4

Let $X$ be a set with topology with $\tau$. Suppose that $A \subset X$ such that for all $a \in A$ there is an open subset $U_{a}$ containing $a$ and contained in $A$. Prove that $A$ is open.

I think that $A$ can be written as the union of all $a$'s in $A$, thus is contained in the union of $U_{a}$'s. But that alone does not prove that $A$ is open. What in addition do I need to prove such a statement?

Henno Brandsma
  • 242,131
youngeAn
  • 309
  • Very good start! You should also prove that the union of the $U_a$'s is contained in $A$ :) – diracdeltafunk Nov 08 '21 at 03:51
  • Let $x \in \cup U_{a}$ be arbitrary, thus $x \in U_{a}$ for some a; Thus $x\in U_{a} \subset A$. This way we proved that all elements of $\cup U_{a}$ are in A, proving the other direction of containment? – youngeAn Nov 08 '21 at 03:56
  • If $A$ is open the property trivially holds for $A$, so it's necessary and sufficient. – Henno Brandsma Nov 08 '21 at 11:55

1 Answers1

2

Since $U_a \subset A$ for every $a\in A$, we have $\bigcup_{a\in A} U_a \subset A$. You have already shown that $A \subset \bigcup_{a\in A} U_a$. Thus, $$A = \bigcup_{a\in A} U_a$$

stoic-santiago
  • 13,705