Prove that the intersection of an arbitrary collection of compact sets in R is compact.
Proof:
Let, $\{K_\alpha\}$ be a collection of compact sets in $\mathbb{R}$. This implies that the sets are closed and bounded. Then, the sets are also closed by the heine borel theorem. Then, since intersection of arbitrary number of closed sets is closed, this implies, that $\bigcap K_\alpha$ is closed. But since, all the sets in $\{K_\alpha\}$ are bounded, then this too, would be bounded as it will be a subset of all these sets. Then this implies, that the intersection is compact.
Can anyone verify this? Also, does arbitrary mean that we can take infinite many sets and this theorem would hold?
Thank you.
The proof is correct, but it has a sentence which is doing nothing there: “Then, the sets are also closed by the heine borel theorem”.
A somewhat different proof can be obtained by fixing one $\alpha$ and noting that, since $\bigcap K_\alpha$ is a closed subset of $K_\alpha$ and since $K_\alpha$ is compact, then $\bigcap K_\alpha$ must be compact too.