I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt:

Suppose $Y \subset X$, where $X$ and $Y$ are metric spaces. Prove that a subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.

My Proof:

Suppose $E$ is open relative to $Y$. Then, for each $e \in E$ there is some $r$ s.t. $B_r(e) \cap Y \subset E$ where $B$ is an open ball in $X$. Now, let $G$ be the union of all $B_r(e)$ for every $e \in E$, and so $G \cap Y \subset E$. We can see that $B_r(e)$ is an open subset of $X$, and therefore $G$ is a collection of open subsets of $X$ and is thus an open subset of $X$. So $Y \cap G \subset E$ for an open subset $G$ of $X$. For any $e \in E$, $e \in B_r(e)$ for its associated open ball and since $E \subset Y \subset X$, $e \in Y$ and $e \in X$. So $e \in B_r(e) \cap X = G_e$ and so $e \in G$ and $E \subset G$. Thus, $E \subset Y \cap G$ and so $E = Y \cap G$.

Now suppose $E = Y \cap G$ for some open subset $G$ of $X$. For every $e \in E$ there is some $r$ such that $q \in G$ if $d(p,q) < r$. If $q$ is also an element of $Y$, then $q \in Y \cap G$. So, for every $e \in E$ there is some $r$ such that $q \in Y \cap G = E$ if $d(p,q) < r$ and $q \in Y$, so $E$ is open relative to $Y$.

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    $\begingroup$ Are $X$ and $Y$ metric spaces? Your question should probably say this. $\endgroup$ May 5 '20 at 20:05
  • $\begingroup$ Yes they are! I added that to the question. $\endgroup$ May 5 '20 at 21:09
  • $\begingroup$ Also, what is your definition of relative topology? Usually, what you're trying to prove is simply taken as the definition. $\endgroup$ May 6 '20 at 12:02
  • $\begingroup$ Hmm, I am not sure. Here, I am trying to prove Theorem 2.30 in Rudin's Mathematical Analysis (it also has a proof in the book, although that proof is slightly different than mine). $\endgroup$ May 6 '20 at 14:17

The proof is correct, but it seems a bit too wordy to me. I'd shorten the first part of your proof to

Suppose $E$ is open relative to $Y$. Then, for each $e \in E$, there is some $r_e \in \mathbb{R}_+$ s. t. $B_{r_e}(e) \cap Y \subseteq E$, where $B_{r_e}(e)$ is a subset of $X$. Then $G = \bigcup_{e \in E} B_{r_e}(e)$ is an open set in $X$ fulfilling $E = G \cap Y$.

Since this part makes use of open balls, I would prefer to use the same idea for the converse as well, e.g.

Suppose $E = Y \cap G$ with $G \subseteq X$ open. Choosing for each $g \in G$ radii $r_g \in \mathbb{R}_+$ as before such that $B_{r_g}(g) \subseteq G$, we can write $E = Y \cap \left(\bigcup_{g \in G} B_{r_g}(g)\right) = \bigcup_{g \in G} (B_{r_g}(g) \cap Y)$, so $E$ is open in $Y$ by definition of the relative topology.


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