# A subset of a topological space is open if and only if it is a neighbourhood to all of its points

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Links to other pages, gosh for something so simple it's long! Also verification of proof

## Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]\mathcal{O}\in\mathcal{P}(X)[/ilmath] be any subset of [ilmath]X[/ilmath]. Then [ilmath]\mathcal{O} [/ilmath] is open if and only if [ilmath]\mathcal{O} [/ilmath] is a neighbourhood[Note 1] to all of its points. Symbolically:

• [ilmath]\forall\mathcal{O}\in\mathcal{P}(X)\left[(\mathcal{O}\in\mathcal{J})\iff\left(\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]\right)\right][/ilmath][Note 2]

## Proof

This theorem comprises of two parts:

1. An open set is a neighbourhood to all of its points (the [ilmath]\implies[/ilmath] direction)
2. If a set is a neighbourhood to all of its points then it is open (the [ilmath]\impliedby[/ilmath] direction)

(These pages may or may not redirect here)

### ([ilmath]\implies[/ilmath]): an open set is a neighbourhood to all of its points

We wish to show, for a topological space, [ilmath](X,\mathcal{ J })[/ilmath] that: [ilmath]\forall\mathcal{O}\in\mathcal{P}(X)\left[(\mathcal{O}\in\mathcal{J})\implies(\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}])\right][/ilmath]

• Clearly this can be distilled to just: [ilmath]\forall \mathcal{O}\in\mathcal{J}\forall p\in \mathcal{O}\underbrace{\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]}_{\mathcal{O}\text{ is a neighbourhood of }p}[/ilmath] as we only care about [ilmath]\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}][/ilmath] being true when we have [ilmath]\mathcal{O}\in\mathcal{J} [/ilmath] (by the nature of logical implication if [ilmath]\mathcal{O}\notin\mathcal{J} [/ilmath] we do not care if the RHS is true or false)
This proof is slightly more involved if you use the metric space definitions (rather than considering the metric space as a topological space)
This proof is extremely easy, it only spans several lines here because it is done formally. The reader should have no trouble doing this himself.
• Let [ilmath]\mathcal{O}\in\mathcal{J} [/ilmath] be given
• Let [ilmath]p\in\mathcal{O} [/ilmath] be given. We must show that there exists an open [ilmath]U[/ilmath] such that [ilmath]p\in U[/ilmath] and [ilmath]U\subseteq\mathcal{O} [/ilmath]
• Choose [ilmath]U:=\mathcal{O}[/ilmath]
• Clearly [ilmath]p\in\mathcal{O} [/ilmath] (by definition)
• Clearly [ilmath]\mathcal{O}\subseteq\mathcal{O} [/ilmath] (a set is a subset of itself)
• This completes the logic of the proof
• Since [ilmath]p\in\mathcal{O} [/ilmath] was arbitrary this is true for all [ilmath]p\in\mathcal{O} [/ilmath]
• Since [ilmath]\mathcal{O}\in\mathcal{J} [/ilmath] was arbitrary this is true for all [ilmath]\mathcal{O}\in\mathcal{J} [/ilmath]

### ([ilmath]\impliedby[/ilmath]): if a set is a neighbourhood to all of its points then it is open

We wish to show for a topological space, [ilmath](X,\mathcal{ J })[/ilmath], that: [ilmath]\forall\mathcal{O}\in\mathcal{P}(X)\left[(\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}])\implies(\mathcal{O}\in\mathcal{J})\right][/ilmath]

Outline of proof
• Let [ilmath]\mathcal{O}\in\mathcal{P}(X)[/ilmath] be given and suppose [ilmath]\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}][/ilmath] (that it is a neighbourhood to all the points inside of it)
• We wish to show that [ilmath]\mathcal{O} [/ilmath] is an open set. As we lack really any properties to try and show [ilmath]\mathcal{O} [/ilmath] has that would imply it was open we must proceed by trying to construct [ilmath]\mathcal{O} [/ilmath] as an open set.
• As [ilmath]\mathcal{J} [/ilmath] is a topology we know it is closed under arbitrary union (the union of absolutely any collection of things in [ilmath]\mathcal{J} [/ilmath] is in [ilmath]\mathcal{J} [/ilmath]) and finite intersection.
• We see that by hypothesis, for each [ilmath]p\in\mathcal{O} [/ilmath] there exists an open set containing [ilmath]p[/ilmath] fully contained in [ilmath]\mathcal{O} [/ilmath].
• If we take the union (over all [ilmath]p\in\mathcal{O} [/ilmath]) of theses [ilmath]U[/ilmath]s (that are open) that each contain their [ilmath]p[/ilmath], we get an open set that contains every point in [ilmath]\mathcal{O} [/ilmath] - this is how we'll prove the statement.
• Let [ilmath]\mathcal{O}\in\mathcal{P}(X)[/ilmath] be given and suppose [ilmath]\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}][/ilmath] (that it is a neighbourhood to all the points inside of it)
• Let [ilmath]p\in\mathcal{O} [/ilmath] be given
• By hypothesis, [ilmath]\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}][/ilmath] - there exists an open set containing [ilmath]p[/ilmath] fully contained in [ilmath]\mathcal{O} [/ilmath]
• Call this set [ilmath]U_p[/ilmath]
• Now we have a collection: [ilmath]\{U_p\}_{p\in\mathcal{O} }\subseteq\mathcal{J}[/ilmath] of open sets
• As the union of any collection of open sets is an open set (by the definition of topology) we see:
• $\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J}$ notice that [ilmath]\mathcal{O}\subseteq \bigcup_{p\in\mathcal{O} }U_p[/ilmath], as [ilmath]p\in U_p[/ilmath] and the union is over all points in [ilmath]\mathcal{O} [/ilmath] and [ilmath]p\in U_p[/ilmath]. Notice additionally that [ilmath]\bigcup_{p\in\mathcal{O} }U_p\subseteq\mathcal{O} [/ilmath] as each [ilmath]U_p[/ilmath] is contained in [ilmath]\mathcal{O} [/ilmath].
• If these are true we would see [ilmath]\bigcup_{p\in\mathcal{O} }U_p=\mathcal{O}[/ilmath]. As [ilmath]\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J} [/ilmath] we must have [ilmath]\mathcal{O}=\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J}[/ilmath], ie [ilmath]\mathcal{O}\in\mathcal{J} [/ilmath]
• We now must show [ilmath]\bigcup_{p\in\mathcal{O} }U_p=\mathcal{O}[/ilmath]. We shall do this in two parts:
1. [ilmath]\bigcup_{p\in\mathcal{O} }U_p\subseteq\mathcal{O} [/ilmath], by the implies-subset relation this is simply: [ilmath]\forall x\in\bigcup_{p\in\mathcal{O} }U_p[x\in\mathcal{O}][/ilmath].
• Let [ilmath]x\in\bigcup_{p\in\mathcal{O} }U_p[/ilmath] be given. We wish to show [ilmath]x\in\mathcal{O} [/ilmath]
• By definition of union, to have [ilmath]x\in\bigcup_{p\in\mathcal{O} }U_p[/ilmath] means [ilmath]\exists p\in\mathcal{O}[x\in U_p][/ilmath]
• As [ilmath]U_p\subseteq\mathcal{O} [/ilmath] (by definition of each [ilmath]U_p[/ilmath]) and the implies-subset relation we see that:
• [ilmath]x\in U_p\implies x\in\mathcal{O} [/ilmath]
• Thus we have shown if [ilmath]x\in \bigcup_{p\in\mathcal{O} }U_p[/ilmath] then [ilmath]x\in\mathcal{O} [/ilmath]. As required
2. [ilmath]\mathcal{O}\subseteq\bigcup_{p\in\mathcal{O} }U_p [/ilmath], by the implies-subset relation this is simply: [ilmath]\forall x\in\mathcal{O}[x\in\bigcup_{p\in\mathcal{O} }U_p][/ilmath]
• Let [ilmath]x\in\mathcal{O} [/ilmath] be given. We wish to show that this means [ilmath]x\in\bigcup_{p\in\mathcal{O} }U_p[/ilmath]
• Recall by the definition of the [ilmath]U_p[/ilmath]s that [ilmath]\forall p\in\mathcal{O}[p\in U_p][/ilmath].
• In order to show [ilmath]x\in\bigcup_{p\in\mathcal{O} }U_p[/ilmath] we'd have to show [ilmath]\exists p\in\mathcal{O}[x\in U_p][/ilmath] (by definition of union)
• Choose [ilmath]p:=x[/ilmath], then [ilmath]x\in U_x[/ilmath] so [ilmath]x\in\bigcup_{p\in\mathcal{O} }U_p[/ilmath]
• Thus we have shown that if [ilmath]x\in\mathcal{O} [/ilmath] then [ilmath]x\in\bigcup_{p\in\mathcal{O} }U_p[/ilmath] as required.
• Combining these we see and as [ilmath]\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J} [/ilmath] we have [ilmath]\mathcal{O}\in\mathcal{J} [/ilmath] - the very definition of an open set