I'm trying to follow a proof in this video, #20 in the ThoughtSpaceZero topology series. I get the first part, but have a problem with second part, which begins at 8:16.(adsbygoogle = window.adsbygoogle || []).push({});

Let there by a topological space [itex](X,T)[/itex]. Let [itex]x[/itex] denote an arbitrary element of [itex]X[/itex].

Definition 1.Topological base. A set [itex]B \subseteq T[/itex] such that [itex](\forall T_i \in T) (\exists C \subseteq B) [T_i = \cup_C C_j][/itex].

Definition 2.Neighbourhood base for[itex]x[/itex]. A subset [itex]\beta [x][/itex] of [itex]V[x][/itex], the neighbourhoods of [itex]x[/itex], such that [itex](\forall V_i \in V[x])(\exists B_i \in \beta [x])[B_i \subseteq V_i][/itex].

Theorem. Let there be a topological space [itex](X,T)[/itex]. Let [itex]B \subseteq 2^X[/itex]. Let [itex]\beta [x] = \left \{ B_i \in B \;|\; x \in B_i \right \} \subseteq B[/itex]. Then [itex]B[/itex] is a topological base for T if and only if, for all [itex]x[/itex], the set [itex]\beta [x][/itex] is a neighbourhood base for [itex]x[/itex].

Proof. Assume [itex]B[/itex] is a base for [itex]T[/itex]. [...]

I understand that part; but I don't follow his proof of the converse. Paraphrasing here: (My comments in square brackets.)

What if [itex]U_x \notin T[/itex]? Since [itex]U_x[/itex] is a neighbourhood of [itex]x[/itex], we could replace it with a subset of itself which is open, but how would we know that this subset of [itex]U_x[/itex] is in [itex]B[/itex]? Assume that, [itex]\forall x[/itex], [itex]\beta [x][/itex] is a neighbourhood base for [itex]x[/itex]. Let [itex]U \in T[/itex]. Then [itex](\forall x \in U) (\exists U_x \in \beta [x])[U_x \subseteq U][/itex], by the definition of a neighbourhood base. [Because [itex]U[/itex], as an open set, is a neighbourhood of [itex]x[/itex], being a superset of itself.]But remember that the neighbourhood base is a subset of the base, by definition: [itex]\beta [x] \subseteq B[/itex].[By definition of what? Of the suggestively labelled set [itex]\beta [x][/itex]? Or was this part of the definition of a neighbourhood base? I'm guessing that "base (unqualified) = topological base" here, and that the reference to [itex]B[/itex] might be an accidental anticipation of the conclusion yet to be proved.] So [itex]U_x \in B[/itex]. So [itex]U = \cup_{x \in U} U_x[/itex], so [itex]B[/itex] is a [topological] base.

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# Topological and neighbourhood bases

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