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I am looking to prove the statement "$U \subset X$ is open iff for every $y \in U$, $U \in N_y$.

We prove the former implies the latter:

$\Rightarrow$: If $U$ is open, for every $y \in U$, we have $y \in U \subset U \Rightarrow U \in N_y$

We look to prove the other way.

$\Leftarrow$: If for every $y \in U, U \in N_y$, then for every $y \in U$ there exists an open set with $y \in U_y \subset U$. From this point we are told that U = $\cup_{y \in U}$ $U_y$. Now, I think I can see why we'd have $\cup_{y \in U}$ $U_y$ $\subset$ $U$, as take $y \in$ $\cup_{y \in U}$ $U_y$, then $y$ is in these sets, so it being in the union of all of them makes sense and each one of these is a subset of U so the union is a subset of U.I think this is correct (please tell me if I'm wrong). But I really cannot see at all why $U \subset$ $\cup_{y \in U}$ $U_y$.

Any help?

KeynesianSpaceman
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1 Answers1

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(I will assume that $N_y$ means the set of all neighborhoods of $y$.)

You've almost done: if $a\in U$ then $a\in U_a\subset\bigcup\limits_{y\in U} U_y$, thus every element $a$ of $U$ belongs to the latter union. Isn't it the inclusion you're looking for?

jp boucheron
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