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conditions for a collection of subsets to be a basis for some topology
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(Proof)
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Not just any collection of subsets of $X$ can be a basis for a topology on $X$ . For instance, if we took $\mathcal{C}$ to be all open intervals of length $1$ in $\mathbb{R}$ , $\mathcal{C}$ isn't the basis for any topology on $\mathbb{R}$ :
$(0,1)$ and $(.5, 1.5)$ are unions of elements of $\mathcal{C}$ , but their intersection $(.5,1)$ is not. The collection formed by arbitrary unions of members of $\mathcal{C}$ isn't closed under finite intersections and isn't a topology.
We'd like to know which collections $\mathcal{B}$ of subsets of $X$ could be the basis for some topology on $X$ . Here's the result:
Theorem 1 A collection $\mathcal{B}$ of subsets of $X$ is a basis for some topology on $X$ if and only if:
- Every $x\in X$ is contained in some $B_x\in \mathcal{B}$ , and
- If $B_1$ and $B_2$ are two elements of $\mathcal{B}$ containing $x\in X$ , then there's a third element $B_3$ of $\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2$ .
Proof. First, we'll show that if $\mathcal{B}$ is the basis for some topology $\mathcal{T}$ on $X$ , then it satisfies the two conditions listed.
$\mathcal{T}$ is a topology on $X$ , so $X\in \mathcal{T}$ . Since $\mathcal{B}$ is a basis for $\mathcal{T}$ , that means $X$ can be written as a union of members of $\mathcal{B}$ : since every $x\in X$ is in this union, every $x\in X$ is contained in some member of $\mathcal{B}$ . That takes care of the first condition.
For the second condition: if $B_1$ and $B_2$ are elements of $\mathcal{B}$ , they're also in $\mathcal{T}$ . $\mathcal{T}$ is closed under intersection, so $B_1\cap B_2$ is open in $\mathcal{T}$ . Then $B_1\cap B_2$ can be written as a union of members of $\mathcal{B}$ , and any $x\in B_1\cap B_2$ is contained by some basis element in this union.
Second, we'll show that if a collection $\mathcal{B}$ of subsets of $X$ satisfies the two conditions, then the collection $\mathcal{T}$ of unions of members of $\mathcal{B}$ is a topology on $X$ .
- $\emptyset \in \mathcal{T}$ : $\emptyset$ is the null union of zero elements of $\mathcal{B}$ .
- $X\in \mathcal{T}$ : by the first condition, every $X$ is contained in some member of $\mathcal{B}$ . The union of all the members of $\mathcal{B}$ is then all of $X$ .
- $\mathcal{T}$ is closed under arbitrary unions: Say we have a union of sets $T_{\alpha}\in \mathcal{T}$ ...
Since that's a union of elements of $\mathcal{B}$ , it's also a member of $\mathcal{T}$ .
- $\mathcal{T}$ is closed under finite intersections: since a collection of sets is closed under finite intersections if and only if it is closed under pairwise intersections, we need only check that the intersection of two members $T_1, T_2$ of $\mathcal{T}$ is in $\mathcal{T}$ .
Any $x\in T_1\cap T_2$ is contained in some $B_x^1\subset T_1$ and $B_x^2\subset T_2$ . By the second condition, $x\in B_x^1\cap B_x^2$ gets us a $B_x^3$ with $x\in B_x^3 \subset B_x^1\cap B_x^2 \subset T_1\cap T_2$ . Then $$ T_1\cap T_2 = \bigcup_{x\in T_1\cap T_2} B_x^3$$
which is in $\mathcal{T}$ .

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"conditions for a collection of subsets to be a basis for some topology" is owned by waj.
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| Keywords: |
teaching proofs, characterization of a basis, what a basis looks like |
This object's parent.
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Cross-references: zero elements, null, open, contained, finite, closed under, intersection, unions, length, open intervals, topology, basis, subsets, collection
There is 1 reference to this entry.
This is version 1 of conditions for a collection of subsets to be a basis for some topology, born on 2004-05-10.
Object id is 5845, canonical name is ConditionsForACollectionOfSubsetsToBeABasisForSomeTopology.
Accessed 3096 times total.
Classification:
| AMS MSC: | 54A99 (General topology :: Generalities :: Miscellaneous) | | | 54D70 (General topology :: Fairly general properties :: Base properties) |
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Pending Errata and Addenda
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