Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Some caveats: Let $K$ be non-empty and compact, $F$ be non-empty and closed, $X, Y \subset \mathbb{R}^n$.

Define $d(K, F) = \inf\{ d(x,y), x \in K, y \in F\}$, where $d(x,y)$ is one of the metrics $d_1$, $d_2$, or $d_\infty$ on $\mathbb{R}^n$.

Show that $d(a,b) = d(K,F)$ for some $a\in K$, $b\in F$.

It seems like the solution would use some properties of compactness and perhaps the Heine-Borel theorem, but I can't figure out where to start. Any help much appreciated.

share|cite|improve this question
@t.b. Thanks for this! I think the questions are different, though: this question doesn't require disjoint sets, and is more about attaining the inf instead of finding that it is greater than 0. – dhz Dec 2 '11 at 12:06
I just noticed that I didn't prove exactly the statement you wanted to see. The point is that $d(\cdot,F)$ is continuous, hence it assumes its minimum at some point $k \in K$ because $K$ is compact. Now choose $f_n \in F$ such that $d(k,f_n) \to d(k,F)$. Since the intersection of $F$ with the closed ball of radius $d(k,F) + \varepsilon$ around $k$ is compact, we may extract a convergent subsequence of the $f_n$. The limit point $f$ will satisfy $d(k,f) = d(k,F) = d(K,F)$ and $f \in F$ because $F$ is closed. – t.b. Dec 2 '11 at 12:08
got it. thanks! Brilliant. – dhz Dec 2 '11 at 13:02
You start off with X and Y and seem to change them to K and F... – Daniel Freedman Dec 2 '11 at 14:56
up vote 7 down vote accepted

For the sake of having an answer, here's a slightly different argument.

For each $n$ there are $k_n \in K$ and $f_n \in F$ such that $d(K,F) \leq d(k_n,f_n) \leq d(K,F) + 1/n$ by definition of the infimum. Since $K$ is compact, we may pass to a subsequence $k_{n_i} \to k \in K$. For $\varepsilon \gt 0$ we can ensure for $i$ large enough that $d(k,f_{n_i}) \leq d(k,k_{n_i})+d(k_{n_i},f_{n_i}) \leq d(K,F) + \varepsilon$. Since the intersection $F'$ of $F$ with the closed ball of radius $d(K,F)+\varepsilon$ around $k$ is compact and $f_{n_i} \in F'$ we can pass to a further subsequence such that $f_{n_i}$ converges to some $f \in F' \subset F$. Then $$d(K,F) \leq d(k,f) = \lim_{i \to \infty} d(k_{n_i},f_{n_i}) \leq \lim_{i \to \infty} \left(d(K,F) + \frac{1}{n_i} \right)= d(K,F)$$ proves that $d(k,f) = d(K,F)$ as desired. (this works for any of the metrics you're interested in)

Notice that we used compactness crucially twice. First for passing to a subsequence $k_{n_i}\to k$ and then we used compactness of closed balls to conclude that the intersection of some closed ball with the closed set $F$ is compact and thus we could pass to a further subsequence.

Without assuming that either one of $F$ or $K$ is compact, the statement you ask about becomes wrong. For instance the subsets $F = \mathbb{N}$ and $K = \{n+1/n: n\in\mathbb{N}, n \geq 2\}$ of $\mathbb{R}$ have $d(K,F) = 0$ but as $F \cap K = \emptyset$ there are no points $f \in F$ and $k \in K$ such that $d(k,f) = 0 = d(K,F)$.

Here are three related threads that you might find interesting:

The first one is the one I erroneously identified as a duplicate, the second one shows that the distance function $d(\cdot,A)$ from a non-empty set is continuous and the third one shows in particular that for answering your question the ingredient that closed and bounded sets are compact (a special feature of $\mathbb{R}^n$, also called the Heine-Borel property) is crucial, too.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.