Let $X$ be a set, and $\mathcal{S}$ a set of subsets of $X$. Let $\tau'(\mathcal{S})$ be the set of finite intersections of $\mathcal{S}$, and $\sigma(\tau'(\mathcal{S}))$ the set of arbitrary unions of $\tau'(\mathcal{S})$. Then $\mathcal{T}=\sigma(\tau'(\mathcal{S}))$ is a topology on $X$. My question is, if we reverse the operations $\tau'$ and $\sigma$, do we still get a topology on $X$ ? That is, if we first form arbitrary unions of $\mathcal{S}$ and then their finite intersections, do we get a topology on $X$ ? (In other words, is $\tau'(\sigma(S))$ a topology on $X$ ?) I think I can show $\tau'(\sigma(S))\subseteq\mathcal{T}$, but is it an equality? My guess is no, but is there a counterexample?
Reversing finite intersection and arbitrary union operations on a subbasis
general-topology
Related Solutions
First of all, I think you have some confusion about what $\tau$ is, based on your claim that $\tau$ is countable (finite, even!). $\tau$ is definitely uncountable: any uncountable set has uncountably many countable subsets - e.g. the set of singleton subsets is uncountable.
I think you may be conflating $\tau$ - which is a set of subsets of $X$ - with individual subsets of $X$.
Now as to the different types of union, it might help to consider a concrete example. Let $X=\mathbb{R}$, and consider the set $[0,1]$. Then:
$[0,1]$ is a union of sets in $\tau$, namely $$[0,1]=\bigcup_{x\in[0,1]}\{x\},$$ so $[0,1]$ had better be in $\tau$ if $\tau$ is to be a topology.
But $[0,1]$ isn't all of $X$ and it also isn't countable, so it's not in $\tau$. What's going on here is that while we can write it as a union of a bunch of sets in $\tau$ (as above), we can't write it as a union of only countably many sets in $\tau$.
The difference between countable unions and arbitrary unions is just how many sets we're allowed to "union together." In a countable union, we're taking the union of only countably many sets; in an arbitrary union, we're taking the union of as many sets as we want. For example, a countably union of countable sets is countable (briefly, "$\aleph_0\times\aleph_0=\aleph_0$" if you're familiar with $\aleph$-notation), but an arbitrary union of countable sets can be as big as you want: indeed, any set is an arbitrary union of one-element sets via $$S=\bigcup_{s\in S}\{s\}.$$
As a further exercise, thinking along the lines above we can show:
If $\tau$ is a topology on a set $X$ which contains every one-element subset of $X$, then in fact $\tau$ contains every subset of $X$ (we say $\tau$ is the discrete topology in this case).
Once you're comfortable with the proof of this fact, I think you'll understand the issues above perfectly.
Yes, it is guaranteed that the complement of any closed set in $(E,\tau)$ is in $\tau$, since, by definition, asserting that $A$ is closed means that $A^\complement\in\tau$.
On the other hand, it makes no sense to assert that $\tau$ is “closed under … of closed sets”. But that is not the question. There are in fact two questions:
- the set of closed subsets of $(E,\tau)$ is closed under finite unions;
- the set of closed subsets of $(E,\tau)$ is closed under arbitrary intersections.
Best Answer
No, not in general. Take $\Bbb R$ with subbase $\mathcal{S}$ all sets of the form $(x,\rightarrow)=\{y\in \Bbb R: y > x\}$ and $(\leftarrow,x)=\{y \in \Bbb R: y < x\}$. This is the standard subbase for any linearly ordered topological space (LOTS), as $\Bbb R$ is.
$\sigma(\mathcal{S})$ contains just $\mathcal{S} \cup \{\Bbb R\}$, plus "gap sets" like $(\leftarrow, x) \cup (y,\rightarrow)=\Bbb R \setminus [x,y]$ for $x < y$.
The finite intersections of these will not give all open subsets of $\Bbb R$: e.g. the complement of the Cantor middle third set won't be formed with finitely many intersections.