A Set is Open if Every Element in the Set is Contained in an Open Subset of the Set

Let $X$ be a topological space and $A \subseteq X$ a set. If for every $x \in A$, we can find an open set $U$ containing $x$ such that $U \subseteq A$, then $A$ is open in $X$. I think this is a trivial statement, but the proof that I have in mind does not seem to appear anywhere online (it's really just two lines):

Since $U \subseteq A \subseteq X$ is open, we have a basis element $B \subseteq U \subseteq A$ such that $x \in B$. Therefore, $A$ is open by definition of the topology generated by the basis (using the fact that every topology is generated by itself).

The "standard proof" seem to have just used the fact that the union of the open subsets is $A$ and the definition of a topology (involving less concepts, which is nice). However, I am wondering if this proof is correct or am I missing something?

Solution 1:

There is no result in e.g. Munkres involving a base that I can see that immediately implies $A$ is open directly from "$\forall x \in A: \exists B \in \mathcal B: x \in B \subseteq A$", where $\mathcal B$ is a base for $X$. If there is, please state it.

Otherwise just use the standard prooflet: denote for $x$ each open set (or base set, if you prefer) from the statement, by $O_x$. Then $A = \{O_x\mid x \in A\}$, as all $O_x$ are a subset of $A$ (so their union too) and every $x \in A$ is in its own $O_x$ (so $A$ is covered). So $A$ is a union of open sets hence open.

There is absolutely no shortcut or improvement by using bases in this statement. The proof stays the same regardless.

Added after comments discussion:

There is a way using the text (Munkres, 2nd ed.) as reference: let $\mathcal{B}:=\mathcal{T}$. This is a base for $(X,\mathcal{T})$ by 13.2 trivially. Conditions 1 and 2 from the start of the paragraph 13 are easily checked as $\mathcal{B}$ is closed under finite intersections and contains $X$. So it generates some topology $\mathcal{T}'$ by the rule following the definition: the set of all $A$ so that "$\forall x \in A: \exists B \in \mathcal B: x \in B \subseteq A$" indeed. But as $\mathcal{B} \subseteq \mathcal{T}$ and by 13.1 $\mathcal{T'}$ also consists of all unions from subfamilies of $\mathcal{T}$ and as $\mathcal{B}=\mathcal{T}$ is already closed under unions we get $\mathcal{T}' = \mathcal{T}$ and so $A$ is in $\mathcal{T}$ as well.

In fact a later exercise, 5, tells us that the topology generated by a (sub)base is the minimal topology that contains the given (sub)base and with that knowledge $\mathcal{T'}=\mathcal{T}$ is also immediate.

All this because Munkres wants to keep distinguishing between "a base for some topology" and a "base for a given topology". Things need not be so hard IMO. We get a lot of questions on this site about this paragraph as students are confused by it.