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Let A and B be nonempty compact subsets of $\mathbb{R}^n$ with $A \cap B = \emptyset$. Then there exists a $\delta > 0 $ such that for all $a \in A$ and $b \in B$, $|a-b|> \delta$.

I am having trouble with this problem because I do not see where the compactness comes in. To me it seems that since $|a-b|$ is a metric we have that it must be greater than or equal to zero. Therefore if we assume the negative, i.e $|a-b|=0$ then by defintion of a metric we must have $a=b$ which is a contradiction to the fact that A and B are disjoint. What is wrong and how should I go about this. Thank you!

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    To see what can happen without compactness consider the two parts of the graph of $x \mapsto 1/x^2$.2012-12-13
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    Or just let $n=1$ and $A=\{0\}$ and $B=(0,1)$.2012-12-13
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    I see from those examples we need compactness but where did I do something wrong in my explanation. It must be wrong since I never used compactness. Can you give me a hint on how to start the proof for this? Thank you!2012-12-13
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    The negative would be that however small you choose $\delta$ there will be points in $A$ and $B$ that are less than $\delta$ apart. That does not mean that their distance is zero.2012-12-13

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HINT: The distance function is continuous, continuous functions assume ther minimum on compacta.

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Since $A \cap B = \varnothing$ each pair of points from $A$ and $B$ has a positive distance between them and therefore the sets

$$ U_n = \{ (a, b) \in A \times B \mid |a - b| > \tfrac{1}{n} \} $$

cover $A \times B$. What properties do you need from $A \times B$ and $U_n$ to finish this proof?

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Here is what might confuse you: you say that the negative of there exists a $\delta > 0$ such that, for any $a \in A, b \in B$, you have $|a - b| > \delta$ is $|a - b| = 0$. This would only be true if $a, b$ were fixed elements; however, as stated, you have a uniform $\delta$ for any two points you can pick in your spaces. In particular, as $\mathbb{R^{n}}$ is metric, you can choose a sequence of elements for which this inequality will hold. Similar to the comments above, this wouldn't hold for, eg, $[-1, 0) \cup (0, 1]$ (choose a sequence $a_n$ converging to zero from above; and one $b_n$ converging to zero from below). It's quite obvious that the statement would not be true: for any $\delta$, you can find sequence elements which are $|a_n - b_n| < \delta$.

As a hint how to use compactness:

(1) fix a point $x_0$ in one set $A$, and a neighborhood $N_x$ around it. For each $y$ in the other set $B$ choose a neighborhood that does not intersect $N_x$. Then, by compactness of $B$, you can find a finite cover $\cup U_i$ of $B$ that does not intersect N_x.

(2) Do this for every point $x$ in A, and use compactness again, this time for the set $A$. Fill in the details.

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    Please forgive my slowness on the matter. I have been teaching it to myself and I find some of it pretty hard. First off assume the negative (hopefully this is the negative now) For all $\delta >0 \;\; |a-b|< \delta$. Let $a_1 \in N_{\epsilon}(a)$, $a_2 \in N_{\epsilon/2}(a)$, and $a_n \in N_{\epsilon/n}(a)$ Therefore there is a sequence is a seq in A converging to $a$. but since $|a-b|<\delta$ we have that $\{a_n\}$ converges to $b$ also. Since A is compact we have that $b \in A$ which contradicts the fact A and B are disjoint. Is that somewhat right?2012-12-13
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    I'm out and about (on phone), so it's a bit hard to type, but that's not the negative: the negative is that for any $\delta > 0$, there is an $a \in A, b \in B$, such that $|a - b| \leq \delta$. From cursory reading outside, this should show that the negative doesn't imply your construction.2012-12-13
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    As to apologizing, no need: everyone learning math is confused about what he currently learns. It will get clear over time, some chapters later. To get predicate logic right is a key part of you learning. It too will come with time. Make sure you learn this from a good book. I warmly recommend Munkres, Topology - assuming you had about 1 year of a decent analysis (not calculus) class.2012-12-13