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!)