# Open set

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## Definition

### Topological space

In a topological space [ilmath](X,\mathcal{J})[/ilmath] we have:

• [ilmath]\forall S\in\mathcal{J} [/ilmath] that [ilmath]S[/ilmath] is an open set. [ilmath]\mathcal{J} [/ilmath] is by definition the set of open sets of [ilmath]X[/ilmath]

### Metric space

In a metric space [ilmath](X,d)[/ilmath] there are 2 definitions of open set, however it will be shown that they are equivalent. Here [ilmath]U[/ilmath] is some arbitrary subset of [ilmath]X[/ilmath].

I claim that the following definitions are equivalent:

#### Definition 1

• A set [ilmath]U\subseteq X[/ilmath] is open in [ilmath](X,d)[/ilmath] (or just [ilmath]X[/ilmath] if the metric is implicit) if [ilmath]U[/ilmath] is a neighbourhood to all of its points[1], that is to say:
• [ilmath]U\subseteq X[/ilmath] is open if $\forall x\in U\exists\delta_x>0[B_{\delta_x}(X)\subseteq U]$ - (recall that [ilmath]B_r(x)[/ilmath] denotes the open ball of radius [ilmath]r[/ilmath] centred at [ilmath]x[/ilmath]) or
• For all [ilmath]x[/ilmath] in [ilmath]U[/ilmath] there is an open ball centred at [ilmath]x[/ilmath] entirely contained within [ilmath]U[/ilmath]

#### Definition 2

• A set [ilmath]U\subseteq X[/ilmath] is open in [ilmath](X,d)[/ilmath] (or just [ilmath]X[/ilmath] if the metric is implicit) if[2]:

It is easy to see that these definitions are very similar to each other (these are indeed equivalent is claim 1)

## Immediate results

It is easily seen that:

• [ilmath]\emptyset[/ilmath] is open (claim 2)
• [ilmath]X[/ilmath] itself is open (claim 3)
• [ilmath]\text{Int}(U)[/ilmath] is open
(see interior for a proof of this. The claim is the same as [ilmath]\text{Int}(\text{Int}(U))=\text{Int}(U)[/ilmath] as by the first claim we can use either definition of open, so we use [ilmath]U[/ilmath] is open if [ilmath]U=\text{Int}(U)[/ilmath])

## Proof of claims

Claim 1: A set, [ilmath]U\subseteq X[/ilmath] is open according to definition [ilmath]1[/ilmath] [ilmath]\iff[/ilmath] [ilmath]U[/ilmath] is open according to definition [ilmath]2[/ilmath]

TODO: Trivial proof, be bothered to do it

Claim 2: [ilmath]\emptyset[/ilmath] is open

TODO: Trivial proof, be bothered to do it

Claim 3: [ilmath]X[/ilmath] itself is open

TODO: Trivial proof, be bothered to do it

# Old page

Here $(X,d)$ denotes a metric space, and $B_r(x)$ the open ball centred at $x$ of radius $r$

## Metric Space definition

"A set $U$ is open if it is a neighborhood to all of its points"[1] and neighborhood is as you'd expect, "a small area around".

### Neighbourhood

A set $N$ is a neighborhood to $a\in X$ if $\exists\delta>0:B_\delta(a)\subset N$

That is if we can puff up any open ball about [ilmath]x[/ilmath] that is entirely contained in [ilmath]N[/ilmath]

## Topology definition

In a topological space the elements of the topology are defined to be open sets

### Neighbourhood

A subset [ilmath]N[/ilmath] of a Topological space [ilmath](X,\mathcal{J})[/ilmath] is a neighbourhood of [ilmath]p[/ilmath][2] if:

• $\exists U\in\mathcal{J}:p\in U\wedge U\subset N$

## References

1. Bert Mendelson, Introduction to Topology - definition 6.1, page 52
2. Introduction to topology - Third Edition - Mendelson