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?
 A: 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 $\varepsilon>0$ take a rational point $q\in\mathbb{Q}\cap [0,3]$ with $2-\varepsilon<q<2$. 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 $Y$.
A: If you are asking: interior point of $Y$ with respect to $X$'s topology, the answer is no.
A proof by contradiction is a good idea.
So assume there is $r>0$ such that
$$
\{y\in X \;;\; |y-2|<r\}=\mathbb{Q}\cap[0,3]\cap(2-r,2+r)\subseteq Y\subseteq [2,3].
$$
By density of the rationals, we can find $y$ rational in $(2-r,2)\cap [0,3]$.
Such a number will be in the lhs set, but not in the rhs set.
Contradiction.
A: Technically speaking, the question is unanswerable as you've quoted it, since it doesn't specify the metric on $X$.  However, it seems reasonable to assume that $X$ is meant to be equipped with the usual Euclidean metric $d(a,b) = |a-b|$, in which case the answer is "No."
To show this, observe that every open ball in $X$ centered on $2$ is of the form $(2-r,\, 2+r) \cap X$ for some positive real number $r$, and thus contains the non-empty set $(2-r,\, 2) \cap X$ which is disjoint with $Y$.  Thus, $Y$ contains no open ball centered on $2$ in $X$, and so $2$ is not an interior point of $Y$ in $X$.
Note that, per the definition you've quoted at the beginning of your post, we only need to examine open balls centered on $2$.  However, if you prefer, it's easy to extend the argument above to any open balls in $X$ containing $2$ by noting that they're all of the form $(a,b) \cap X$ for some $a < 2 < b$, and thus contain the non-empty set $(a,2) \cap X$ which is disjoint with $Y$.  Or we can just note that any open ball (or, indeed, any open set at all) containing a point $x$ in a metric space must include, as a subset, an open ball centered on $x$, and then apply the earlier argument above.
A: 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 $Y$.
A: I am giving this as another answer because I stick by my original answer even though the other answers  appear to disagree with me. 
This message was tagged general topology and metric spaces. Let me point out that the definition people seem to be using is NOT the standard one for either of those concepts.  Some examples: We have $\mathbb R$ is a metric space containing  $X$, then $X$ is a metric space with metric induced by the metric of $\mathbb R$ with no interior because for any interval about a point   $x\in X$ there is an irrational number in any ball about $x$. We now have a topological space $X$ whihc is not open in its own topology because we know some bigger space that happens to contain it. Moreover the topological definition of a function $f:U\rightarrow V$ where $U$ and $V$ are topological spaces has definition of continuity "$f$ is continuous iff $f^{-1}[S]$ is open for every open subset $S\subset V$". So the function $f:X\rightarrow {\mathbb R}$, $f(x)= x$, is not a continuous function of topological spaces? 
