# Topological and neighbourhood bases

1. Jul 21, 2011

### 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.)

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 $B$?

Last edited: Jul 21, 2011
2. Jul 21, 2011

### micromass

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).

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

3. Jul 21, 2011

### Fredrik

Staff Emeritus
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.

Last edited: Jul 21, 2011
4. Jul 21, 2011

### micromass

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!!

5. Jul 21, 2011

### Fredrik

Staff Emeritus
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).

6. Jul 21, 2011

### 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!)