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Here's a fairly easy fact from point-set topology that I'm having trouble finding a reference for. Say $X$ is a total order satisfying the least-upper bound property, and $S$ is a closed subset of it. Then the subspace topology on $S$ and the order topology on $S$ coincide.

Does anyone know what would be a reference I can cite for this fact so I don't have to take up space in my paper to include a proof?

Actually, I'm pretty certain I've seen a proof of this statement given either here or else on math.stackexchange, but I haven't been able to find it. Those aren't ideal references, obviously, but still seem preferable to taking up space including a proof myself...

Thank you all!

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  • $\begingroup$ I think you can handle this with the "it is easily seen that..." stuff. $\endgroup$ Aug 21 at 20:33
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    $\begingroup$ @AlessandroDellaCorte along with many other famous last words... $\endgroup$ Aug 21 at 20:50
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    $\begingroup$ Subspaces of ordered spaces are often called GO-spaces (after General Ordered). That may be the keyword you are missing. Hope this helps. $\endgroup$ Aug 21 at 20:57
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    $\begingroup$ @François G. Dorais of course! One of my professors, 20 years ago, used to repeat: "ok, this is trivially true. Is it also true?" $\endgroup$ Aug 21 at 23:19
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While I agree that it's pretty direct to show, I was unable to find a reference for a proof of this fact myself (I thought it was in Willard, but I thumbed through my copy and failed to find it). So here's a proof in any case.

Let $S$ be a closed subset of a linear order $X$ with the least upper bound property, that is, every subset of $X$ with an upper bound has a least upper bound, and consequently, every subset of $X$ with a lower bound has a greatest lower bound. Note that any least upper bound or greatest lower bound for a subset of $S$ must belong to $S$ due to it being closed. Subbasic open sets for the subspace $S$ are given by $U_b=(\leftarrow,b)\cap S$ and $V_a=(a,\rightarrow)\cap S$ for $a,b\in X$; we will only show $U_b$ is open in $S$'s restricted order topology, since the proof for $V_a$ is identical (with orders reversed).

If $U_b=S$ or $U_b=\emptyset$ we are done; otherwise note that $U_b$ is non-empty and bounded above, so it has a least upper bound $b'\in S$, and likewise $S\setminus U_b=[b,\rightarrow)\cap S$ is nonempty and bounded below, so it has a greatest lower bound $b''\in S$. If $b'\not\in U_b$, then $U_b=(\leftarrow,b')\cap S$ and we're done. If $b'\in U_b$, then $b'<b\leq b''$, and $(b',b'')\cap S=\emptyset$. Thus $U_b=(\leftarrow,b']\cap S=(\leftarrow,b'')\cap S$, finishing the proof.

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  • $\begingroup$ Oh, that's actually quite a bit nicer than than the proofs I'd seen before or came up with myself. :) $\endgroup$ Aug 22 at 21:27

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