# 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?

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.

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.
• I might be mixing up definitions, but isn’t this “result” the definition of a topology generated by the basis $\mathcal{B}$? i.e. how the open sets are defined in this topology? Jan 5 at 10:13
• In section 13 right below the definition of basis Munkres give the definition of the topology generated by a basis $\mathcal{B}$: for each $x \in U$, there is a basis element $B \in \mathcal{B}$ such that $x \in B$ and $B \subseteq U$. I think what you stated is Lemma 13.1, which is often used as an alternative definition of the basis generated topology. Jan 5 at 10:25
• @Mathematics_Beginner in that definition it follows, when you take $\mathcal B = \mathcal T$. You still need this argument somewhere to show that the topology generated by this trivial base is exactly the topology we start with, see? There are several equivalent definitions and Munkres is a confusing writer sometimes. Jan 5 at 10:30