Topology generated by a basis

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. ∀U,V∈B[U∩V≠∅⟹∀x∈U∩V∃B∈B[x∈W∧W⊆U∩V]]^{[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 B be a collection of sets, then (⋃B,{⋃A | A∈P(B)}) is a topological space if and only if ∀U,V∈B ∀x∈U∩V ∃W∈B[x∈W⊆U∩V]
- This is just condition $2$ from above, clearly $1$ isn't needed as $\bigcup\mathcal{B}=\bigcup\mathcal{B}$ (obviously/trivially)

Proof

Grade: A

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

This is actually an important proof

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:
- $\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]$
Jump up ↑
Suppose that U,V∈B are given but disjoint, then there are no x∈U∩V to speak of, and x∈W may be vacuously satisfied by the absence of an X, however:
- x∈W⊆U∩V is taken to mean x∈W and W⊆U∩V, so we must still show ∃W∈B[W⊆U∩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

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

[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)

[7] $x\in W\subseteq U\cap V$ 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)

[8] 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)

[Note 1]

[Note 2]

[Note 3]

Topology generated by a basis

Contents

Statement

Proof

Notes

References

Navigation menu

Views

Personal tools

Navigation

Search

Tools