Are basis elements of a topology always open? I'm currently going through Munkres' Topology, and his discussion of bases for topologies has me a little confused. He defines a basis $\mathfrak{B}$ for a topology $\mathscr{T}$ on a set $X$ as a collection of subsets that satisfy the given requirements. These elements of $\mathfrak{B}$ are not defined as open, but in the following lemma which claims that $\mathscr{T}$ equals the collection of all unions of elements from $\mathfrak{B}$, he claims in the proof of this lemma that the elements of $\mathfrak{B}$ are in $\mathscr{T}$, and therefore all of the basis elements are open.
Why is this true? I know the basis elements are open in the topology generated by $\mathfrak{B}$, but the lemma refers to an arbitrary topology on some set X. Why should I assume that these are open sets? I appreciate any help. Thanks.
 A: Here is your confusion: what Munkres starts out with is a set $X$ and a collection $\mathfrak B$ of subsets of $X$, which satisfy the given properties. Only then do we define the topology $\mathscr T$ using this basis, so until this is done it doesn't make sense to talk about the basis elements of $\mathfrak B$ as being "open".
Now, once we do define our topology, recall that we define a subset $U$ of $X$ to be open if given $x\in U$ there exists $B\in\frak B$ such that $x\in B\subseteq U$. Since this certainly holds when $U=B\in\frak B$, we see that all elements of $\frak B$ are open.
This is the precise statement of the lemma you are referring to:

Let $X$ be a set; let $\frak B$ be a basis for a topology $\mathscr T$ on $X$. Then $\mathscr T$ equals the collection of all unions of elements of $\frak B$.

So when Munkres says this, he is doing as we say above: that is, he is starting with some collection of subsets $\frak B$ that generate a topology, and he is calling this topology $\scr T$. So by what we said above, all basis elements are open in $\scr T$ and there should be no issue.
