# Prove that a subset is open relative to another subset (Proof Verification)

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 $$Y$$.

• Are $X$ and $Y$ metric spaces? Your question should probably say this. – Ben Steffan May 5 '20 at 20:05
• Yes they are! I added that to the question. – user10360304 May 5 '20 at 21:09
• Also, what is your definition of relative topology? Usually, what you're trying to prove is simply taken as the definition. – Ben Steffan May 6 '20 at 12:02
• 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). – user10360304 May 6 '20 at 14:17

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 $$E = G \cap Y$$.
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.