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.

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    $\begingroup$ 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? $\endgroup$ – Jack Lee Dec 6 '18 at 18:53
  • $\begingroup$ @JackLee the set of unions, but I don't know how to write that as a set. $\endgroup$ – Garmekain Dec 6 '18 at 18:56
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    $\begingroup$ Also you seem to be using $b$ for two different things. Perhaps you want $b'$ in "I need to prove...". $\endgroup$ – user113102 Dec 6 '18 at 19:00
  • $\begingroup$ See my edit to the question. $\endgroup$ – Jack Lee Dec 6 '18 at 19:00
  • $\begingroup$ No, now it's correct. Intuitively, I need to find an element which is in the "spanning set" of all the other elements. $\endgroup$ – 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 $X = \mathbb{R}$.)

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  • $\begingroup$ I'm sorry, but my question is about an arbitrary topology, Is it true for at least one topology on a given set? $\endgroup$ – Garmekain Dec 6 '18 at 19:07
  • $\begingroup$ @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. $\endgroup$ – stochastic randomness Dec 6 '18 at 19:11
  • $\begingroup$ @Garmekain Your question is really about bases and topologies. If the base contains a set which is not the smallest open neighbourhood of any of its points, then there will be a set in the base having the desired property. I've added some details above. $\endgroup$ – stochastic randomness Dec 6 '18 at 19:24

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