Topology generated by a basis/Statement

Grade: A

This page requires references, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:

I could do this now but I can't be bothered!

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

Jump up ↑
By the implies-subset relation ∀x∈X∃B∈B[x∈B] really means X⊆⋃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 ∀B∈B[B∈P(X)] which is the same as (by power-set and subset definitions) ∀B∈B[B⊆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.
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

[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] $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ 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 $X$ .

[3] 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:
$\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]$

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

[Note 1]

[Note 2]

Topology generated by a basis/Statement

Contents

Statement

Notes

References

Navigation menu

Views

Personal tools

Navigation

Search

Tools