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This is the problem:

Let $X$ be a metric space with metric $d$ and $K\subset X$ compact and $F\subset X$ closed and $K\cap F=\varnothing$. Let $x\in F,y\in K$. Prove that there exists $\delta>0$ such that for every $x\in F$ and $y\in K$, $d(x,y)> \delta.$

My idea was to use sequences,once I know that in a compact set we have a subsequence converging to an element in $K$ and how $F$ is closed there's a sequence converging to an element in $F$.

But ,then I got stuck...I don't know what to do now.Any hint?Much appreciated!

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2 Answers

up vote 2 down vote accepted

Assume by contradiction, then for every $n$ you can find $x_n\in K, y_n\in F$ such that $d(x_n,y_n)<\frac1n$.

Since $K$ is compact there is a convergent subsequence $x_{n_k}$, whose limit is $x$. By triangle inequality $$d(x,y_{n_k})\leq d(x,x_{n_k})+d(x_{n_k},y_{n_k})$$

Juggle $\varepsilon$'s a bit and derive that $x\in F$, and the contradiction.

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I thought about triangle inequality...but I looked at it and seemed to be wrong...couldn't go forward.Thanks!!! –  Charlie Aug 23 '12 at 0:53
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I assume that you want to prove that there exists $\delta > 0$ such that for every $x\in F$ and $y\in K$, $d(x,y) > \delta$. (The order of quantifiers is opposite in your question, which makes it trivial.)

Hint: prove that there is a sequence $(x_k, y_k)$ s.t. $d(x_k,y_k) < 1/k$. Find a converging subsequence $y_{k_i}$. Let $y_{k_i} \to y \in K$ as $i\to \infty$. Prove that $x_{k_i} \to y \in K$. Conclude that $K$ and $F$ are not disjoint.

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Thanks a lot!It's really useful! –  Charlie Aug 23 '12 at 0:53
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