Terminology; Basis for a Topology In section 13 of Munkres's topology, he defines a basis $\mathcal{B}$ containing subsets $B$ of a set $X$. One of the first things he proves is that the resulting collection, $\mathcal{T}$, is in fact a topology.
Here is my confusion. He uses "is an element of $\tau$" and "is open" interchangeably. But after we have a topology, we can call its elements open sets, so this seems circular to me. Is the idea that elements generated by a basis are open, and then we check this is actually a topology? Otherwise,  it seems to me that we should be talking about membership in $\mathcal{T}$ and only use the notion of openness after checking that it satisfies the axioms for a topology.
 A: The definition of "being open" is "being in $\mathcal{T}$". A topology is nothing more than the collection of the open subsets that are defined on a set. In this case, even though it is not proven (yet) that $\mathcal{T}$ is indeed a topology, as some have already pointed out in the comments, membership in $\mathcal{T}$ and openness are really the same thing, since openness is not an absolute property of a set (something that is or is not true regardless of the context), but rather a relative property. Hence, it doesn't make sense to say that it is false to say that a set is open, even if $\mathcal{T}$ hasn't been proven to be a topology yet. So for me there is no problem whatsoever.
A: You are entirely right. Munkres is abusing terminology a little bit by calling the sets in $\mathcal{T}$ "open" before we actually know that $\mathcal{T}$ is a topology.
That said, you don't need to fear any kind of circular reasoning. Munkres never uses any facts about open sets in his proof, he only uses the name as an abbreviation for $U \in \mathcal{T}$. For instance, here:

Now let us take an indexed familiy $\{ U_\alpha \}_{\alpha \in J}$, of elements of $\mathcal{T}$ and show that
$$U = \bigcup_{\alpha \in J} U_\alpha$$
belongs to $\mathcal{T}$. Given $x \in U$, there is an index $\alpha$ such that $x \in U_\alpha$. Since $U_\alpha$ is open, there is a basis element $x \in B \subset U_\alpha$. Then $x \in B$ and $B \subset U$, so that $U$ is open, by definition.

Munkres is showing that axiom $(2)$ for a topology holds, by checking that the union of elements of $\mathcal{T}$ is again an element of $\mathcal{T}$. Recall $V \in \mathcal{T}$ if and only if for each $x \in V$, there is a $B$ in the basis with $x \in B \subset V$. He uses the word "open", but you can see that making the replacements

*

*"Since $U_\alpha$ is open" $\rightsquigarrow$ "Since each $U_\alpha \in \mathcal{T}$"

*"so that $U$ is open" $\rightsquigarrow$ "so that $U \in \mathcal{T}$
doesn't change the proof at all.
In fact, in the next paragraph (where he proves axiom $(3)$ holds), he uses this (slightly clearer) terminology instead.

I hope this helps ^_^
