Difference between revisions of "Topology generated by a basis"

Latest revision as of 21:14, 15 January 2017

Statement

Let $X$ be a set and let $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ be any collection of subsets of $X$, then:

• $(X,\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})$ is a topological space with $\mathcal{B}$ being a basis for the topology $\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\}$

if and only if

• we have both of the following conditions:
  1. $\bigcup\mathcal{B}=X$ (or equivalently: $\forall x\in X\exists B\in\mathcal{B}[x\in B]$^{[Note 1]}) and
  2. $\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big]$^{[Note 2]}
     • Caveat:$\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V]$ is commonly said or written; however it is wrong, this is slightly beyond just abuse of notation.^{[Note 3]}

Note that we could also say:

• Let $\mathcal{B}$ be a collection of sets, then $(\bigcup\mathcal{B},\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})$ is a topological space if and only if $\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V]$
  • This is just condition $2$ from above, clearly $1$ isn't needed as $\bigcup\mathcal{B}=\bigcup\mathcal{B}$ (obviously/trivially)

Proof

Notes

1. Jump up ↑ By the implies-subset relation $\forall x\in X\exists B\in\mathcal{B}[x\in B]$ really means $X\subseteq\bigcup\mathcal{B}$, as we only require that all elements of $X$ be in the union. Not that all elements of the union are in $X$. However:
   • $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ by definition. So clearly (or after some thought) the reader should be happy that $\mathcal{B}$ contains only subsets of $X$ and he should see that we cannot as a result have an element in one of these subsets that is not in $X$.
   Thus $\forall B\in\mathcal{B}[B\in\mathcal{P}(X)]$ which is the same as (by power-set and subset definitions) $\forall B\in\mathcal{B}[B\subseteq X]$.
   • We then use Union of subsets is a subset of the union (with $B_\alpha:\eq X$) to see that $\bigcup\mathcal{B}\subseteq X$ - as required.
2. Jump up ↑ We could of course write:
   • $\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]$
3. Jump up ↑ Suppose that $U,V\in\mathcal{B}$ are given but disjoint, then there are no $x\in U\cap V$ to speak of, and $x\in W$ may be vacuously satisfied by the absence of an $X$, however:
   • $x\in W\subseteq U\cap V$ is taken to mean $x\in W$ and $W\subseteq U\cap V$, so we must still show $\exists W\in\mathcal{B}[W\subseteq U\cap V]$
     • This is not always possible as $W$ would have to be $\emptyset$ for this to hold! We do not require $\emptyset\in\mathcal{B}$ (as for example in the metric topology)

References