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Let $K$ be a compact set and let {$U_{\alpha}$}$_{\alpha \in A}$ be an open covering of $K$. Prove that there is an $\epsilon$ $> 0$ such that if $k \in K$ then the interval $(k - \epsilon, k + \epsilon)$ lies in some $U_{\alpha}$.

Note that the same $\epsilon$ $> 0$ must work for every $k \in K$.

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Hint: Start by finding a finite subcover. –  Arturo Magidin Feb 23 '12 at 18:22
See here; or google "Lebesgue number". –  David Mitra Feb 23 '12 at 18:31
When asking a homework question, you should explain what you have done so far. –  Nate Eldredge Feb 23 '12 at 18:50

1 Answer 1

Suppose there was no such $\epsilon$. In that case, for every $\epsilon>0$, there exists a point $k$ and an interval $(k-\epsilon, k+\epsilon)$ not contained in any $U_\alpha$. So, for each $n \in \mathbb{N}$, we can choose a $k_n\in K$ such that no $U_\alpha$ contains $(k_n-1/n, k_n+1/n)$ . Now, $X$ is compact so there exists a subsequence $(k_{n_i})$ of the sequence of points $(k_n)$ that converges to some $l\in K$. But since $\mathcal{U}$ is an open cover, there is some $\delta$ such that $(l-\delta, l+\delta) \subseteq U_\beta$ for some $\beta\in A$. Pick $i$ large enough so that $|k_{n_i}-l|< \delta/2$ and $1/n_i < \delta/2$. Use the triangle inequality to show that $(k_{n_i}-1/n_i, k_{n_i}+1/n_i)\subseteq U_\beta$:

$$\left|k_{n_i}+\frac{1}{n_i}-l\right|\leq \left|k_{n_i}-l\right|+\left|\frac{1}{n_i}\right| < \frac{\delta}{2}+\frac{\delta}{2} = \delta$$

$$\left|k_{n_i}-\frac{1}{n_i}-l\right|\leq \left|k_{n_i}-l\right|+\left|\frac{1}{n_i}\right| < \frac{\delta}{2}+\frac{\delta}{2} = \delta$$

Then $(k_{n_i}-1/n_i, k_{n_i}+1/n_i)\subseteq U_\beta$, which is a contradiction since no $U_\alpha$ was supposed to contain these intervals.

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