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I'm trying to solve this question but yet I don't manage to find an answer...

Does it exist a topology with cardinality $\alpha, \> \forall \> \alpha \ge 1$ ?

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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 $a_i\in S$
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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 $\alpha$.

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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 $S$.

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I believe that there is some choice in showing that for infinite sets $S,$ we have that $S$ has the same cardinality as the set of its finite subsets. Countable choice is sufficient, iirc, but it may not be necessary. I'll have to think on it. – Cameron Buie Aug 27 '13 at 17:02
Thanks, @CameronBuie . Really wasn't sure, but it seemed weaker than Well Ordering. You really only need that $S\times S$ has the same cardinality as $S$ when $S$ infinite, and that $\mathbb S\times N$ is the same cardinality as $S$ to prove the above, which follows since $\mathbb N$ has smaller cardinality than $S$. – Thomas Andrews Aug 27 '13 at 17:05
Thomas, @Cameron: In order to prove that for every infinite $S$, $[S]^{<\omega}\sim S$ one has to use the axiom of choice. No less. – Asaf Karagila Aug 27 '13 at 17:08
Actually, that $S\times S$ is equinumerous to $S$ for all infinite $S$ is equivalent to AC, so I think you actually need it all. – Cameron Buie Aug 27 '13 at 17:08
Really? That's fascinating. @CameronBuie Thanks. – Thomas Andrews Aug 27 '13 at 17:09

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