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$X⊂𝐑^n$,任何$X→𝐑$的连续函数是有界的,证明$X$列紧

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hbghlyj posted 2022-8-3 13:16 |Read mode
Let $X$ be a subset of $\mathbf{R}^n$ such that every continuous function $f: X → \mathbf{R}$ is bounded. Show that $X$ is sequentially compact.

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Czhang271828 posted 2022-8-3 15:52
Last edited by Czhang271828 2022-8-7 11:32Hint: 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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original poster hbghlyj posted 2022-8-6 14:11
Czhang271828 发表于 2022-8-3 08:52
If $X$ is not seq compact, then $\exists\{x_n\}_{n\geq 1}\subset X\subset $ s.t. $\|x_n\|\geq n$ for each $n\in \mathbb Z_{\geq 1}$.
I think “$X$ is not compact” doesn't imply “$X$ is unbounded”.
$X$ can be bounded but not compact, for example $(0,1)$.

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Czhang271828 posted 2022-8-6 14:36
hbghlyj 发表于 2022-8-6 14:11
I think “$X$ is not compact” doesn't imply “$X$ is unbounded”.
$X$ can be bounded bu ...
The answer in 2# is updated just now.

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original poster hbghlyj posted 2022-8-7 02:56

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Czhang271828 posted 2022-8-7 11:39
hbghlyj 发表于 2022-8-7 03:01
I think it should be $\overline {B_\delta (x_0)}\subset \mathbb R^n\setminus \{x\}$

I ha ...
Already fixed.

There is no need to determine exact values of Lipschitz coeffitient. The existence is quite obvious, since the $\mathbf{grad}$ of $f(t)=\dfrac{1}{\|x-t\|}$ remains bounded as $t\in X\cap B_\delta(x_0)$.

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