# Help proving that the order topology is a topology

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.

• 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. – Shaun Apr 9 at 19:24
• Thanks for the advice is there any way to edit my question? @shaun – spazmferret Apr 9 at 19:25
• Use the edit button. – Shaun Apr 9 at 19:26
• found it thanks – spazmferret Apr 9 at 19:26
• What definition of order topology are you using? – 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.

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