Topology's cardinality I'm trying to solve this question but yet I don't manage to find an answer...
Does it exist a topology with cardinality $\alpha, \> \forall \> \alpha  \ge 1$ ?
 A: Let $S$ be a set of cardinality $\alpha$ and (using $\alpha\ge1$) pick $s\in S$. Invoking the axiom of choice, let $\le $ be a wellorder on $X:=S\setminus\{s\}$ and extend it to a wellorder on $S$ by declaring $x\le s$ for all $x\in S$.
For $x\in S$ let $[x,\infty):=\{\,y\in S\mid x\le y\,\}$
and let $$\mathcal T=\{\,[x,\infty)\mid x\in S\,\}.$$
Then $\mathcal T$ is in obvious bijection with $S$, hence of cardinality $\alpha$. One verifies that $(X,\mathcal T)$ is a topological space:


*

*$\emptyset = [s,\infty)\in\mathcal T$

*$[a,\infty)\cap [b,\infty)=[\max\{a,b\},\infty)\in\mathcal T$ for $a,b\in S$

*$\bigcup_{i\in I}[a_i,\infty)=[\min\{\,a_i\mid i\in I\,\},\infty)\in\mathcal T$ for $I\ne\emptyset$ with $a_i\in S$

A: For well-orderable cardinalities, at least, this is true (so holds for all non-zero cardinalities assuming the Axiom of Choice). For any Von-Neumann ordinal $\alpha,$ $\alpha\cup\{\alpha\}$ is a topology on $\alpha$ of cardinality $|\alpha|+1.$ This gets us all finite non-zero cardinalities, and all well-orderable infinite cardinalities, since $|\alpha|+1=|\alpha|$ for infinite $\alpha$.
A: For $S$ infinite, the co-finite topology on $S$ is has the same cardinality as the set of finite subsets of $S$. I'm not sure if that requires Axiom of Choice, but well-ordering certainly does.
(The cofinite topology on $S$ has as closed sets $S$ and any finite subset of $S$.)
That would leave the finite sets $S$.
