Difference between revisions of "Topology generated by a basis/Statement"

Revision as of 21:21, 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. $\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V]$^{[Note 2]}

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:
   • $\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]$

References