1

Let $(X,d)$ be a metric space and $K \subset X$ a compact subset. We define $$\text{dist}_K(x) := \inf_{y \in K} d(x,y).$$ Show that

  1. for every $x \in X \setminus K$ there exists disjoint open subsets $U,V \subset X$ so that $x \in U$ and $K \subset V$. Note: I know this is very similar to this question and this one but here, I have a different metric.

  2. $\text{dist}_K(x) = 0 \iff x \in K$.

My Proof

  1. Let $x \in X \setminus K$. Then we have $\text{dist}_K(x) > 0$ and therefore $U := B\left(x, \frac{1}{2}\text{dist}_K(x)\right) \subset X$ (which is the open ball centered around $x$ with radius $ \frac{1}{2}\text{dist}_K(x)$) which is an open subset containing $x$. Now, define $V := \bigcup_{k \in K} B\left(k, \frac{1}{2}\text{dist}_K(x)\right)$ which is open in $X$ as union of sets which are open in $X$. Now, to show $\tilde{U} := U \cap V = \emptyset$, we assume there exists an $u \in \tilde{U}$ and seek a contradiction. By definition we have \begin{equation} d(u,x) < \text{dist}_K(x) \quad \forall x \in X \setminus K \qquad \text{and} \qquad d(u,k) < \text{dist}_K(x) \quad \forall k \in K \end{equation} Adding both equations, for all $x \in X \setminus K$ and all $k \in K$ we have \begin{equation} \inf_{y \in K} d(x,y) > d(x,u) + d(u,k) \overset{\triangle \neq}{\ge} d(x,k), \end{equation} which is a contradiction.

  2. "$\implies$": If $\text{dist}_K(x) = 0$ we have $I := \inf_{y \in K} d(x,y) = 0$ By the definition of the infimum there exists a sequence $(x_n)_{n \in \mathbb{N}} \subset K$ with $\lim_{n \to \infty} x_n = x$. But since $K$ is compact it's also closed and therefore we know that $x \in K$.

"$\impliedby$": If $x \in K$, we have \begin{equation} \text{dist}_K(x) = \inf_{y \in K} d(x,y) = d(x,x) = 0, \end{equation} because $d$ is a norm and therefore positive definite on $X$.

ViktorStein
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  • In $2\implies$ you wrote $I := \inf_{y \in K} d(x,y)$ and $I\in K.$ Note that $I \in\mathbb R.$ Moreover, what is monotonic sequence of elements of $K?$ – mfl Mar 15 '19 at 15:26
  • @mfl You're right. Is it correct now? – ViktorStein Mar 15 '19 at 15:30
  • No. You say $\lim_{n \to \infty} x_n = I.$ The limit belongs to $K$ (if it exists) and $I$ is a real number. Use that $(x_n)$ has a subsequence such that $x_n\to x\in K.$ – mfl Mar 15 '19 at 15:32
  • @mfl Why do I need a subsequence. Isn't my mistake just that I wrote $(x_n) \to I$ where as should be $(x_n) \to x$ and since $K$ is compact, $x \in K$? Note: I edited accordingly. – ViktorStein Mar 15 '19 at 15:50
  • Well, why are you sure the sequence converges? You need to show it. But what is sure is that there exists a subsequence which converges to a point in $K.$ – mfl Mar 15 '19 at 16:11
  • I thought that by definition for any set $A$ there always exists a sequence $(x_n)$ which converges to $\sup(A)$ and one that converges to $\inf(A)$. – ViktorStein Mar 15 '19 at 16:23
  • There is a sequence $(x_n)$ such that $\lim_n d(x_n,x)=d(x,K).$ But this doesn't mean that $(x_n)$ converges. Consider $K=\mathbb S^1$ and $x=(0,0)$ on the plane. It is always $d(x_n,x)=1.$ But you can consider the sequence $(0,-1), (0,1), (0,-1), (0,1), \cdots$ which is not convergent. Note that in $K$ we don't have any order. So, what is $\sup K?$ It makes no sense for general metric space. – mfl Mar 15 '19 at 16:28
  • You are right, but I am not using a sequence that is converging to inf(K), which wouldn't make any sense but instead a sequence with converges to $\inf { d(x,y): y \in K}$. – ViktorStein Mar 15 '19 at 16:43
  • You need to show that that sequence exist. In other case the proof is not complete. – mfl Mar 15 '19 at 16:56

2 Answers2

3

Your proof of $1$ isn't correct because until you prove $2$, you don't know that $d(x, K) \gt 0$.

In any event, but I'd simplify and generalize the proof so that it works in any Hausdorff space.

For each $y \in K~ \exists U_y, V_y~(y \in U_y, x \in V_y, U_y \cap V_y = \emptyset)$. The $U_y$ are an open cover of $K$ so choose finitely many of them that cover $K$, $U_{y_1}, \ldots, U_{y_n}.$ Then $V= \bigcap_{k=1}^n V_{y_k}$ is a finite intersection of open sets, so it's open, and by construction $V \cap U_{y_k} = \emptyset$ for all $k$, so let $U= \bigcup_{k=1}^n U_{y_k}.$ Then $x \in V, U \cap V = \emptyset$ and $K \subseteq U.$

Edited to correct comment on proof in original post.

Robert Shore
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  • If I would just prove 2 before 1, everything would be valid, since I don't need anything from 1 for the proof of 2, right? – ViktorStein Mar 15 '19 at 16:25
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    Yes, but $2$ is where the action is. Notice that you don't seem to be using the fact that $K$ is compact, merely that it's closed. In fact, that's all you need in a metric space but in a more general topological space this property (you can use open sets to separate a point and a closed set) means that the space is "regular" and not all Hausdorff spaces are regular. The proposition is still true for compact sets even if the underlying topological space isn't regular, as long as it's Hausdorff. – Robert Shore Mar 15 '19 at 16:41
1

For every $k\in K$ consider $B(k, \frac13d(k,x))$ and define $V=\bigcup_{k\in K} B(k, \frac 13 d(k,x)).$ Define $U=B(x,\frac 13 d(x,K)).$ (Note that $d(x,K)$ is shown in $2$.)

Assume $y\in U\cap V.$ So, there exists $k\in K$ such that $y\in B(k, \frac13 d(k,x)).$ But then

$$d(x,K)\le d(x,k)\le d(x,y)+d(y,k)\le \frac 13 d(x,K)+\frac 13 d(x,K)=\frac 23d(x,K)$$ which is not possible unless $d(x,K)=0.$ Since $d(x,K)>0$ we have shown that $U\cap V=\emptyset.$

In the second case, assume that $d(x,K)=0.$ So, there is a sequence $(x_n)$ of points of $K$ such that $\inf_n d(x,x_n)=0.$ Since $K$ is compact there exists a subsequence of $(x_n)$ which converges to a point $x_0\in K.$ Let's denote the subsequence by $(y_n).$ We have that

$$d(x,x_0)\le d(x,y_n)+d(y_n,x_0).$$ Since $\inf_n d(x,x_n)=0$ and $\lim_n y_n=x_0$ we get that $d(x,x_0)=0.$ That is $x\in K.$

mfl
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  • You proposed another proof for 1. Is mine not correct? The same goes to 2 $\implies$: I have now edited: is it now still not correct? – ViktorStein Mar 15 '19 at 15:53
  • Can you explain why is $\begin{equation} \inf_{y \in K} d(x,y)

    d(x,u) + d(u,k)

    \overset{\triangle \neq}{\ge} d(x,k), \end{equation}?$

     – mfl Mar 15 '19 at 16:15
  • I have $d(x,u) < \frac{1}{2} \text{dist}K(x)$ and $d(u,k) < \frac{1}{2} \text{dist}_K(x)$. If I add both equations, I get $d(x,u) + d(u,k) < \frac{1}{2} \text{dist}_K(x) + \frac{1}{2} \text{dist}_K(x) = \text{dist}_K(x) \overset{\text{Def}}{=} \inf{y \in K} d(x,y)$. By the triangle inequality we have $d(x,k) \le d(x,u) + d(u,k)$. – ViktorStein Mar 15 '19 at 16:21
  • I think your proof of $1$ is correct once one have shown that $d(x,K)>0.$ – mfl Mar 15 '19 at 16:31