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

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

Proof

This theorem comprises of two parts:

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

(These pages may or may not redirect here)

($\implies$): an open set is a neighbourhood to all of its points

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

• Clearly this can be distilled to just: $\forall \mathcal{O}\in\mathcal{J}\forall p\in \mathcal{O}\underbrace{\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]}_{\mathcal{O}\text{ is a neighbourhood of }p}$ as we only care about $\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]$ being true when we have $\mathcal{O}\in\mathcal{J}$ (by the nature of logical implication if $\mathcal{O}\notin\mathcal{J}$ 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 $\mathcal{O}\in\mathcal{J}$ be given
  • Let $p\in\mathcal{O}$ be given. We must show that there exists an open $U$ such that $p\in U$ and $U\subseteq\mathcal{O}$
    • Choose $U:=\mathcal{O}$
      • Clearly $p\in\mathcal{O}$ (by definition)
      • Clearly $\mathcal{O}\subseteq\mathcal{O}$ (a set is a subset of itself)
    • This completes the logic of the proof
  • Since $p\in\mathcal{O}$ was arbitrary this is true for all $p\in\mathcal{O}$
• Since $\mathcal{O}\in\mathcal{J}$ was arbitrary this is true for all $\mathcal{O}\in\mathcal{J}$

($\impliedby$): if a set is a neighbourhood to all of its points then it is open

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

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Outline of proof

• Let $\mathcal{O}\in\mathcal{P}(X)$ be given and suppose $\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]$ (that it is a neighbourhood to all the points inside of it)
  • We wish to show that $\mathcal{O}$ is an open set. As we lack really any properties to try and show $\mathcal{O}$ has that would imply it was open we must proceed by trying to construct $\mathcal{O}$ as an open set.
    • As $\mathcal{J}$ is a topology we know it is closed under arbitrary union (the union of absolutely any collection of things in $\mathcal{J}$ is in $\mathcal{J}$) and finite intersection.
    • We see that by hypothesis, for each $p\in\mathcal{O}$ there exists an open set containing $p$ fully contained in $\mathcal{O}$.
    • If we take the union (over all $p\in\mathcal{O}$) of theses $U$s (that are open) that each contain their $p$, we get an open set that contains every point in $\mathcal{O}$ - this is how we'll prove the statement.

• Let $\mathcal{O}\in\mathcal{P}(X)$ be given and suppose $\forall p\in\mathcal{O}\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]$ (that it is a neighbourhood to all the points inside of it)
  • Let $p\in\mathcal{O}$ be given
    • By hypothesis, $\exists U\in\mathcal{J}[p\in U\subseteq\mathcal{O}]$ - there exists an open set containing $p$ fully contained in $\mathcal{O}$
    • Call this set $U_p$
  • Now we have a collection: $\{U_p\}_{p\in\mathcal{O} }\subseteq\mathcal{J}$ 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 $\mathcal{O}\subseteq \bigcup_{p\in\mathcal{O} }U_p$, as $p\in U_p$ and the union is over all points in $\mathcal{O}$ and $p\in U_p$. Notice additionally that $\bigcup_{p\in\mathcal{O} }U_p\subseteq\mathcal{O}$ as each $U_p$ is contained in $\mathcal{O}$.
      • If these are true we would see $\bigcup_{p\in\mathcal{O} }U_p=\mathcal{O}$. As $\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J}$ we must have $\mathcal{O}=\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J}$, ie $\mathcal{O}\in\mathcal{J}$
  • We now must show $\bigcup_{p\in\mathcal{O} }U_p=\mathcal{O}$. We shall do this in two parts:
    1. $\bigcup_{p\in\mathcal{O} }U_p\subseteq\mathcal{O}$, by the implies-subset relation this is simply: $\forall x\in\bigcup_{p\in\mathcal{O} }U_p[x\in\mathcal{O}]$.
       • Let $x\in\bigcup_{p\in\mathcal{O} }U_p$ be given. We wish to show $x\in\mathcal{O}$
         • By definition of union, to have $x\in\bigcup_{p\in\mathcal{O} }U_p$ means $\exists p\in\mathcal{O}[x\in U_p]$
         • As $U_p\subseteq\mathcal{O}$ (by definition of each $U_p$) and the implies-subset relation we see that:
           • $x\in U_p\implies x\in\mathcal{O}$
       • Thus we have shown if $x\in \bigcup_{p\in\mathcal{O} }U_p$ then $x\in\mathcal{O}$. As required
    2. $\mathcal{O}\subseteq\bigcup_{p\in\mathcal{O} }U_p$, by the implies-subset relation this is simply: $\forall x\in\mathcal{O}[x\in\bigcup_{p\in\mathcal{O} }U_p]$
       • Let $x\in\mathcal{O}$ be given. We wish to show that this means $x\in\bigcup_{p\in\mathcal{O} }U_p$
         • Recall by the definition of the $U_p$s that $\forall p\in\mathcal{O}[p\in U_p]$.
         • In order to show $x\in\bigcup_{p\in\mathcal{O} }U_p$ we'd have to show $\exists p\in\mathcal{O}[x\in U_p]$ (by definition of union)
         • Choose $p:=x$, then $x\in U_x$ so $x\in\bigcup_{p\in\mathcal{O} }U_p$
       • Thus we have shown that if $x\in\mathcal{O}$ then $x\in\bigcup_{p\in\mathcal{O} }U_p$ as required.
  • Combining these we see and as $\bigcup_{p\in\mathcal{O} }U_p\in\mathcal{J}$ we have $\mathcal{O}\in\mathcal{J}$ - the very definition of an open set