Notes:Basis for a topology/Attempt 2

Overview

Last time I tried to merge the definitions, which got very confusing! This time I shall treat them as separate things and go from there.

Here we will use GBasis for a generated topology (by a basis) and TBasis for a basis of an existing topology.

Let $X$ be a set and $\mathcal{B}\subseteq\mathcal{P}(X)$ be a collection of subsets of $X$ . Then we say:

B is a GBasis if it satisfies the following 2 conditions:
1. $\forall x\in X\exists B\in\mathcal{B}[x\in B]$ - every element of $X$ is contained in some GBasis set.
2. $\forall B_1,B_2\in\mathcal{B}\forall x\ B_1\cap B_2\exists B_3\in\mathcal{B}[B_1\cap B_2\ne\emptyset\implies(x\in B_3\subseteq B_1\cap B_2)]$ ^{[Note 1]}^{[Note 2]}

Then $\mathcal{B}$ induces a topology on $X$ .

Let $\mathcal{J}_\text{Induced}$ denote this topology, then:

$\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}_\text{Induced}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big]$

Suppose $(X,\mathcal{ J })$ is a topological space and $\mathcal{B}\subseteq\mathcal{P}(X)$ is some collection of subsets of $X$ . We say:

B is a TBasis if it satisfies both of the following:
1. $\forall B\in\mathcal{B}[B\in\mathcal{J}]$ - all the basis elements are themselves open.
2. $\forall U\in\mathcal{J}\exists\{B_\alpha\}_{\alpha\in I}[\bigcup_{\alpha\in I}B_\alpha=U]$

If we have a TBasis for a topological space then we may talk about its open sets differently:

$\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}\iff\big(\forall p\in U\exists B\in\mathcal{B}[p\in B\wedge B\subseteq U]\big)\big]$

Jump up ↑
Note that x∈B3⊆B1∩B2 is short for:
- $x\in B_3\wedge B_3\subseteq B_1\cap B_2$
Jump up ↑ Note that if $B_1\cap B_2$ is empty (they do not intersect) then the logical implication is true regardless of the RHS of the $\implies$ } sign, so we do not care if we have $x\in B_3\wedge B_3\subseteq B_1\cap B_2$ ! Pick any $x\in X$ and aany $B_3\in\mathcal{B}$ !