Let $H \subseteq \Bbb R^n$.

Prove that $H$ is compact $\iff$ every cover $\{E_{\alpha}\}_{\alpha \in A}$ where $E_{\alpha}$'s are relatively open in $H$ has a finite subcovering.

$\bf{Solution \ trial:}$

For $\Rightarrow$

Suppose $H$ is compact.

Suppose $\{E_{\alpha}\}$ are relatively open covering of $H$. Since $\{E_{\alpha}\}$ are relatively open covering of $H$,

$\exists$ open set $U_{\alpha}$ such that $U_{\alpha} \cap H= E_{\alpha}$

Then,$\ U_{\alpha}$ is open covering of $H$

Since $H$ is compact, $\exists $ finite subset $A_0 \subset A$ such that $$H\subseteq \bigcup_{\alpha \in A_0} \{U_{\alpha}\}$$

Then, $\{E_{\alpha}\}_{\alpha\in A_0} $ is a finite subcovering of $\{E_{\alpha}\}_{\alpha\in A} $

For $\Leftarrow$

Since $\{E_{\alpha}\}_{\alpha\in A} $ is relatively open subcovering of H,

$$\{E_{\alpha}\cap H\}_{\alpha\in A}$$ is relatively open covering.

$\exists$ a finite subset $A_0 \subset A$ such that $\{ V_{\alpha} \cap H\}_{\alpha \in A_0}$ covers H.

$$\{ V_{\alpha} \}_{\alpha \in A_0}$$ covers H.

i.e H is compact.

Is the proof enough? Does There exist any mistake or missings in the detail of the solution?

Please correct them. Thank you.

  • 2
    $\begingroup$ This solution-verification is important for me. I am a new learner. And this is self-studying. Thus, I want to learn properly. Thank you:) $\endgroup$ – Bstr Jul 26 '13 at 21:24
  • $\begingroup$ That's actually the definition of compact: If $\{E_i\mid i\in I\}$ is a set of open subsets of $H$ whose union equals $H$, then finitely many suffice to cover $H$. But no matter how you define it, the equivalence is still useful. $\endgroup$ – Stefan Hamcke Jul 26 '13 at 21:27
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    $\begingroup$ For the second implication, do you mean to start with an arbitrary cover $\{V_\alpha\}$ of $H$ by open subsets $V_\alpha$ of $\mathbf{R}^n$? $\endgroup$ – Keenan Kidwell Jul 26 '13 at 21:28
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    $\begingroup$ Dear @Stefan, You forgot to add that the $E_i$ are open. $\endgroup$ – Keenan Kidwell Jul 26 '13 at 21:28
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    $\begingroup$ @KeenanKidwell: Thanks :-) $\endgroup$ – Stefan Hamcke Jul 26 '13 at 21:32

Assuming the first sentence of the second implication is meant to be ``let $\{V_\alpha\}$ be a covering of $H$ by open subsets of $\mathbf{R}^n$," your argument is correct.

  • $\begingroup$ Except for this, everthing is correct. Is it? Does the details of the answer is totally correct and enough really? $\endgroup$ – Bstr Jul 26 '13 at 21:34
  • $\begingroup$ Agreed. But the fact that this proof works for $H \subseteq X$, where $X$ is an arbitrary topological space, makes me think it was not what the exercise was asking. $\endgroup$ – Pedro M. Jul 26 '13 at 21:35
  • $\begingroup$ Dear @Pedro, I think this seems like a reasonable problem for a first course in analysis where one studies only metric spaces and not general topological spaces. $\endgroup$ – Keenan Kidwell Jul 26 '13 at 21:35
  • $\begingroup$ Dear @Bstr, Yes. I guess you should write $\{U_\alpha\}$ in the first implication, because it is the collection of $U_\alpha$ that constitutes the cover, but I assume this is what you intended. Do you understand why you should start with a cover of $H$ by open sets in $\mathbf{R}^n$ for the second implication? $\endgroup$ – Keenan Kidwell Jul 26 '13 at 21:36
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    $\begingroup$ @Keenan Kidwell: I agree. Nevertheless, in these courses the definition of compact set is usually "bounded closed set" (in $\mathbb{R}^n)$ or "a set where every sequence has a convergent subsequence whose limit lies in the set" (in a metric space). That leads me to think that it might be a harder problem, that of showing equivalence of definitions in $\mathbb{R}^n$. $\endgroup$ – Pedro M. Jul 26 '13 at 21:40

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