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Let $A$ be a set, $\mid A \mid$=$\aleph_0$ (assume that A$ \subseteq$$\mathbb{R}$) ,

and let $\chi$=$\{\tau\subseteq$$P(A)$$\mid$$($A,$\tau$$)$is a Hausdorff space, $\mid \tau \mid$=$\aleph$$\}$ .
($P(A)$ is the power set of $A$)

What is the "value" of $\mid \chi \mid$ $?$

The question is basically how many Hausdorff topologies $\tau$ ($\mid \tau \mid=\aleph$) (up to isomorphism) are exist on a set $A$, $\mid A \mid=\aleph_0$.

Correct me if I'm wrong, but I know that there is at least one topology as I mentioned before:

For example:

The set is $\mathbb{Q}$ and the topology $\tau$ is Subspace topology of the euclidean metric on $\mathbb{R}$ (its Hausdorff).

Obviously $\mid \mathbb{Q} \mid$=$\aleph_0$, and we know that the base of $\tau$ in $\mathbb{R}$ is all the open intervals, hence $\mid \tau \mid$ on $\mathbb{Q}$ is $\aleph$.

Hence $\chi \neq \emptyset$.

What is $\mid \chi \mid$?

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  • $\begingroup$ $\aleph$ is not a definite cardinal? What do you mean, just that $\tau$ is infinite ? $\endgroup$ – Henno Brandsma Aug 22 at 8:40
  • $\begingroup$ @HennoBrandsma I suppose that by $\aleph$ he means the cardinality of the real numbers. $\endgroup$ – Cronus Aug 22 at 8:44
  • $\begingroup$ @Henno: I think that you should have seen the notation $\aleph$ for the cardinality of the reals already. It's been used quite extensively in tags you seem to be participating in. $\endgroup$ – Asaf Karagila Aug 22 at 8:55
  • $\begingroup$ The title does not exactly fit the question.: are you interested in the number of topologies up to isomorphism, or just the number of topologies ? (The former question may be harder to answer, although I suspect it's the same number) $\endgroup$ – Maxime Ramzi Aug 22 at 10:00
  • $\begingroup$ @AsafKaragila I think I’ve seen it before, but it’s old-fashioned. Most of the modern books I have use $\mathfrak{c}$. $\endgroup$ – Henno Brandsma Aug 22 at 14:32
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There are at most $\kappa:=2^{2^{\aleph_0}}$ topologies on $A$. Also, there are that many non-equivalent free ultrafilters on $A$, and each of those gives rise that a mutually non-homeomorphic Hausdorff topology on $A$ (by using it as the set of neighbourhoods of a unique non-isolated point). So there are $\kappa$ many Hausdorff topologies on $A$ (all of size of the reals). The answer is thus the size of the power set of $\Bbb R$.

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