Neighbourhood base about a point $p$ of a topological group I am reading topological groups from Van der Waerden. The conventions followed in this book are these. An open set that contains the point $p$ is called an open neighbourhood of $p$. Any set which contains an open neighbourhood of $p$ is called a neighbourhood of $p$.
Now in order to define a neighbourhood base around a point $p$, we assign certain sets $U(p)$, which satisfy the following conditions:
a) For each $p$ there are basis sets $U(p)$ and each such set contains $p$,
b) For any two basis sets $U(p)$ and $V(p)$, there is a basis set $W(p)$ such that it is contained in both of them.
c) Each basis set $U(p)$ contains a basis set $V(p)$ such that for each point $q$ of $V(p)$, there is a basis set $W(q)$ contained in $U(p)$.
Then he defines open sets $M$ in terms of these basis sets as those sets which with each of their points $p$ contain an entire basis set $U(p)$.
Now using (c), he claims that $U(p)$ is a neighbourhood of $p$. What he says is this : Since (c) is satisfied, a set $U'$ can be defined in $U(p)$ which consists of all points $q$ such that a basis set $W(q)\subset U(p)$. Then he says $U'$ is open and $p\in U'$. So, $U(p)$ contains an open nbd of $p$, and so $U(p)$ is a nbd. of $p$. 
I don't understand why $U'$ is open. There is some $V(p)\subset U'$. But if $q\in U'$, should there not be some $W(q)\subset U'$? How can I resolve this? Any help will be appreciated.
 A: So for fixed $p$ and $U(p)$ we define $U' = \{q: \exists V(q): V(q) \subset U(p)\}$.
It is quite clear $p$ is in $U'$: $U(p)$ itself witnesses that. Also, any $q \in U'$ is clearly in $U(p)$, as for the promised $V(q)$ we have $q \in V(q) \subset U(p)$, so $U' \subset U(p)$.
Now, we want to prove that $U'$ is open, so for every $q \in U'$ we want to see that there is a $W(q)$ in the local base at $q$, such that $W(q) \subset U'$.
First we get $V(q)$ in the local base at $q$ such that $V(q) \subset U(p)$, by definition of being in $U'$. Then we apply axiom c) to $q$ and this $V(q)$ and we get $W(q)$ in the local base at $q$ such that for each $r \in W(q)$ there is a $W'(r) \subset V(q)$. Note that we renamed the points in c), but the statement is still the same!
The claim is that $W(q) \subset U'$: for every $r \in W(q)$ we find $W'(r)$ such that $W'(r) \subset V(q) \subset U(p)$, so $W'(r)$ witnesses that $r \in U'$, as required.
So $W(q) \subset U'$ and this holds for all $q \in U'$ so $U'$ is open.
So the proof is just essentially applying axiom c) (but the points are renamed, as $p$, $U(p)$ are already taken) and checking the definitions. 
