Lemma 3.3.3 ([Proposition 1.2.1, Bou95]).label Let $(X,\Tc)$ be a topological space, then $U\suf X$ is open if and only if $U\in\cali{N}_{\Tc}(x)$ for all $x\in U$.
Proof. Suppose $U\in\Tc$ then $U\in\cali{N}_{\Tc}(A)$ for any $A\suf U$. Conversely for each $x\in U$, there exissts $V_{x}\in\Tc$ such that $x\in V_{x}\suf U$. Thus $U=\cups{}{x\in U }V_{x}\in\Tc$.$\square$
Post a Comment