I have been looking into topology recently and have run into a brick wall with the order topology. A lot of the sources that I have been looking at have brushed over the proof that the order topology meets the axioms of a topology. I can prove that the whole space and the empty set are in the topology, but I have trouble proving the finite intersection and arbitrary union are in the topology. It seems easy with the properties of a metric space but without this idea of a metric I can't visualize how this would work. Any help would be greatly appreciated.

  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Apr 9 at 19:24
  • $\begingroup$ Thanks for the advice is there any way to edit my question? @shaun $\endgroup$ – spazmferret Apr 9 at 19:25
  • $\begingroup$ Use the edit button. $\endgroup$ – Shaun Apr 9 at 19:26
  • $\begingroup$ found it thanks $\endgroup$ – spazmferret Apr 9 at 19:26
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    $\begingroup$ What definition of order topology are you using? $\endgroup$ – Chessanator Apr 9 at 19:31

Check that the set $\mathcal{B}$ is a base for a topology on $(X,<)$, having at least two points, where $\mathcal{B}$ is given by $$\{(x,y)\mid x < y, x,y \in X\} \cup \{[\min(X),x)\mid x \in X\} \cup \{(x,\max(X)]\mid x \in X\}$$

where the latter two are only used if $\min(X)$ resp. $\max(X)$ exist.

So suppose both exist (that's the most "work", if neither exist we only have the open intervals to deal with, which is a bit easier), and let $p \in X$. If $p=\min(X)$ then we have some $p\neq q \in X$, and we know that $p \in [\min(X),q) \in \mathcal{B}$. If $p=\max(X)$ then we have some $p \neq q \in X$ and we have $p \in (q,\max(X)]\in \mathcal{B}$. If neither of these cases holds, $p$ is neither the minimal nor the maximum of $(X,<)$ and so we have $q_1,q_2 \in X$ with $q_1 < p$ and $p < q_2$. But then $p \in (q_1, q_2) \in \mathcal{B}$.

This shows that $\bigcup \mathcal{B}=X$, which is the first condition to be a base.

Now, let $B_1,B_2$ be two members of $\mathcal{B}$ and let $p \in B_1 \cap B_2$. If $p = \min(X)$ then $B_1$ must be of the form $[\min(X), q_1)$ (as no other types in $\mathcal{B}$ can contain $\min(X)$, as those imply an element strictly smaller than $\min(X)$ which cannot be), and likewise $B_2$ has to be of the form $[\min(X), q_2)$. But then $x \in B_3:=[\min(X), \min(q_1,q_2)) \subseteq B_1 \cap B_2$ and we are done checking the condition for intersections in this case. The $p=\max(X)$ is quite similar, check it.

So we can assume $p \notin \{\min(X),\max(X)\}$ and $B_1 = (q_1, q_2)$, say, and $B_2 = (r_1, r_2)$ for some $q_1 < q_2, r_1 < r_2$ in $X$ (even either is of one of the two special types, we could safely remove the min/max endpoint and have an open interval instead). Then take $B_3 = (\max(q_1,r_1), \min(q_2,r_2))$ and verify that $p \in B_3 \subseteq B_1 \cap B_2$, finishing the intersection condition check.

This makes the order topology's definition well-defined.

I usually prefer to take all sets $$\mathcal{S} = \{(x,\rightarrow):=\{y \in X: y > x\}\mid x \in X\}\cup \{(\leftarrow,x):=\{y \in X: y < x\}\mid x \in X\}$$

and define that as a subbase (so that the order topology is the smallest topology that contains $\mathcal{S}$) and the induced base then becomes exactly $\mathcal{B}$, as can easily be checked as well.

  • $\begingroup$ Thanks a bunch, this is super helpful. $\endgroup$ – spazmferret Apr 9 at 21:56

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