$(-\sqrt 2, \sqrt 2) \cap \mathbb Q$ in order topology. What will be it's bases?

Question

$(-\sqrt 2, \sqrt 2) \cap \mathbb Q$ in order topology. What will be its bases?

My try: [ $(a,b)$ where $a,b \in (-\sqrt2, \sqrt2) \cap \mathbb Q$ and $(-\infty , b)$ , $(a, \infty)$ {because supremum and infimum of $(-\sqrt 2, \sqrt 2) \cap \mathbb Q$ do not exist}]

Can anyone please correct me if I go wrong anywhere?

Answer 1

Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.

I would put it this way:

One basis for the order topology on $X:=\left(-\sqrt{2},\sqrt{2}\right)\cap\Bbb Q$ is $$\bigl\{(a,b)\cap X\mid a,b\in X\bigr\}\cup\bigl\{(-\infty,b)\cap X\mid b\in X\bigr\}\cup\bigl\{(a,\infty)\cap X\mid a\in X\bigr\}.$$

At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.

Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.

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Added: As an exercise, you should try to prove that $\bigl\{(a,b)\cap X\mid a,b\in X\bigr\}$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).

Answer 2

Let r = $\sqrt 2.$ A simple base is
{ (a,b) $\cap$ Q : a,b in (-r,r) }.

Exercise. Show a subbase is
{ (-r,a) $\cap$ Q, (a,r) $\cap$ Q : a in (-r,r) }.

In both definitions a,b can be futher restricted to rationals.