How to show that a point is not an interior point? I understand that in order to show that a point, $x$, is an interior point of some set $A \subset B$, where $(B,d)$ is a metric space you just need to show that you can have an open ball around $x$ which is contained in $A$.
I was asked a question today, which just showed I don't really understand what an interior point is.

Let $X=(\mathbb{Q}\cap[0,3])$ be a metric space. Define $Y=\{y \in\mathbb{Q}:2\leq y \leq 3\}$
Is $2$ an interior point of $Y$?

I believe that it's not. The only way I thought of perphaps proving this is by taking an arbitrary open ball which is contained in $Y$ and show that 2 cannot be inside it. This however seems like too much work. Is there perhaps a shorter way of thinking and proving this?
Note that an open $\varepsilon$-radius ball $B(2,\varepsilon)$ around $2$ is of the form $(2-\varepsilon,2+\varepsilon)\cap\mathbb{Q}\cap [0,3]$. For any $\varepsilon&gt;0$ take a rational point $q\in\mathbb{Q}\cap [0,3]$ with $2-\varepsilon&lt;q&lt;2$. Now since $q\in B(2,\varepsilon)$ and $q\notin Y$, then $B(2,\varepsilon)\not\subseteq Y$. Hence $2$ is not an interior point of $Y$.
Here $B=X$ and $A=Y$ in your definition of open set, and the topology is the induced topology on $Y$. Hence the $(\ 1,\ 3\ ) \cap Y$ is an open subset of $Y$ with the induced topology and it contains $2$, so $2$ is an interior point of $Y$.