Topological and neighbourhood bases

[Rasalhague]

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.

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

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

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

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

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

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

Assume that, \forall x, \beta [x] is a neighbourhood base for x. Let U \in T. Then (\forall x \in U) (\exists U_x \in \beta [x])[U_x \subseteq U], by the definition of a neighbourhood base. [Because U, as an open set, is a neighbourhood of x, being a superset of itself.] But remember that the neighbourhood base is a subset of the base, by definition: \beta [x] \subseteq B. [By definition of what? Of the suggestively labelled set \beta [x]? 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 B might be an accidental anticipation of the conclusion yet to be proved.] So U_x \in B. So U = \cup_{x \in U} U_x, so B is a [topological] base.

What if U_x \notin T? Since U_x is a neighbourhood of x, we could replace it with a subset of itself which is open, but how would we know that this subset of U_x is in B?

 

[micromass]

But remember that the neighbourhood base is a subset of the base, by definition:

What he meant to say is that \beta(x)\subseteq B. Calling it a base is wrong (since that's what we wanted to prove).

What if U_x\notin T?

Well noticed! Indeed, one should take B a subset of T from the very beginning. That is, the theorem should be

Let (X,\mathcal{T}) be a topological space and let \mathcal{B}\subseteq \mathcal{T}. Then \mathcal{B} is a base if and only if
\beta (x)=\{B\in \mathcal{B}~\vert~x\in B\}
is a neighbourhood base for all x
​

 

[Fredrik]

I wrote this before micromass replied.

I don't have time to study the proof in detail, but I have some notes on this that I can (almost) just copy and paste, so I'll do that. Maybe it will help, maybe it won't.

Theorem: Suppose that (X,\tau) is a topological space. The following conditions on a set \mathcal B\subset\mathcal P(X) are equivalent.

(a) Every non-empty open set is a union of members of \mathcal B.
(b) If E is open and x\in E, there's a B\in\mathcal B such that x\in B\subset E.Proof:
(a) \Rightarrow (b): Let E be an arbitrary non-empty open set, and let x\in E be arbitrary. By assumption, there's a set \{B_i\in\mathcal B|i\in I\} such that E=\bigcup_{i\in I} B_i. This means that there's a j\in I such that x\in B_j\subset\bigcup_{i\in I} B_i=E.

(b) \Rightarrow (a): Let E be an arbitrary non-empty open set. For each x\in E, choose B_x\in B such that x\in B_x\subset E. The fact that x\in B_x for all x\in E implies that E\subset\bigcup_{x\in E} B_x. The fact that B_x\subset E for all x\in E implies that \bigcup_{x\in E} B_x\subset E. So \bigcup_{x\in E} B_x=E.

Definition: If (X,\tau) is a topological space, a set \mathcal B\subset\tau that satisfies the equivalent conditions of the theorem is said to be a base for the topology \tau.

I wasn't even aware that there's a term for a collection of subsets that satisfies (a) and another one for a collection of subsets that satisfies (b).

Edit: I fixed a mistake in the statement of the theorem. I had left out the word "non-empty" from condition (a). It should definitely be there.

 

[micromass]

Edit: I see that I need to change something because of the empty set. The problem is that (a) implies that \emptyset\in\mathcal B, but (b) doesn't. So the two statements can't be equivalent. I'm thinking about what to change now.

That is not a problem because of some trickery. In fact, (a) doesn't imply that \emptyset\in \mathcal{B}. We can always write

\emptyset = \bigcup \{B\in \mathcal{B}~\vert~B\neq B\}

The statement B\neq B isn't really important, the importance is to take an empty union! The fun thing is that the empty union is always empty. So if we demand that every open set is a union of sets, then this is always true for the empty set!

 

[Fredrik]

That is not a problem because of some trickery. In fact, (a) doesn't imply that \emptyset\in \mathcal{B}. We can always write

\emptyset = \bigcup \{B\in \mathcal{B}~\vert~B\neq B\}

That's a fun trick. I certainly didn't think of that. But I think I prefer to just add the word "non-empty" to (a).

 

[Rasalhague]

Thanks all for clearing that up!

(I've just made some edits of my own to the OP. Must have been too intent on getting the LaTeX right to notice that I wrote "neighbourhood basis for x if and only if [...] is a neighbourhood basis for x". Yikes!)