# Prove Theorem 2.30 from baby Rudin.

I want to prove that

Suppose $Y\subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E=Y\cap G$ for some open subset $G$ of $X$.

The following statement is a bit confusing to me

Conversely, if $G$ is open in $X$ and $E=G\cap Y$, every $p\in E$ has a neighborhood $V_{p}\subset G$. Then, $V_{p}\cap Y\subset E$, so that $E$ is open relative to $Y$.

for the meaning of $V_{p}\cap Y\subset E$ is not straightforward. This lecture note also has the same proof, but with a different wording:

Thus the relative neighborhood $V_{p}\cap Y$ is contained in $G\cap Y=E$.

In conclusion, the intention was to say that

• every $V_{p}$ has $V_{p}\cap Y$ as its narrower version, and
• $V_{p}\cap Y$ is contained in both $E$ and $Y$.

However, it seems clear to me that $V_{p}\cap Y$ is not necessarily a neighborhood of $p$, for it might not contain all the points within the distance of $r$ (truncated by $Y$). Then, how does it qualify as a proof?

• The points that aren't in Y ... are not in Y. All the points within r that are in Y are in the set. So the set is open in Y. It doesn't need to be open in any space "larger" than Y. – fleablood Jun 19 '16 at 6:49
• What you want to prove is generally taken as the definition of the subspace topology on Y, rather than a consequence of other statements. – DanielWainfleet Jun 19 '16 at 7:19

$V_p\cap Y$ is, in fact, not necessarily a neighbourhood of $p$ in the ambient space $X$, but it is, by definition, a neighbourhood of $p$ in $Y$, since it is the intersection of an open set with $Y$ and it contains $p$. So it qualifies as proof since a neighbourhood of $p$ relative $Y$ has been found.
To illustrate this, look at $[0,1]\subset \mathbb{R}$. Then $(1-\varepsilon, 1] = (1-\varepsilon, 1+\varepsilon)\cap [0,1]$ is a neighbourhood of $1$ in $[0,1]$, even though it is certainly not in $\mathbb{R}$