# A question on metric spaces which does not have Lebesgue covering property

Let $X$ be a metric space , $\{U_{\alpha}\}$ be an open cover of $X$ which has no Lebesgue number . So for every $r>0$ , there is an open ball of radius $r$ which is not contained in any open set of the open cover , in particular , for every $n>1 , \exists x_n \in X$ such that $B(x_n , 1/n)$ is not contained in any of the open sets of the open cover . Now how do I show that $\{x_n\}$ has no cluster point i.e. no convergent subsequence ? Please help . Thanks in advance . ( I am planning a proof that if every real valued continuous function on a metric space is uniformly continuous , then the metric space has Lebesgue covering property , so subsequently any continuous function from $X$ to any metric space is uniformly continuous answering this If every real valued continuous function on $X$ is uniformly continuous , then is every continuous function to any metric space uniformly continuous? question of mine )

## 1 Answer

Assume for a contradiction that $x \in X$ is a cluster point of the $x_n$. Since the $U_{\alpha}$ cover $X$, there is some $\alpha$ such that $x \in U_{\alpha}$, and since $U_{\alpha}$ is open, for some $\epsilon > 0$, $B(x, \epsilon) \subseteq U_{\alpha}$. Now choose $N$ such that $1/N < \epsilon/2$. As $x$ is a cluster point of the $x_n$, $0 < d(x, x_n) < 1/n$ for infinitely many $n$, implying that for some $n > N$, we have $0 < d(x, x_n) < 1/n < 1/N < \epsilon/2$. But then (using the triangle inequality for the first inclusion), we have $B(x_n, 1/n) \subseteq B(x, \epsilon) \subseteq U_{\alpha}$.