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Need to add: A function is continuous if and only if the pre-image of every basis element is open - Alec (talk) 18:35, 17 December 2016 (UTC)

Definition

Let $(X,\mathcal{ J })$ be a topological space and let $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ be any collection of subsets of $X$ ^{[Note 1]}. We say $\mathcal{B}$ is a basis for the topology $\mathcal{J}$ if both of the following are satisfied:

$\forall B\in\mathcal{B}[B\in\mathcal{J}]$ - every element of $\mathcal{B}$ is an open set of $(X,\mathcal{ J })$
$\forall U\in\mathcal{J}\exists\{B_\alpha\}_{\alpha\in I}\subseteq\mathcal{B}[\bigcup_{\alpha\in I}B_\alpha=U]$ - every open set in $(X,\mathcal{ J })$ is the union of some arbitrary family of basis elements^{[Note 2]}

The elements of $\mathcal{B}$ are called basis elements.

Basis criterion

Let $(X,\mathcal{ J })$ be a topological space and let $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ be a topological basis for $(X,\mathcal{ J })$ . Then^[1]:

$\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}\iff\underbrace{\forall p\in U\exists B\in\mathcal{B}[p\in B\subseteq U]}_{\text{basis criterion} }\big]$ ^{[Note 3]}

If a subset $U$ of $X$ satisfies^{[Note 4]} $\forall p\in U\exists B\in\mathcal{B}[p\in B\subseteq U]$ we say it satisfies the basis criterion with respect to $\mathcal{B}$ ^[1]

Topology generated by a basis

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 5]}) and
2. ∀U,V∈B[U∩V≠∅⟹∀x∈U∩V∃B∈B[x∈W∧W⊆U∩V]]^{[Note 6]}
  - 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 7]}

Notes

Jump up ↑
We could say something else instead of B∈P(P(X)):
- Let $\mathcal{B}\in\mathcal{P}(\mathcal{J})$ - so $\mathcal{B}$ is explicitly a collection of open sets, then we could drop condition $1$ . Or!
- Let $\mathcal{B}\subseteq\mathcal{J}$ . But it is our convention to not say "let $A\subseteq B$ " but "let $A\in\mathcal{P}(B)$ " instead. To emphasise that the power-set is possibly in play.
We do not do these because it (sort of) violates the Doctrine of Least Surprise, we usually deal with subsets of the space not subsets of the set system on that space.
That is a weird way of saying if we have a structure (eg topological space, measurable space, so forth) say (A,B) we usually deal with (collections of) subsets of A and specify they must be in B.
Jump up ↑ The elements of $\mathcal{B}$ are called basis elements. This is mentioned later in the article
Jump up ↑ Note that when we write $p\in B\subseteq U$ we actually mean $p\in B\wedge B\subseteq U$ . This is a very slight abuse of notation and the meaning of what is written should be obvious to all without this note
Jump up ↑ This means "if a $U\in\mathcal{P}(X)$ satisfies...
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

↑ ^{Jump up to: 1.0} ^1.1 Introduction to Topological Manifolds - John M. Lee

OLD PAGE

Definition

Let $X$ be a set. A basis for a topology on $X$ is a collection of subsets of $X$ , $\mathcal{B}\subseteq\mathcal{P}(X)$ such that^[1]:

$\forall x\in X\exists B\in\mathcal{B}[x\in B]$ - every element of $X$ belongs to at least one basis element.
$\forall B_1,B_2\in\mathcal{B},x\in X\ \exists B_3\in\mathcal{B}[x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)]$ ^{[Note 1]} - if any 2 basis elements have non empty intersection, there is a basis element within that intersection containing each point in it.

Note that:

The elements of $\mathcal{B}$ are called basis elements^[1]

Topology generated by $\mathcal{B}$

If $\mathcal{B}$ is such a basis for $X$ , we define the topology $\mathcal{J}$ generated by $\mathcal{B}$ ^[1] as follows:

A subset of X, U⊆X is considered open (equivalently, U∈J) if:
- $\forall x\in U\exists B\in\mathcal{B}[x\in B\wedge B\subseteq U]$ ^{[Note 2]}

[Expand]
Claim: This $\mathcal{(J)}$ is indeed a topology

Notes

Jump up ↑
This is a great example of a hiding if-and-only-if, note that:
- (x∈B3∧B3⊆B1∩B2)⟹x∈B1∩B2 (by the implies-subset relation) so we have:
  - $(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)$
- Thus $(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\iff x\in B_1\cap B_2$
This pattern occurs a lot, like with the axiom of extensionality in set theory.
Jump up ↑ Note that each basis element is itself is open. This is because $U$ is considered open if forall x, there is a basis element containing $x$ with that basis element $\subseteq U$ , if $U$ is itself a basis element, it clearly satisfies this as $B\subseteq B$

TODO: Make this into a claim

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 Topology - Second Edition - James R. Munkres

[1] Jump up ↑ We could say something else instead of $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ :
Let $\mathcal{B}\in\mathcal{P}(\mathcal{J})$ - so $\mathcal{B}$ is explicitly a collection of open sets, then we could drop condition $1$ . Or!

Let $\mathcal{B}\subseteq\mathcal{J}$ . But it is our convention to not say "let $A\subseteq B$ " but "let $A\in\mathcal{P}(B)$ " instead. To emphasise that the power-set is possibly in play.
We do not do these because it (sort of) violates the Doctrine of Least Surprise, we usually deal with subsets of the space not subsets of the set system on that space.
That is a weird way of saying if we have a structure (eg topological space, measurable space, so forth) say $(A,\mathcal{B})$ we usually deal with (collections of) subsets of $A$ and specify they must be in $\mathcal{B}$ .

[2] Let $\mathcal{B}\in\mathcal{P}(\mathcal{J})$ - so $\mathcal{B}$ is explicitly a collection of open sets, then we could drop condition $1$ . Or!

[3] Let $\mathcal{B}\subseteq\mathcal{J}$ . But it is our convention to not say "let $A\subseteq B$ " but "let $A\in\mathcal{P}(B)$ " instead. To emphasise that the power-set is possibly in play.

[2] Jump up ↑ The elements of $\mathcal{B}$ are called basis elements. This is mentioned later in the article

[4] Jump up ↑ Note that when we write $p\in B\subseteq U$ we actually mean $p\in B\wedge B\subseteq U$ . This is a very slight abuse of notation and the meaning of what is written should be obvious to all without this note

[5] Jump up ↑ This means "if a $U\in\mathcal{P}(X)$ satisfies...

[6] 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.

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

[9] 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.

[7] Jump up ↑ We could of course write:
$\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]$

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

[8] 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)

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

[14] 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)

[ITTMJML-3] {Jump up to: 1.0} ^1.1 Introduction to Topological Manifolds - John M. Lee

[10] Jump up ↑ This is a great example of a hiding if-and-only-if, note that:
$(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2$ (by the implies-subset relation) so we have:
$(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)$

Thus $(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\iff x\in B_1\cap B_2$
This pattern occurs a lot, like with the axiom of extensionality in set theory.

[17] $(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2$ (by the implies-subset relation) so we have:
$(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)$

[18] $(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\implies x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)$

[19] Thus $(x\in B_3\wedge B_3\subseteq B_1\cap B_2)\iff x\in B_1\cap B_2$

[11] Jump up ↑ Note that each basis element is itself is open. This is because $U$ is considered open if forall x, there is a basis element containing $x$ with that basis element $\subseteq U$ , if $U$ is itself a basis element, it clearly satisfies this as $B\subseteq B$

TODO: Make this into a claim

[Top-9] {Jump up to: 1.0} ^1.1 ^1.2 Topology - Second Edition - James R. Munkres

[Note 1]

[Note 2]

[1]

[Note 3]

[Note 4]

[Note 5]

[Note 6]

[Note 7]

[1]

[Note 1]

[Note 2]

Basis for a topology

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Basis for a topology

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Topology generated by $\mathcal{B}$