View source for A set is open if and only if every point in the set has an open neighbourhood contained within the set

{{Stub page|grade=A*|msg=Flesh out and clean up, then '''demote to grade B'''<br/>See page 25 in Lee's top manifolds, there's a bunch of equivalent conditions. However I do think there's a more general [[neighbourhood]] one, if I could just be bothered to prove, "''a set is open if and only if it is neighbourhood to all of its points''" or something}}
__TOC__
==Statement==
Let {{Top.|X|J}} be a [[topological space]], then{{rITTMJML}}:
* A set, {{M|A\in\mathcal{P}(X)}} is [[open set|open]] {{iff}} every point of {{M|A}} has an [[open neighbourhood]] contained in {{M|A}}
==Proof==
{{Requires proof|grade=B|easy=true}}
==References==
<references/>
{{Theorem Of|Topology}}