Can we prove that a bounded closed subset of $\mathbb R^n(n \ge 1)$ is compact without using Axiom of Choice?
This is a related question which was closed.
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Sign up to join this communityCan we prove that a bounded closed subset of $\mathbb R^n(n \ge 1)$ is compact without using Axiom of Choice?
This is a related question which was closed.
First, closed and bounded intervals in $\Bbb R$ are compact: It suffices to prove for $[0,1]$. Let $\mathcal B$ be an arbitrary open cover of $[0,1]$, simply consider $$x=\sup\{y\in[0,1]\mid [0,y]\text{ has a finite subcover in }\mathcal B\},$$ deduce that $[0,x]$ is finitely covered as well, and then argue that we have to have $x=1$ (by the same reason).
Next, show that the product of finitely many compact sets is compact (done by induction, and the only interesting case is the case for product of two compact sets, the argument is quite straightforward by considering an open cover of the product and finding a finite subcover). Therefore closed and bounded boxes in $\Bbb R^n$ are compact, as products of closed intervals.
Finally, closed subsets of a compact space are compact. The proof is the same as with the axiom of choice.
Now we have that every closed and bounded set in $\Bbb R^n$ can be bounded by a product of closed intervals. So it is a closed subset of a compact set, so it is compact.