Definition 1 : The set $A\subset \mathbb R$ is compact if it's closed and bounded.

Definition 2 : The set $A\subset \mathbb R$ is compact if for all cover $\mathcal U$ of open set there is $U_1,...,U_n\in\mathcal U$ s.t. $$A\subset \bigcup_{i=1}^n U_i.$$

I have to show that both definition are equivalent.

My attempt

for Def 1 $\implies $ Def 2 : Let $(x_n)$ a convergent sequence of $A$. Note $\ell$ it's limit. Since $A$ is close $\ell\in A$. How can I continue ?

For the reciprocally, Let $\mathcal U$ a cover of open and $U_1,...,U_n\in \mathcal U$ such that $$A\subset \bigcup_{i=1}^n U_i.$$ Let $(x_n)$ a convergent serie. How can I show that the limit is in $A$ ? And that $A$ is bounded ?

thank you :-)

  • $\begingroup$ What are open sets on $\mathbb{R}$? That should help find your cover. $\endgroup$ – IAmNoOne Sep 26 '15 at 20:53
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem $\endgroup$ – user84413 Sep 26 '15 at 21:12
  • $\begingroup$ Why would you use limits? To proof that a closed and bounded subspace in $\mathbb{R}$ is compact, you could use that there is a $F_0 =[-a,a]$ (which is closed and bounded) which contains your closed and bounded subspace. Closed subspaces of a compact set are compact. Thus it is sufficient to show that $F_0$ is compact. $\endgroup$ – Jan Sep 26 '15 at 21:12

Just something to think about. As others mentioned, this is called Heine-Borel theorem.

Def1 $\Rightarrow $ Def 2. Proof by contradiction. Let's construct: $$I_{0} = A$$ $$I_{1} = I_{0} - U_{1}=A-U_{1}$$ $$I_{2} = I_{1} - U_{2}=A-(U_{1} \bigcup U_{2})$$ $$...$$ $$I_{k} = I_{k-1} - U_{k}=A-(\bigcup_{i=1}^{k} U_{i})$$ $$...$$ Where $$I_{k}\neq \varnothing, \forall k$$ otherwise we have finite sub-cover. Each $I_{k}$ is closed and bounded (because $U_{k}$ is open), and $$I_{0} \supseteq I_{1} \supseteq I_{2} \supseteq ... \supseteq I_{k} \supseteq ...$$

Cantor's theorem (https://en.wikipedia.org/wiki/Cantor%27s_intersection_theorem) says that $\bigcap I_{k} \neq \varnothing$, but $\bigcap I_{k}=\bigcap (A-(\bigcup_{i=1}^{k} U_{i}))=A-\bigcup (\bigcup_{i=1}^{k} U_{i})=A-\bigcup U_{k}=\varnothing $. So one of $I_{k}$ must be $\varnothing$.

Def2 $\Rightarrow $ Def 1. Left as a home exercise. Besides, you can find this proof in (almost) any mathematical analysis book.


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