# Open sets relative to a metric space

Let $X$ be a metric space. Let $E\subset Y\subset X$. By an example, it is possible to find a set $E$ which is open relative to $Y$, but not open relative to $X$. The classic one is the segment $(a,b)$, which, considered as a subset of $\Re^2$ is not open, but as a subset of $\Re\times{0}$, it is.

I am wondering the converse: is it possible to find a set $E$, open relative to $X$ (the $bigger$ space) but not open relative to $Y$? My intuition is that it is not possible. That is: if $E$ is open relative to $X$, then $E$ is open relative to $Y\subset X$. My proof by contradiction is this:

Suppose $E$ is open relative to $X$ and $E$ is not open relative to $Y\subset X$.

Since $E$ is open relative to $X$, then $\forall p\in E$, $p$ is an interior point. On the other hand, if $E$ is not open relative to $Y\subset X$, there exists a point $q\in E$ which is not interior and that's how I obtain a contradiction.

Could someone tell me if that is correct? Otherwise, any hints would be greatly appreciated.

• aren't the open sets relative to $Y$ not exactly the intersection of open sets in $X$ with $Y$? Commented Aug 5, 2015 at 22:35
• Yes, that's true. I was looking for a set not open relative to $Y$, but open relative to $X$. That cannot happen. Commented Aug 5, 2015 at 22:41

We're assuming $E \subset Y \subset X$? If $E$ is open relative to $X$, for any point $p$ in $E$, there's an open ball of center $p$ and radius $\varepsilon$ entirely contained within $E$. But $E \subset Y$, so $E$ is also open in $Y$.
• Yes, we are assuming $E\subset Y\subset X$. Perfect, a simple, straightforward proof. Thanks! Commented Aug 5, 2015 at 22:34