Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 $\frac{1}{n}>\frac{1}{n+1}$ –  Dimitrios Ntalampekos 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 : math.stackexchange.com/questions/109167/… –  Dimitrios Ntalampekos Apr 21 '13 at 22:38
But now $0\in A$. –  Stefan Hamcke Apr 21 '13 at 22:38
show 1 more comment

2 Answers 2

up vote 3 down vote accepted

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 $x_0\not\in A$

share|improve this answer
add comment

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

share|improve this answer
add comment

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.