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I know that a previous proof of this question can be found here, but I want to try to show the above statement using the fact that a set is compact if it is both closed and bounded.

We know that any intersection of closed sets is closed. So we need to show that any intersection of bounded sets is bounded.

A set is bounded if it is contained in some ball, $B(0, R)$ about $0$ (or equivalently in a ball about any point).

Since each set $S$ is bounded, there exists a ball $B(0,R)$ such that $S\subset B(0,R)$.

Is it sufficient to say that any intersection of these bounded sets is also bounded since the intersection is a subset of each of its sets (which are bounded)?

Therefore, the intersection of infinitely many compact sets is compact since is it closed and bounded.

J. W. Tanner
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Raoul Duke
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    compact means closed and bounded in $\mathbf R^n,$ according to the Heine-Borel theorem – J. W. Tanner Dec 19 '19 at 03:11
  • Yes, it's that clear (in $\mathbb{R}^n$). The intersection will live inside one of its terms, but each of its terms is already bounded. – Randall Dec 19 '19 at 03:11

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Yes you can say the intersection of an arbitrary collection of bounded sets $\{ A_\alpha\}$is a bounded. This is actually quite simple to see. Choose any $ A \in \{ A_\alpha\}$. $\forall x \in \cap\{ A_\alpha\}, x\in A$. By the definition of boundedness, $\forall x \in A, |x| \le M$ for some $M > 0$. So therefore $\forall x \in \cap\{ A_\alpha\}, x < M$.

Ziad Fakhoury
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