Distance between closed and compact sets.

Question

This question is (1-21)(b) from M. Spivak's Calculus on Manifolds.

Question: If $A$ is closed, $B$ is compact, and $A \cap B = \emptyset$, prove that there is $d > 0$ such that $||y - x|| \geq d$ for all $y \in A$ and $x \in B$.

Now, I interpret this as an instruction to find a single $d$ that works for all $y \in A$ and $x \in B$. However, I can't see why the following is not a counter-example:

Consider the set $$A_0 = (-\infty, 0) \cup \left[\bigcup_{n=1}^{\infty} \left(\frac{1}{n + 1}, \frac{1}{n}\right)\right] \cup (1, \infty)$$ where $(a,b)$ denotes the open interval as usual. Since $A_0$ is a union of open sets, it too is open. Thus $$A = \mathbb{R} - A_0 = \left\{ \frac{1}{n} \quad \colon \quad n \in \mathbb{N}\right\}$$ is closed. The set $$B = [-1, 0]$$ is certainly compact. Moreover, $A \cap B = \emptyset$. However, for all $d > 0$, there exists a $y \in A$ such that $$||0 - y|| = ||y|| < d$$

I must be overlooking something somewhere. Any help spotting where will be appreciated.

The complement is not ${1/n\mid n\in\mathbb N}$. You should also change the end points of the intervals as they are not in the correct order. — Stefan Hamcke, Apr 21 '13 at 22:32
$0\in A_0$? or maybe it is in $A$...? Also correct the definition of $A_0$, since $\frac{1}{n}>\frac{1}{n+1}$ — Dimitris, Apr 21 '13 at 22:36
@StefanH. I believe I have corrected the mistakes you mention. — providence, Apr 21 '13 at 22:36
Check an answer to the question here : http://math.stackexchange.com/questions/109167/if-a-is-compact-and-b-is-closed-show-da-b-is-achieved — Dimitris, Apr 21 '13 at 22:38
@StefanH. Ah, I see where this is going. I believe you have in fact answered my question. — providence, Apr 21 '13 at 22:41

score 9 · Accepted Answer · answered Apr 21 '13 at 22:46

9

Here we prove the result of the book:

Recall that the function $x\mapsto d(x,A)$ is continuous and that (since $A$ is closed): $$x\in A\iff d(x,A)=0$$

$$d=\inf_{x\in B}d(x,A)$$

The function

$$f:B\to \mathbb{R}\quad,\quad x\mapsto d(x,A)$$ is continuous on the compact $B$ then it's bounded and there's $x_0\in B$ s.t $$f(x_0)=\min_{x\in B}f(x)=d=d(x_0,A)>0$$ since $x_0\not\in A$

answered Apr 21 '13 at 22:46

Really clean proof, but how do you show that $d(x,A)$ is continuous? – grayQuant Feb 19 '15 at 04:52
We prove that it's lipschitzian function using the triangle inequality. – Feb 19 '15 at 05:56

score 4 · Answer 2 · answered Jun 20 '18 at 12:15

Spivak includes this exercise in a section of the book that precedes the definitions of functions and continuity. He gives a hint for solving it simply from the definition of compactness, and using a previous result, that the distance between a closed set and a single point in its complement is positive. That result follows immediately from the fact that the complement of a closed set is open, and a point in an open set in $\Bbb R^n$ has a ball around it that's contained in the open set.

Anyway, here's a proof, using Spivak's hint: for every $x$ in our compact set $B$, we have a distance $d_x$ so that $\|x-y\|>2d_x$ for all $y$ in our closed set $A$. Let $U_x$ be the ball of radius $d_x$ centered at $x$. Doing this for every $x\in B$, we obtain an open cover. This open cover has the property that, for each $x$, every point in $U_x\cap B$ has distance at least $d_x$ from $A$. (Triangle inequality)

Since $B$ is compact, the open cover $\{U_x\mid x\in B\}$ admits a finite subcover: $\{U_{x_i}\mid i=1,\ldots,n\}$. Let $d=\min\{d_{x_i}\}_{i=1\ldots n}$. This positive number satisfies the requirement that $\|x-y\|>d$ for all $x\in B, y\in A$.

score 1 · Answer 3 · answered Apr 21 '13 at 22:43

1

The counter-example fails as the set $A$ contains $0$ so $A \cap B \ne \emptyset$. I had overlooked this fact for some reason.

answered Apr 21 '13 at 22:43

providence

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Distance between closed and compact sets.

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