Is the set of rationals between $\sqrt{2}$ and $\sqrt{3}$ open or closed in $\mathbb Q$?

Question

Consider the set of all rationals, $\mathbb Q$ as a subset of the set of all reals $\mathbb R$. Assign $\mathbb Q$ the subspace topology induced by the standard topology on $\mathbb R$.

Consider now the set $(\sqrt{2}, \sqrt{3})$ subset of $\mathbb Q$. Is the set open in $\mathbb Q$? Is the set closed in $\mathbb Q$ ? Prove your answer?

Solution

This is what I have thought till now:
To prove $X = (\sqrt{ 2}, \sqrt{3})$ is open. Let $p$ belong to $X$.
To prove: $p$ is an interior point of $X$ i.e. there exists a epsilon neighborhood of $p$ which lies in $X$.
I am not sure how to proceed from here.

Answer 1

Note that in ${\bf Q}$, $V\cap {\bf Q}$ is open when $V$ is open in ${\bf R}$.

So let $V=(\sqrt{2},\sqrt{3})$. So $V\cap {\bf Q}$ is open in ${\bf Q}$

Answer 2

This set is open in relation to $\mathbb{Q}$. Remember that a set $A\subset\mathbb{R}$ is open in $\mathbb{Q}$ iff there is an open set $B\subset \mathbb{R}$ such that $A=\mathbb{Q}\cap B$. Given that every open interval is an open set of $\mathbb{R}$, $(\sqrt2,\sqrt3)$ is open, which means that $(\sqrt2,\sqrt3)\cap\mathbb{Q}$ is open in $\mathbb{Q}$.