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Show that $\beta\mathbb Z_+$ (The Stone Čech compactification of the positive integers) has cardinality at least as great as $I^I$ where $I=[0,1]$.

I know that I is compact Hausdorff and so $I^I$ is compact Hausdorff. I was given a hint that $I^I$ has a countable dense subset but I do not see how that helps. I have also been told that I^I is a compactification of $\mathbb Z_+$ but I don't see how, or how to prove that it is.

Thanks for the help!!!

Martin Sleziak
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Borislav
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3 Answers3

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You already have a nice hint. Use it to constuct a surjection from $\beta\mathbb Z_+$ to $I^I$.

Alex Ravsky
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  • Can I say this: Let D be a countable dense subset of I^I. Then there exists a bijection f:Z+ -> D. We can extend to f: Z+ -> I^I. Since I^I is compact we can extend f to a function g:B(Z+) -> I^I. Since the image is compact and D is in the image and dense, B(Z+) maps onto I^I by g. – Borislav Dec 11 '13 at 04:21
  • @Borislav Yes, this is OK. – Alex Ravsky Dec 11 '13 at 04:28
  • But I thought that this only works if I^I is a compactification of Z+. – Borislav Dec 11 '13 at 04:29
  • @Borislav Not necessarily, because a Stone-Čech compactification of a space $X$ has the extension property for all continuous maps of $X$, not only for its embeddings. The trick in this answer is that it seems to be harder to show that $I^I$ is separable than to show that $|\beta\mathbb Z_+|\ge 2^\frak c$. :-) In fact, $|\beta\mathbb Z_+|=2^\frak c$ because $|X|\le 2^{2^{d(X)}}$ for each Hausdorff space $X$. – Alex Ravsky Dec 11 '13 at 04:41
  • Moreover, each infinite closed subset $F$ of $|\beta\mathbb Z_+|$ should contain a copy of $|\beta\mathbb Z_+|$ and therefore $F$ should have a cardinality $2^\frak c$ (see Theorem 3.6.14 in “General Topology” by Ryszard Engelking) – Alex Ravsky Dec 11 '13 at 04:48
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Any space $X$ has at most $2^{2^{|X|}}$ ultrafilters, then, if $X$ is a Tychonoff space, $|\beta X|\leq 2^{2^{X}}$. For any regular space, one has $\operatorname{w}(X)\leq 2^{\operatorname{d}(X)}$. Thus, for any Tychonoff space $X$, $\operatorname{w}(\beta X)\leq 2^{|X|}$. Now let $\kappa \geq\omega$ be a cardinal and denote the discrete space of cardinality $\kappa$ as $D(\kappa)$. By the Hewitt-Marczewski-Pondiczery theorem, $I^{2^\kappa}$ has a dense subspace of cardinality $\kappa$; therefore, exists a continuous injection $f: D(\kappa)\to I^{2^{\kappa}}$ such that $f[D(\kappa)]$ is dense in $I^{2^{\kappa}}$. Now, extend $f$ to $\overline{f}:\beta D(\kappa)\to I^{2^\kappa}$. As $\beta D(\kappa)$ is compact, $\overline{f}$ is a continuous surjection from a compact space to a Hausdorff space and therefore $|\beta D(\kappa)|\geq |I^{2^{\kappa}}|$ and $\operatorname{w}(\beta D(\kappa))\geq \operatorname{w}(I^{2^\kappa}) = {2^{\kappa}}$.

As $|\mathbb{Z}_+| = \omega$ and $\operatorname{w}(\mathbb{Z}_+) = \omega$, one has $|\beta \mathbb{Z}_+| = 2^\mathfrak{c}$ and $\operatorname{w}(\beta\mathbb{Z}_+) = \mathfrak{c}$.

Paulo Henrique
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HINT:

Let $X$ be any non-empty set, $\tau_d$ the discrete topology on $X$, then for every topology $\tau$, and every function $f\colon X\to X$, $f$ is continuous as a function from $(X,\tau_d)$ to $(X,\tau)$.

Use the fact that $I^I$ has a countable dense set to conclude there is a continuous surjection from $\beta\Bbb Z_+$ onto $I^I$.

Asaf Karagila
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