In measure theory, we know there is a (unique) minimal $\sigma$-algebra generated by a generator. I am wondering whether this applies to topology and its basis. There are two directions to consider here.
- Given a topology $\mathcal T$ of a set $X$, is there a unique basis for $\mathcal T$, please? If yes, how to prove it and if not, is there a simple counter-example, please?
- Given a basis, is the topology generated by this basis unique and minimal, please? If yes, how to prove it and if not, is there a simple counter-example, please?
It seems to me that this question is basically ignored in the book I am reading. But it seems that some proofs implicitly assume something like the above. Could anyone explain this to me or direct me to useful references, please? Thank you!
Suppose we have a collection $\mathcal{B}$ of subsets of $X$, satisfying the usual axioms:
Then we define $\mathcal{T} = \{ \cup \mathcal{O}: \mathcal{O} \subset \mathcal{B} \}$, the family of all unions of subfamilies of $\mathcal{B}$. This is a topology:
a. $\emptyset = \cup \emptyset$ and $\emptyset \subset \mathcal{B}$ and $X = \cup \mathcal{B}$ by axiom 1.
b. If $O_1 = \cup \mathcal{O}_1 \in \mathcal{T}$ and $O_2 = \cup \mathcal{O}_2 \in \mathcal{T}$, then for each $x \in O_1 \cap O_2$ we find $B_1 \in \mathcal{O}_1 \subset \mathcal{B}$ such that $x \in B_1$, and $B_2 \in \mathcal{O}_2 \subset \mathcal{B}$ such that $x \in B_2$. These $B_1, B_2$ are in $\mathcal{B}$ and $x$ is in both, so axiom 2. finds us a $B_3(x)$ such that $x \in B_3(x) \subset B_1 \cap B_2 \subset O_1 \cap O_2$. Having $B_3(x)$ for all $x \in O_1 \cap O_2$, we then define $\mathcal{O_3} = \{B_3(x): x \in O_1 \cap O_2\}$ and we see that $O_1 \cap O_2 = \cup \mathcal{O}_3$ (all members are subsets of $O_1 \cap O_2$ and each $x$ is covered by its own $B_3(x)$ at least). So $O_1 \cap O_2 \in \mathcal{T}$.
c. Suppose $I$ is an index set and for each $i \in I$ we have a set $O_i \in \mathcal{T}$, so $O_i = \cup \mathcal{O}_i \subset \mathcal{B}$. Then define $\mathcal{O} = \cup_{i \in I} \mathcal{O}_i \subset \mathcal{B}$, and we see that $\cup_{i \in I} O_i = \cup \mathcal{O}$ by associativity of unions, so $\cup_{i \in I} O_i \in \mathcal{T}$.
So we started with a collection of subsets satisfying two axioms, and we can define a topology from that. And moreover, this collection $\mathcal{B}$ does indeed form a base for this newly defined topology $\mathcal{T}$; to see this depends slightly on how you define base: by definition, all open subsets are unions from the base, or otherwise it's clear that for each $O \in \mathcal{T}$, and every $x \in O= \mathcal{O} \subset \mathcal{B}$, $x$ must by definition be in one of the members of $\mathcal{B}$ from $\mathcal{O}$. So there indeed exists $B \in \mathcal{B}$ with $x \in B \subset O$.
Also, there is no choice: if $\mathcal{T}'$ is any topology that has $\mathcal{B}$ as a base, then for any $O \in \mathcal{T'}$ we can write $O$ as a union of members of $\mathcal{B}$ (this is what being a base means!) and so $O$ is then already in our defined $\mathcal{T}$. Hence $\mathcal{T}' \subset \mathcal{T}$. The other side is also obvious: all sets in $\mathcal{B'}$ are in $\mathcal{T'}$ (base sets for a topology are themselves open in that topology) and so all unions of their subfamilies, i.e. $\mathcal{T}$, are also in $\mathcal{T}'$, as topologies are closed under unions. So $\mathcal{T} \subset \mathcal{T}'$, and we have equality. So the topology defined by a (pregiven) base $\mathcal{B}$ is uniquely determined by that base, in the way we define above.
On the other hand, given a topology $\mathcal{T}$, there are in general lots of bases for it. Of course $\mathcal{T}$ is itself a base (trivially). But often we cam omit sets from a base, and still have a base:
Let $X$ be a $T_1$ space (meaning that all singletons are closed sets). If $\mathcal{B}$ is a base for $X$, and $O \in \mathcal{B}$ has more than 1 point, then $\mathcal{B} \setminus \{O\}$ is also a base for $X$. First we show that $O$ can be written as a union of proper subsets, all from $\mathcal{B}$: pick $p \neq q$, both in $O$, which can be done by assumption. Then $O \setminus \{p\} = O \cap (X \setminus \{p\})$ is open in $X$ and thus a union of members from $\mathcal{B}$, and all of these miss $p$, so they are proper subsets of $O$. Similarly $O \setminus \{q\}$ can be so written and then we combine these two families to get what we want. So if $U$ is any open set of $X$, we can write it as a union of members of $\mathcal{B}$. If we do not use $O$, we are done, otherwise replace $O$ in this union by the proper subsets of it (from $\mathcal{B}$ as well) that union up to it, as we know we can. In all cases, we have written $U$ as a union of sets from $\mathcal{B}\setminus\{O\}$.
So from any base for $\mathbb{R}$, e.g., we can remove any open set (either such a set is empty, or it has uncountably many points) and still have a base for $\mathbb{R}$. And we can go on doing that and remove any finite number of them. Countable need not work (as there are countable bases for this topology, and we cannot remove all of them, of course...)
If the topology of $X$ is closed under all (not just finite) intersections, as happens trivially in finite spaces, but in others as well, there is a canonical minimal base: all sets $B_x = \cap \{O : O \text{ open and } x \in O \}$ must be in the base and a base consisting of all of these cannot be thinned out any further. For a discrete space e.g. this is just the base consisting of all singletons. But even in these cases, a base is not unique, but there happens to be a unique minimal one, which makes reasoning about such spaces a bit more convenient.