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I want to query, whether I'm right. (I'm sorry if don't use the correct words in my translation, please feel free to correct, and give me hints.)

I have a metric space $(X,d)$ which is sequentially compact (that means every sequence has a convergent subsequence). I want to show: by a given open cover $\mathfrak{U}$ of $X$, there is a $\varepsilon>0$, such that $K_\varepsilon(x):=\{y\in X : d(x,y)<\varepsilon\}\subseteq U$, for some $U\in\mathfrak{U}$.

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We assume there is no $\varepsilon>0$. For every $n\in\mathbb{N}$ we define $\varepsilon_n:=\frac{1}{n}$. Now it exists for every $\varepsilon_n$ a $x_n\in X$ such that $K_{\varepsilon_n}(x_n)\nsubseteq U$ for all $U\in\mathfrak{U}$ by assumtion. The sequence $(x_n)_{n\in\mathbb{N}}$ has a convergent subsequence, such that $x_{n_\nu}\to x$ for $\nu\to\infty$. Now, we choose $U\in\mathfrak{U}$ such that $x\in U$. $U$ is open, hence there exists a $\delta>0$ such that $K_\delta(x)\subseteq U$. Because our subsequence converge there is a $N\in\mathbb{N}$ such that $x_{n_\nu}\in K_{\frac{\delta}{2}}(x)$ for all $\nu\geq N$. Let $\varepsilon:=\frac{\delta}{2}$. So we get $K_\varepsilon(x_{n_\nu})\subseteq K_\delta(x)\subseteq U$ für all $\nu\geq N$ However, $\varepsilon_n$ is convergent with $\lim_{n\to\infty}\varepsilon_n=0$ and $K_\varepsilon(x)$ is open. Hence there exists a $N'\geq N$ such that $K_{\varepsilon_n}(x_n)\subseteq K_\varepsilon(x)\subseteq U$, this is our contradiction.

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    $\begingroup$ The idea of your proof seems to be correct, but maybe some last steps should be done more accurately. It seems that a sequentially compact metric space is compact, and each open cover of it has non-zero Lebesgue number (which you essentialy proved), so I think you may accept your answer. $\endgroup$ – Alex Ravsky Nov 28 '13 at 11:09

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