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Last edited by Czhang271828 at 2022-8-7 11:32:00Hint: In metric spaces, seq compact $\Leftrightarrow$ compact.
An unsuccessful attempt (the previous answer) is hidden by the white font: If $X$ is not seq compact, then $\exists\{x_n\}_{n\geq 1}\subset X\subset $ s.t. $\|x_n\|\geq n$ for each $n\in \mathbb Z_{\geq 1}$. As a result, the continuous function $\|\cdot \|:\mathbb R^n\to \mathbb R,x\mapsto \|x\|$ is unbounded.
Solution. For the sake of contradictory, we assume that $X$ is not compact, that is, not seq compact.
If$X$ is unbounded, then each norm function is unbounded either. Since norm functions are always continuous, it contradicts our assumption.
Otherwise $X$ is bounded, hence is not closed. There exists $x\in \overline X\setminus X\,(\neq \emptyset)$ such that the function $t\mapsto \dfrac{1}{\|x-t\|}$ is well defined on $X$ and unbounded. In order to lead a contracdiction, we shall prove its continuity as follows:
For each $x_0\in X$ and arbitrary $\varepsilon >0$, there exists $\delta>0$ such that $\overline {B_\delta (x_0)}\subset \mathbb R^n\setminus \{x\}$. It is obvious that $t\mapsto\dfrac{1}{\|x-t\|}$ is continuous on $X\cap B_\delta (x_0)$ with Lipschitz coefficient $\dfrac{1}{(\sqrt{\|x-x_0\|^2-\delta^2})^2}$. As a result, $t\mapsto\dfrac{1}{\|x-t\|}$ is continuous on $\forall x_0\in X$, thus is continuous on $X$. |
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Czhang271828
Posted at 2022-8-3 15:52:20
