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According to Willard,

If $(X,\tau)$ is a topological space, a base for $\tau$ is a collection $\mathscr{B} \subset \tau$ such that $\tau=\{ \bigcup_{B \in \mathscr C} : \mathscr C \subset \mathscr B\}$. Evidently, $\mathscr B$ is a base for $X$ iff whenever $G$ is an open set in $X$ and $p \in G$ there is some $B \in \mathscr B$ such that $p \in B \subset G$.

Question 1: Is it safe to assume that in the sentence beginning with "Evidently" it is assumed that $\mathscr B \subset \tau$, for otherwise the iff statement is not true.

Question 2: I've been told that not all basic sets are open, but it seems by the above definition that they are defined to be open.

Comment on Question 2: There is also a definition of being a base for "a" topology. Is this what was meant by not all basic sets are open? Is this just a semantic issue, i.e. the basic sets are open in the topology that the base is a base for but not open in general? Or can there be a base for a topology where the basic sets are not open in that same topology?

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+1 for reading Willard : ) – Rudy the Reindeer Oct 17 '12 at 19:16
One can also consider bases of neighborhoods, and then its elements are not necessarily open. – Mariano Suárez-Alvarez Oct 17 '12 at 19:26
up vote 1 down vote accepted

Question 1: Yes, Willard was still talking about collections $\mathscr{B}\subseteq\tau$. However, with a minor change in wording he wouldn’t have to be, because we could deduce that $\mathscr{B}\subseteq\tau$. Specifically, suppose that $\langle X,\tau\rangle$ is a topological space, and that $\mathscr{B}$ is a family of subsets of $X$ such that

$(*)\quad$ for each $G\subseteq X$, $G\in\tau$ iff for each $p\in G$ there is some $B_p\in\mathscr{B}$ such that $p\in B_p\subseteq G$.

If $B\in\mathscr{B}$, the condition is satisfied if for each $p\in B$ we let $B_p=B$, so $B\in\tau$, and therefore $\mathscr{B}\subseteq\tau$. Thus, he could have said that a family $\mathscr{B}$ of subsets of a space $X$ is a base for the topology of $X$ iff $(*)$ holds. He wouldn’t have had to specify that the members of $\mathscr{B}$ are open, because that follows from $(*)$.

Question 2: You were misinformed: basic open sets in a topological space are always open sets in that space. If $\mathscr{B}$ is a base for a topology $\tau$ on a set $X$, then $\mathscr{B}\subseteq\tau$.

Question 2': There is a difference between ‘$\mathscr{B}$ is a base for the topology $\tau$ on $X$’ and ‘$\mathscr{B}$ is a base for some topology on the set $X$’. The definition from Willard at the beginning of your question is the definition of the first of these. However, it is also possible to characterize the collections of sets that are bases for some topology on a given set.

Let $X$ be a set, and let $\mathscr{B}$ be a family of subsets of $X$. Then the following are equivalent:

  1. $\mathscr{B}$ is a base for some topology on $X$.
  2. $\left\{\bigcup\mathscr{A}:\mathscr{A}\subseteq\mathscr{B}\right\}$ is a topology on $X$.
  3. $\bigcup\mathscr{B}=X$, and whenever $p\in X$, $B_1,B_2\in\mathscr{B}$, and $p\in B_1\cap B_2$, there is a $B\in\mathscr{B}$ such that $p\in B\subseteq B_1\cap B_2$.

(This is worth trying to prove if you’ve not seen it already.)

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Question 1: Yes. It says so in the first sentence. Unless I am misunderstanding your question.

Question 2: Every base is a subset of the topology it generates. By definition, all sets in a topology are open. Hence all sets in a base also have to be open. Hence "no" to your last question in the comment to question 2.

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Q1: Base sets are open, indeed: consider the one element $\mathscr C$ subsets.

Q2: Lied. Or thought about something else..

A system $\mathscr B$ of subsets of a set $X$ can be a basis for a topology iff $\forall B,C\in\mathscr B\ \forall x\in B\cap C\ \exists D\in\mathscr B$ such that $x\in D\subseteq B\cap C$'. This guarantees that $\tau$ as defined above, will be closed under finite intersection.

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