\infty\frac{|a_k-b_k|}{2^{k+1}}\leq\sum_{k=m+1}^\infty\frac{2}{2^{k+1}}<\varepsilon$$ and continuity follows. Observe that $$\delta$$ is independent of $$x$$, so $$f$$ is uniformly continuous.

Is this good news? No, this is expected, since $$f$$ is continuous on a compact space, thus it must be uniformly continuous. Also, by (2) this gives us a homeomorphism of the cantor set and $$P$$.