# How to prove the following fact about a basis of a topology?

Let $$\tau$$ be a topology on some set $$X$$ and $$\mathcal B$$ a basis of $$\tau$$. Let $$b\in\mathcal B$$ be arbitrary. Now, define $$R_b$$ as the set of all unions of any subset(s) of $$\mathcal B\setminus\{b\}$$, (it can be thought as some sort of a "spanning set" of the basis not including the basis element we are interested), so

$$R_b=\left\{U\in\tau\mid\exists \mathcal S\subseteq(\mathcal B\setminus\{b\}):U= \bigcup _{v\in \mathcal S} v\right\}$$

I need to prove that there exists some $$b'\in\mathcal B$$ such that $$b'\in R_{b'}$$.

I know the definition of a basis but I don't know how to tackle this problem, maybe the fact that for every intersection of two elements of a basis there is another element of the basis contained in it may be of help, but I don't see how. Any help would be appreciated. Thanks.

• Your first paragraph says $R_b$ is the set of all unions of subsets of $\mathcal B\setminus \{b\}$, but the displayed equation says that $R_b$ is just the set of finite unions. Which do you mean? – Jack Lee Dec 6 '18 at 18:53
• @JackLee the set of unions, but I don't know how to write that as a set. – Garmekain Dec 6 '18 at 18:56
• Also you seem to be using $b$ for two different things. Perhaps you want $b'$ in "I need to prove...". – user113102 Dec 6 '18 at 19:00
• See my edit to the question. – Jack Lee Dec 6 '18 at 19:00
• No, now it's correct. Intuitively, I need to find an element which is in the "spanning set" of all the other elements. – Garmekain Dec 6 '18 at 19:03

This is not true in general. Take $$\tau$$ to be the discrete topology on $$X$$, with the basis $$\mathcal{B} = \{ \{ x \} : x \in X \}$$ of all singletons. It is easy to show for each $$b = \{ x \} \in \mathcal{B}$$ that $$R_b = \{ A \subseteq X : x \notin A \}$$. So not only is $$b$$ not an element of $$R_b$$, we actually have that $$b$$ is disjoint from every set in $$R_b$$.
Suppose $$\tau$$ is a topology on $$X$$ with a base $$\mathcal{B}$$ having this property. That is, there is a $$b^\prime \in \mathcal{B}$$ such that $$b^\prime \in R_{b^\prime}$$. This means that for each $$x \in b^\prime$$ there is some other $$b_x \in \mathcal{B} \setminus \{ b^\prime \}$$ with $$x \in b_x \subsetneqq b^\prime$$. In particular $$b^\prime$$ cannot be the smallest open neighbourhood of any of its elements, because $$b_x$$ is a strictly smaller open neighbourhood of $$x \in b$$.
And this should work the other way around. If $$\mathcal{B}$$ is a base for $$\tau$$ which contains some $$b^\prime$$ which is not the smallest open neighbourhood of any of its elements, then $$b^\prime \in R_{b^\prime}$$. (Each $$x \in b^\prime$$ has an open neighbourhood $$V$$ with $$x \in V \subsetneqq b^\prime$$, and so there is a $$b_x \in \mathcal{B}$$ with $$x \in b_x \subseteq V \subsetneqq b^\prime$$, meaning $$b_x \in \mathcal{B} \setminus \{ b^\prime \}$$. Then $$\{ b_x : x \in b^\prime \} \subseteq \mathcal{B} \setminus \{ b^\prime \}$$ and $$\bigcup_{x \in b^\prime} b_x = b^\prime$$.)
In particular, if $$\tau$$ is a topology on $$X$$ such that no point in $$X$$ has a smallest open neighbourhood, then every (nonempty) set in any base for $$\tau$$ will be as desired. (E.g., $$\tau$$ could be the usual (metric/order) topology on $$X = \mathbb{R}$$.)
• @Garmekain It will be true for some topologies. I think that you need to have that at least one $x \in X$ does not have a smallest open neighbourhood. But maybe something stronger. – stochastic randomness Dec 6 '18 at 19:11