Cofinite —> Discrete adding infinite subsets

Question

what is the min number of infinite sized subsets you need to add to a cofinite topology (making it no longer cofinite) to yield the discrete topology given the underlying set is countably infinite?

Answer 1

Write the underlying set $X$ as the disjoint union of two countable sets $X_1$ and $X_2$. (In a bijection with $\Bbb N$, use the image of tje even and odd numbers, e.g.) For every $x\in X$ add $X'_1=X_1\cup \{x\}$ and $X'_2 = X_2 \cup \{x\}$ to the cofinite topology. If we then generate the topology, we get $\{x\}$ as their intersection.

And once all singletons are in a generating set, we get all subsets by unions.

So we only need to add countably many generating infinite sets (2 per element).

And if we add only finitely many sets $\mathcal{F}$ to the cofinite topology $\mathcal{T}$, the resulting generated topology is compact, by Alexander's subbase lemma:

Let $\mathcal{U}$ be an open cover from the subbase $\mathcal{S}=\mathcal{T} \cup \mathcal{F}$. If we have some $O \in \mathcal{U} \cap \mathcal{T}$ then $X\setminus O$ is finite and we can cover all remaining points using finitely many elements from $\mathcal{U}$. So in the remaining case, $\mathcal{U} \subseteq \mathcal{F}$ which was already finite, so we always have a finite subcover for any subbasic cover, hence $X$ in the generated topology by $\mathcal{S}$ is compact, so can never be the discrete topology on an infinite set, which is not compact.

So countably many are necessary and suffice.

Answer 2

Let $\ X\ $ be an infinite countable set. The OP's question is truly and simply about generating the discrete topology by a minimal set of infinite subsets of $\ X.\ $ Indeed, to obtain the discrete topology one must obviously add infinitely many sets to the topology of cofinite sets to generate all singletons.

Actually, we may assume that $$ X:=\mathbb Z\times\{-1\ \ 1\} $$

THEOREM   There exists a countable family $\ \mathbf B\ $ in $\ X\ $ which consists exclusively of infinite sets and which generates the discrete topology.

PROOF   Let $\ e=\pm 1\ $ and

$$ B_{n\ e}\ := \ \big\{(k\ f)\in X:\ \big(\frac 12+k-n\big)\cdot e\cdot f > 0\big\} $$

and $$ \mathbf B\ :=\ \{B_x:\ x\in X\} $$

Then $$ B_{m\ d}\cap B_{n\ e}\ = \ \big\{(k\ f)\in X:\ \big(\frac 12+k-m\big)\cdot d\cdot f > 0 < \big(\frac 12+k-n\big)\cdot e\cdot f \big\} $$ hence $$ B_{m\ -\!1}\cap B_{n\ 1}\ = \ \big\{(k\ f)\in X:\ \big(m-k-\frac 12\big)\cdot f\ >\ 0 \ < \big(\frac 12+k-n)\cdot f \big\} $$ e.g. $\ B_{n\,\ -\!1}\cap B_{n\ 1}\ =\ \emptyset\ $ but that's extra (for fun only). What counts is that

$$ B_{n+1\,\ -\!1}\cap B_{n\ 1}\ = \ \big\{(k\ f)\in X:\ \big(n+\frac12-k\big)\cdot f\ >\ 0 \ < \big(k+\frac 12-n\big)\cdot f \big\} $$ i.e. $$ B_{n+1\,\ -\!1}\cap B_{n\ 1}\ =\ \{(n\ 1)\} $$ is a singleton, and

$$ B_{m\ -\!1}\cap B_{m+1\,\ 1}\ = \ \big\{(k\ f)\in X:\ \big(m-\frac 12-k\big)\cdot f\ >\ 0 \ < \big(k-\frac 12-m)\cdot f \big\} $$ i.e. $$ B_{m\ -\!1}\cap B_{m+1\,\ 1}\ =\ \{(m\,\ -\!1)\} $$

is a singleton again.

Thus, with respect to the topology generated by $\ \mathbf B,\ $ every singleton $\ \{x\}\ $ in $\ X\ $ is isolated, i.e. the topology is discrete.   End of Proof

Answer 3

There was still the issue of the minimality of the addition to the starting topology. The following theorem will address this in full generality.

THEOREM Let $\ (X\ T)\ $ be an arbitrary topological space. Let $\ \mathbf B\ $ be a family which generates the discrete topology in $\ X.\ $ Then cardinality $\ |\mathbf G\setminus T|\ $ is at least as large as the cardinality of the set of the non-isolated points of $\ (X\ T).$

PROOF   Let $\ x\in X\ $ be non-isolated. Then there exists $\ G_x\in\mathbf B\ $ such that $\ x\ $ is the only non-isolated point in $\ G_x.\ $ Obviously, $\ G_x\not\in T,\ $ and equally obviously $\ G_x\cap G_y=\emptyset\ $ whenever $\ x\ne y,\ $ i.e. for different non-isolated points we get different members of $\ \mathbf B\setminus T.\ $   End of Proof