Section 13: Basis for a Topology. Definition and lemma. One of the problem I’m facing while doing topology from James Munkres book is that everything(definition, theorem, proof) is written in words, instead of symbols and quantifiers. Sometimes which(words) makes things ambiguous.
The following is the definition of basis:

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathscr{B}$ of subsets of $X$ (called basis elements) such that
(1)For each $x∈X$, there is at least one basis element $B$ containing $x$
(2)If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis
element $B_3$ containing $x$ such that $B_3⊆B_1∩B_2$.
If $\mathscr{B}$ satisfies these two conditions, then we define the topology $\mathfrak{I}$ generated by $\mathscr{B}$ as follows : A subset $U$ of $X$ is said to be open in $X$( i.e., to be an element of $\mathfrak{I}$ ) if for each $x\in U,$ there is a basis element $B\in \mathscr{B}$ such that $x\in B$ and $B\subset U.$

Above definition could have been written in much elegant and concise way.
Question: (1) “for a topology on X” - Why are we using the word topology? When we don’t know anything about the topology or topological space? We can just defined a set, we’ll call basis, if that set satisfy (1) & (2). Later based on this set(which have some structure), we’ll define topology. Summary: Given $\mathscr{B} \subseteq \wp(X)$(power set), check if $\mathscr{B}$ satisfy two given  conditions and if $\mathscr{B}$ satisfy those conditions, then we’ll call set $\mathscr{B}$ a basis, without mentioning any topology on $X$.
(2) In this book, Mukres uses the word “open set” lots of times. What if we have more than one topology on $X$, then $U$ is open, means what?
The trend follows (3)

Lemma 13.1. Let $X$ be a set; let $\mathscr{B}$ be a basis for a topology $\mathfrak{I}$ on $X$. Then $\mathfrak{I}$ equals the collection of all unions of elements of $\mathscr{B}$.

Now what does “$\mathscr{B}$ be a basis for a topology $\mathfrak{I}$ on $X$” means? Is it topology $\mathfrak{I}$ generated from basis $\mathscr{B}$? If yes, then why not use that wording? Certainly you would agree that we can’t prove the lemma with this “ collection of all unions of elements of $\mathscr{B}$” wording. One need to change from word format to symbols format to prove the lemma. How do i do it?
 A: A topology on $X$ (a set), as you know, is just a set of subsets of $X$ satisfying certain axioms.
Sometimes we can just decsribe this collection in one fell swoop (like the cofinite one, the discrete and the trivial one), but often we define a topology via a basis: a (hopefully smaller and more managable collection of subsets $\mathcal{B}$ of $X$) such that a set is open (i.e. in the topology) iff we can write it as some union of elements from $\mathcal{B}$. This leads to often more workable definitions of open sets, i.e. topologies.
The conditions he presents are necessary and sufficient for a collection $\mathcal{B}$ so that the description of a topology as unions from it actually makes sense and is valid.
So he intends to define a topology using $\mathcal{B}$, but then you first have to check these two conditions on it (for which you only have to know $\mathcal{B}$ itself, and $X$ and no more). After that being done yoy can say you define a topology via the unions (this is what he calls "generating the topology" from $\mathcal{B}$).
OTOH he considers the opposite situation: you already some way are in possession of a topology $\mathcal{T}$ on $X$ and wonder could I have generated it by some base $\mathcal{B}$ as well? Well, $\mathcal{B} = \mathcal{T}$ works trivially, but we'd like a smaller collection somehow that generates $\mathcal{T}$. So he considers $\mathcal{B}$ a base for the topology $\mathcal{T}$ iff

$\mathcal{B} \subseteq \mathcal{T}$ and for every $O \in \mathcal{T}$ and every $x \in O$ we have some $B_x \in \mathcal{B}$ so that $x \in B_x \subseteq O$.

And the lemma you stated in fact says that (in this case) if we had chosen $\mathcal{B}$ as a base on $X$ then $\mathcal{T}$ would have been exactly the generated topology by $\mathcal{B}$. So the two ways of being a base mean the same in the end.
A: Proof of Lemma 13.1.
If $U\in\mathfrak I$ then for every $x\in U$ there is some $B_x\in\mathscr B$ with $x\in B_x\subseteq U$ so that we can write:$$U=\bigcup_{x\in U}B_x$$showing that $U$ can be written as a union of elements of $\mathscr B$.
Conversely every element of $\mathscr B$ is also an element of topology $\mathfrak I$. A topology is closed under unions so that a union of elements of $\mathscr B$ will again be an element of $\mathfrak I$.
