I have studied definitions of Basis and Subbasis from Munkres book. I have a question that is it necessary that if a set S is element of Subbasis then it must be element of basis?
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$\begingroup$ Maybe I do not understand your question but by definition a subbasis is a subset of the topology. $\endgroup$– ScuderiMar 9, 2020 at 13:20
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$\begingroup$ A topology is a basis. A subbase is a subset of a topology. $\endgroup$– user658409Mar 9, 2020 at 13:21
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1$\begingroup$ Not if you start with a fixed subbase and a fixed base. On the other hand: If $S$ is element of a subbase of some topology $\tau$ then $S\in\tau$. And for every $S\in\tau$ we can find a basis $\mathcal B$ such that $S\in\mathcal B$. Actually we could just take $\mathcal B:=\tau$ because $\tau$ is a basis of itself. $\endgroup$– drhabMar 9, 2020 at 13:30
2 Answers
No not necessarily, e.g. $\mathcal{S}=\{(\leftarrow, a): a \in \Bbb R\} \cup \{(a,\rightarrow), a \in \Bbb R\}$ is a subbase for the standard topology on $\Bbb R$. And $\mathcal{B}=\{(a,b): a < b, a,b \in \Bbb R\}$ is a base for that same topology, and no element of $\mathcal S$ is in $\mathcal{B}$.
But what is true, is that any subbase $\mathcal{S}$ for a topology generates a base $\mathcal{B}$ for the same topology, by taking all finite intersections from $\mathcal{S}$ and then $\mathcal{S} \subseteq \mathcal{B}$ as an intersection of a subfamily of size 1 is just the same set. So it depends how you look at it, and how a base and subbase are related to each other.
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$\begingroup$ Thank you so much. I was actually thinking same example that you took. But, I didn't think that this subbasis will create some new basis for standard topology on R. $\endgroup$ Mar 9, 2020 at 13:45
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1$\begingroup$ @user677008 a space can have many subbases and many bases too, some of them related, others not so much. $\endgroup$ Mar 9, 2020 at 13:49
No.
Give $S = (-1,1)$ the subbase
$$\left\{ (-1,r), (r,1) : r\text{ irrational }\right\}\text{
and the base }\left\{ (a,b) : a,b\text{ rational} \right\}.$$