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Take the discrete space $2=\{0,1\}$ and form the infinite product $2^X$ for some infinite index set $X$. The Hewitt-Marczewski-Pondiczery (HMP) theorem (Engelking 2.3.15 for example) says in this case:

Theorem (HMP): If $|X|\le2^\kappa$, then $d(2^X)<=\kappa$.

where $d(Z)$ denotes the density of a space (smallest cardinality of a dense subset). In short: $$d(2^{2^\kappa})\le\kappa$$

HMP only gives an inequality and I am interested to see if we can pinpoint the exact density value depending on the cardinality of $X$. (Assume ZFC.)

The following results seem helpful in that respect. All the cardinals below are infinite cardinals.

Fact 1: $\kappa_1\le\kappa_2$ implies $d(2^{\kappa_1})\le d(2^{\kappa_2})$

(projecting a dense set in $2^{\kappa_2}$ onto $\kappa_1$ chosen coordinates gives a dense set in the smaller space.)

Fact 2: If $d(2^X)=\lambda$, then $|X|\le 2^\lambda$

(This gives a bound for the index set based on the density. See here and https://dantopology.wordpress.com/2009/11/06/product-of-separable-spaces/)

Now based on the results above I am looking at a few cases and wondering if that can be improved. Also could the general case depend on the particular set theory assumptions one could make?

Example 1: $\aleph_0\le\kappa\le 2^{\aleph_0}=\beth_1=\mathfrak{c}$

In this case $d(2^{2^{\aleph_0}})\le\aleph_0$ by HMP and $d(2^{\aleph_0})=\aleph_0$ (Cantor set), so $d(2^\kappa)=\aleph_0$ by Fact 1.

Example 2: $\beth_1<\kappa\le\beth_2=2^{\mathfrak{c}}$

Focusing on the case $\kappa=\beth_2$, let $d(2^{\beth_2})=\lambda$. By HMP, $\lambda=d(2^{2^{\beth_1}})\le\beth_1$. Combining this with Fact 2 we get $\beth_2\le 2^\lambda\le 2^{\beth_1}=\beth_2$. So $\lambda\le\beth_1$ and $2^\lambda=\beth_2$. Can anything more precise ($\lambda=\beth_1$ ?) be deduced in this case about $\lambda$? And anything for other $\kappa$ in this range?

Example 3: $\kappa=\beth_\omega$

My knowledge of cardinal number theory is limited, so please correct anything if necessary. Let $\lambda=d(2^{\beth_\omega})$. If $\lambda<\beth_\omega$, then $\lambda\le\beth_n$ for some $n$. Then by Fact 2 we would have $\beth_\omega\le 2^{\beth_n}=\beth_{n+1}$, which is impossible. Therefore $\beth_\omega\le\lambda$. Also, combining Fact 1 and HMP $d(2^{\beth_\omega})\le d(2^{2^{\beth_\omega}})\le\beth_\omega$. So in this case we can conclude $$d(2^{\beth_\omega})=\beth_\omega$$

PatrickR
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    $2^\kappa\gt\kappa$ for every cardinal $\kappa$ (Cantor's theorem). In particular $2^{\beth_\omega}\gt\beth_\omega$. In fact, by definition, $$\beth_{\omega+1}=2^{\beth_\omega}$$ – bof May 30 '20 at 06:10
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    $2^{\beth_\omega}=\beth_\omega$ --- This should be $2^{\beth_\omega}=\beth_{\omega + 1}$. – Dave L. Renfro May 30 '20 at 06:10
  • I'll update Example 3. – PatrickR May 30 '20 at 07:59
  • What is $\log(\beth_\omega)$? It's by definition $\ge$ all $\beth_n, n \in \omega$, so at least $\sup {\beth_n: n \in \omega} = \beth_\omega$. As $2^{\beth_\omega} > \beth_\omega$ the density must be some $\kappa \in [\beth_\omega, \beth_{\omega+1})$. – Henno Brandsma May 30 '20 at 08:03
  • @HennoBrandsma I added an argument that computes $d(2^{\beth_\omega})$ exactly. Hope it is correct. – PatrickR May 30 '20 at 08:20
  • Actually, as you mention, $2^{\beth_\omega}\ge\beth_\omega$, so $\operatorname{log}(\beth_\omega)$ is at most $\beth_\omega$ and hence equal to $\beth_\omega$, which matches my argument. – PatrickR May 30 '20 at 08:28
  • We're agreed indeed, the case $\kappa=\beth_\omega$ is actually one of the few we can exactly compute in ZFC. – Henno Brandsma May 30 '20 at 08:32

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For cardinals $\kappa$ we define (within cardinal numbers):

$$\log(\kappa)= \min\{\alpha: 2^\alpha \ge \kappa\}$$

and Juhasz shows (in his book(let) Cardinal Functions in Topology, Thm 4.5 (2), referring to 3 papers by Engelking, Hewitt and Pondiczery resp.)

that $$d(2^\kappa) = \log(\kappa)$$

(where HMP indeed provides the upper bound). For the lower bound he (in the end, after unpacking all the arguments; he actually proves something more general in the book) ends up using that for $T_3$ spaces (like $2^\kappa$), we have $w(X) \le 2^{d(X)}$ and so

$$w(2^\kappa)=\kappa \le 2^{d(2^\kappa)}$$

where I showed the first equality here, e.g.

So you'll always have the logarithm to deal with, and this cannot always be completely determined (i.e. in ZFC) for all $\kappa$, I think.

Henno Brandsma
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    Regarding the last paragraph isn't something like $\log(\aleph_2)$ already independent? Could be $\aleph_0$ if CH fails, or $\aleph_1$ if CH holds – Alessandro Codenotti May 30 '20 at 08:12
  • @AlessandroCodenotti You're quite right. Even among beth-numbers it's not quite clear, as we don't know the gaps between them etc. – Henno Brandsma May 30 '20 at 08:28
  • @AlessandroCodenotti If you could add anything in particular about the results for beth-numbers depending on various set theory assumptions, that would be great. – PatrickR May 30 '20 at 08:32
  • The last inequality $\kappa \le 2^{d(2^\kappa)}$ is my Fact 2. Nice. Everything fits. – PatrickR May 30 '20 at 08:39
  • In the case of $\beth_\alpha$ with limit $\alpha$ don't we just have $\log(\beth_\alpha)=\beth_\alpha$? Surely $2^{\beth_\alpha}>\beth_\alpha$ so it cannot be bigger, but it cannot be smaller either because $\beth_\alpha$ is a strong limit cardinal – Alessandro Codenotti May 30 '20 at 08:45
  • Good example. How about $\log(\beth_2)$? – PatrickR May 30 '20 at 08:52
  • @PatrickR it’s undetermined in ZFC ( what aleph it is). – Henno Brandsma May 30 '20 at 09:25
  • In a model where $\aleph_1 < \aleph_2=\beth_0$ and $\beth_1 = \aleph_3$ and $2^{\aleph_1} = \aleph_3$ we have $\log(\beth_2)= \aleph_1 < \beth_1$ while under GCH $\log(\beth_2)= \beth_1$ e.g. – Henno Brandsma May 30 '20 at 09:48