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.

  • $\begingroup$ @AsafKaragila [Many of these questions have been asked before. Please search the site before posting them.] Where is the answer to my question? $\endgroup$ – Makoto Kato Apr 12 '14 at 8:08
  • $\begingroup$ Where is the answer to my question? Below, in case you haven't noticed. $\endgroup$ – Asaf Karagila Apr 12 '14 at 8:15
  • $\begingroup$ An older question: math.stackexchange.com/questions/176646/… $\endgroup$ – Martin Sleziak Apr 12 '14 at 11:14
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    $\begingroup$ The answer follows directly from Shoenfield's absoluteness theorem. $\endgroup$ – Carl Mummert Apr 13 '14 at 0:18
  • $\begingroup$ @CarlMummert Would you please elaborate on your claim as an answer? $\endgroup$ – Makoto Kato Apr 14 '14 at 5:34

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.

  • $\begingroup$ Could you explain how you prove that the product of two compact sets is compact without using Axiom of Choice? $\endgroup$ – Makoto Kato Apr 12 '14 at 8:24
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    $\begingroup$ The same way you prove it in $\sf ZFC$. Pick an open cover, consider its projections, find a finite subcover for the first coordinate, then a finite subcover for the second coordinate. Certainly someone who spent the last two years working on very nontrivial questions can come up with this on their own. $\endgroup$ – Asaf Karagila Apr 12 '14 at 8:26
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    $\begingroup$ Your question doesn't show any effort, why should my answer show effort? Please remember that this is not your answer, and I think that if someone reads an answer and they have to work out some of the details on their own, it's not a big deal. $\endgroup$ – Asaf Karagila Apr 12 '14 at 8:39
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    $\begingroup$ @egreg: Uh, no, it doesn't. You're confusing between the existence of $\sup$ and the fact that it is the limit of a sequence from the set. $\endgroup$ – Asaf Karagila Apr 12 '14 at 9:43
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    $\begingroup$ @AsafKaragila Thanks; that was just a little doubt. On the other hand, it's well known that the proof of Tychonov's theorem for the finite case doesn't require the axiom of choice, although many proofs that can be found around use it (mostly because the same argument can be extended to the infinite case). $\endgroup$ – egreg Apr 12 '14 at 10:12

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