Prove that if $A$ is compact and $B$ is closed and $A\cap B = \emptyset$ then $\text{dist}(A,B) > 0$

Question

Let $X$ be a metric space. For nonempty subsets $A,B\subseteq X$. Define $\text{dist}(A,B) := \inf\{d(x,y) : x\in A, y\in B\}$

a) Prove that if $A$ is compact and $B$ is closed and $A\cap B = \emptyset$ then $\text{dist}(A,B) > 0$

b) Suppose that $X = \mathbb{R}^n$, with the standard metric. Under the same assumptions of (a), prove that $\exists x_0\in A$ and $\exists y_0\in B$ such that dist$(A,B) = ||x_0 - y_0||$.

For a), it seems to me that the crucial point is that $A\cap B = \emptyset$. I can immediately see how this would imply that $\text{dist}(A,B) > 0$. However, I do not understand why it's necessary for $A$ to be compact, and $B$ to be closed. Could you explain why those conditions are necessary?

Prahlad Vaidyanathan · Accepted Answer · 2013-10-18T19:18:45.367

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a) You need $B$ to be closed so that for any $x\notin B, d(x,B) > 0$. You need $A$ to be compact, so that you can restrict your attention to only finitely many $x$'s as follows :

For every $x \in A, \epsilon_x := d(x,B) > 0$, so there is a $\delta_x > 0$ such that $$ d(x,y) < \delta_x \Rightarrow d(y,B) > \epsilon_x/2 $$ Now the balls $B(x,\delta_x)$ cover $A$. Take a finite subcover $\{B(x_i,\delta_{x_i})\}$, and choose $\epsilon$ to be the minimum of the corresponding $\epsilon_{x_i}$'s. Now check that $$ d(A,B) > \epsilon/2 > 0 $$

b) Note that $$ x \mapsto d(x,B) $$ is a continuous function on $A$, and so there is $x_0 \in A$, at which the infimum is attained. ie. $$ d(x_0, B) = \inf\{\|x-y\| : y \in B\} = \inf\{\|x_0 - y\| : y \in B\} $$ Now there is a sequence $(y_n) \in B$ such that $$ d(x_0,B) = \lim \|x_0 - y_n\| $$ Check that $(y_n)$ must be a bouned sequence, and so it must have a convergent subsequence (since you are in $\mathbb{R}^n$). This subsequence must converge to a point $y_0 \in B$. Then $$ d(A,B) = d(x_0,B) = \|x_0 - y_0\| $$

edited Oct 18 '13 at 19:18

answered Oct 18 '13 at 19:11

Prahlad Vaidyanathan

31,554

Isn't the same fact true even if A is only closed, but not Compact? If not, is there a counterexample? – Mateus Sampaio Oct 18 '13 at 19:30
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@MateusSampaio : The graph of $y=1/x$ and the $X$-axis are closed in $\mathbb{R}^2$, but their distance from each other is zero. – Prahlad Vaidyanathan Oct 18 '13 at 19:40
what do you mean by cover and subcover? – JohanLiebert Oct 18 '13 at 21:00
@JohanLiebert, this is standard terminology for a collection of (open) sets whose union contains the set you started with. See http://en.wikipedia.org/wiki/Cover_%28topology%29 – Prahlad Vaidyanathan Oct 19 '13 at 02:42
In a) where does $y$ come from? – kombucza Nov 28 '21 at 21:57

Philippe Malot · Answer 2 · 2013-10-18T21:49:58.073

This does not answer your questions but it's another proof of a).

If $\text{dist}(A,B)=0$, then by definition of this distance there are two sequences $(a_n)$ and $(b_n)$ (with terms in $A$ and $B$ respectively), such that $\lim d(a_n,b_n)=0$.

Since $A$ is compact, you know that there exists a convergent subsequence $(a_{n_k})$ of $(a_n)$ in $A$.

Let $a\in A$ be its limit. $(d(a_{n_k},b_{n_k}))_{k\geqslant 0}$ is a subsequence of the convergent sequence $(d(a_n,b_n))$, so it has the same limit.

Since : $d(a,b_{n_k})\leqslant d(a,a_{n_k})+d(a_{n_k},b_{n_k})$ for all $k$, you find that $\lim b_{n_k}=a$

But $(b_{n_k})$ is a sequence with terms in the closed set $B$, so its limit $a$ is an element of $B$.

Remember that we also have $a\in A$. This proves that $A\cap B\neq\emptyset$.

Consequently, if $A\cap B=\emptyset$ with $A$ compact and $B$ closed, then $d(A,B)>0$.

http://math.stackexchange.com/questions/491179/show-that-mathtt-da-b0-if-a-cap-b-varnothing-a-is-compact-and-b-is?rq=1 — Philippe Malot, Oct 18 '13 at 21:53

score 2 · Answer 3 · answered Oct 18 '13 at 19:20

For an example of why $B$ is needed to be closed consider in $\mathbb R$: $$ A = \{0\}, B=(0,1] $$ clearly $A\cap B=\emptyset$ but $d(A,B)=0$.

The proof of a) and b) is based on the Weierstrass theorem: every continuous function on a compact set has a minimum.

Take the function $$ f(x) = \inf \{d(x,y)\colon y \in B\} $$ this is a continuous function hence has a minimum on $A$. So for some $a\in A$ one has $$ d(A,B) = f(a) = d(a,B). $$ Now notice that $a\not \in B$ and hence $d(a,B) > 0$.

To prove (b) notice that by intersecting $B$ with a bounded closed set you can restrict yourself to the case when also $B$ is compact.

Prove that if $A$ is compact and $B$ is closed and $A\cap B = \emptyset$ then $\text{dist}(A,B) > 0$

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