# A closed subset of a Dedekind-complete order has subspace topology equal to order topology

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!

• I think you can handle this with the "it is easily seen that..." stuff. Aug 21 at 20:33
• @AlessandroDellaCorte along with many other famous last words... Aug 21 at 20:50
• Subspaces of ordered spaces are often called GO-spaces (after General Ordered). That may be the keyword you are missing. Hope this helps. Aug 21 at 20:57
• @François G. Dorais of course! One of my professors, 20 years ago, used to repeat: "ok, this is trivially true. Is it also true?" Aug 21 at 23:19

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 $$b', and $$(b',b'')\cap S=\emptyset$$. Thus $$U_b=(\leftarrow,b']\cap S=(\leftarrow,b'')\cap S$$, finishing the proof.