Difference between revisions of "Basis for a topology"

Revision as of 16:43, 20 March 2016

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]:

1. $\forall x\in X\exists B\in\mathcal{B}[x\in B]$ - every element of $X$ belongs to at least one basis element.
2. $\forall B_1,B_2\in\mathcal{B},x\in X\ \exists B_3\in\mathcal{B}[x\in B_1\cap B_2\implies(x\in B_3\wedge B_3\subseteq B_1\cap B_2)]$^{[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 $X$, $U\subseteq X$ is considered open (equivalently, $U\in\mathcal{J}$) if:
  • $\forall x\in U\exists B\in\mathcal{B}[x\in B\wedge B\subseteq U]$^{[Note 2]}

See also

• Topological space
• Sub-basis for a topology
• Common topological constructs

Notes

1. 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.
2. 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 $B\subseteq B$

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   TODO: Make this into a claim

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References

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