# Rudimentary topology question: interior of a subset of a space is the union of all open sets contained in it.

I'm reviewing topology and just wanted to confirm that I am doing this correctly. Professor Janich has urged the readers of his classic text on topology to prove the following:

Suppose $$(X, \mathcal{O})$$ is a topological space and $$B \subset X$$. Define the interior of the set $$B$$ as $$\hat B$$.

Claim: The interior of $$B,$$ namely $$\hat B$$, is the union of all open sets contained in $$B$$.

Proof of $$\Longrightarrow$$: Let $$x\in \hat B$$. Then by definition, $$B$$ is a neighborhood of $$x$$. Thus $$\exists A \subset B$$ such that $$A$$ is open and $$x \in A$$. Since $$x \in A$$ and $$A \subset B$$, it must be the case that $$x \in \bigcup\{\text{all open sets contained in B}\}$$.

Proof of $$\Longleftarrow$$: Suppose $$x \in \bigcup\{\text{all open sets contained in B}\}$$. Just pick any one of them and call it, say, $$A$$. (Question: did I just use the axiom of choice?) $$x \in A \subset B$$ and $$A$$ is open, so $$B$$ is a neighborhood of $$x$$ and thus, $$x$$ is an interior point of $$B$$.

Is all of that correct? Please feel free to offer any corrections, thank you.

• I prefer single arrows for implications, but you can get the double left arrow with \Leftarrow ($\Leftarrow$). If you want a long one, make it \Longleftarrow. Commented Jan 7, 2021 at 4:53
• @BrianM.Scott. I find it irritating that in Latex we have \to and \gcd but not \from nor \lcm. Commented Jan 7, 2021 at 10:06
• Don't use \implies, that adds unwanted spaces. There is \impliedby for the reverse arrow, but it has the same problem. \Rightarrow and \Leftarrow or \Longrightarrow and \Longleftarrow are better here. Personally I'd use the short versions. Commented Jan 7, 2021 at 10:23
• You don't say what definition of interior you are using. Commented Jan 7, 2021 at 10:24

Recall that if $$x$$ and $$y$$ are sets, $$x \in \textstyle\bigcup y \iff (\exists z \in y) \ x \in z.$$
So, if $$x$$ is in the union of all the open sets contained in $$B$$, by definition of union there exists an open set $$A$$ contained in $$B$$ such that $$x \in A$$. There is no need of choice here. By the way, write \Longleftarrow to obtain $$\Longleftarrow$$.