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

From Maths

**Stub grade: A***

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Flesh out and clean up, then

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, "

**demote to grade B**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## Contents

## Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, then^{[1]}:

- A set, [ilmath]A\in\mathcal{P}(X)[/ilmath] is open
*if and only if*every point of [ilmath]A[/ilmath] has an open neighbourhood contained in [ilmath]A[/ilmath]

## Proof

Grade: B

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).

**This proof has been marked as an page requiring an easy proof**## References