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Let $\tau:=\{A\subseteq \mathbb R: \mathbb R\setminus A \operatorname{finite}\}$

Show that $[0,1]$ is compact.

Let $(A_{n})_{n}$ be an open cover of $[0,1]$

I had two ideas:

$1.$ Since $[0,1]\subseteq\bigcup_{n \in \mathbb N}A_{n}$ there exists an $x \in [0,1]$ and an $\alpha \in \mathbb N$ so that $x \in A_{\alpha}$

I want to claim that since $A_{\alpha}\in\tau\Rightarrow\mathbb R\setminus A_{\alpha}$ is finite and thus $[0,1]\subseteq A_{\alpha}$ but I have not been able to show this directly.

Then I came to second idea:

$2.$ By contradicton, assume that the $A_{\alpha}$ chosen in $1.$ does not contain all of $[0,1]$ and since $\mathbb R \setminus A_{\alpha}$ is finite, $[0,1]$ is at most not covered by $A_{\alpha}$ by finitely many points, and thus we can find a finite subcover.

Any ideas/suggestions/corrections, are well-received.

SABOY
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  • Compactness is defined for arbitrary open covers, not just countable ones; so the notation in your expressions $(A_n)n$, and your use of the notation $n \in \mathbb N$ or $\alpha \in \mathbb N$, must be altered to allow for an arbitrary index set in place of $\mathbb N$. Also, your strategy to prove $[0,1] \subset A\alpha$ is fruitless; I can easily construct an open cover of $[0,1]$ for which no element of the cover contains $[0,1]$, for instance ${U_0,U_1}$ where $U_0=(0,1]$ and $U_1=[0,1)$. – Lee Mosher May 07 '19 at 16:46

1 Answers1

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Let $(A_\alpha)$ be an open cover of $[0,1]$. In particular there exists $A_\beta$ such that $0\in A_\beta$. Since $A_\beta$ is open, $A_\beta$ covers all but finitely many points ${x_1, \dotsc, x_n}$ of $[0,1]$. The remaining finite number of points not covered can be covered by choosing $A_{\beta_{i}}$ such that $x_i\in A_{\beta_{i}}$.

Sri-Amirthan Theivendran
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