Studying General Topology from Munkres, I just read about Stone-Čech compactifications for the first time, tried to solve the exercises provided by the author and I came across the following one:
"Show that β($Z^+$) has cardinality at least as great as $I^I$ where I = [0,1]."
where β(X) is the author's notation for the Stone-Čech compactification of X. My idea for proving the statement was the same as the one of the following proof that I found online:
"The compact Hausdorff space $I^I$ is a compactification of $Z^+$ since it has a countable dense subset (and is not finite). Any compactification of $Z^+$ is a quotient of the β($Z^+$). In particular, $I^I$ is a quotient of β($Z^+$) so |β($Z^+$)| ≥ |$I^I$|."
From what I understand, we need a homeomorphism h from $Z^+$ under the order topology (which is the discrete topology) to that dense subset 0f $I^I$ (call it A) under the subspace topology. There is a bijection between the two sets, but why is there a continuous one between the two topological spaces? If there was one should the image of any one point subset {z} of $Z^+$ not be the intersection of an open in set in $I^I$ with A? Does that not mean that that there is an open set C in $I^I$ which contains only one of the elements of A, h(z)? Since one point sets are closed in $I^I$ should C-{h(z)} not be an open set in $I^I$ which contains none of the elements of A?
Is it that I have not understood the concept of a Stone-Čech compactification yet? Is it not the subspace topology the one I should be considering?
Thanks for taking a look into this.
$I^I$ is separable, let $D= \{d_0,d_1,d_2,\ldots\}$ be a countable dense subset of it. Then mapping $f$ sending $n$ to $d_n \in I^I$ defines a continuous map from $\mathbb{N}$ to $I^I$ (all maps on a discrete space are continuous) and then the extension property of $\beta \mathbb{N}$ guarantees that we can extend it to some continuous $\beta f: \beta{\mathbb{N}} \to I^I$.
But then $\beta f[\beta \mathbb{N}]$ is a compact (hence closed, as $I^I$ is Hausdorff) subset of $I^I$ that contains $D = \beta f[\mathbb{N}] = f[\mathbb{N}]$ and thus must equal $I^I$. This means that $|\beta \mathbb{N}| \ge |I^I|$. (Standard set theory: $|A| \le |B|$ iff there is an injection from $A$ to $B$ iff there is a surjection from $B$ to $A$).
$I^I$ is a compactification of $D$ but not of $\mathbb{N}$, as there cannot be a dense relatively discrete countable subset. So there is no homeomorphism between $\mathbb{N}$ (or $\mathbb{Z}^+$, if you like) and a dense subset of $I^I$. But by the argument from the first two paragraphs we do see that any compactification of a countable space (like $D$) is the continuous image (automatically closed hence quotient by compactness and Hausdorffness) of $\beta \mathbb{N}$, which is enough for the cardinality argument.
Your last argument as to why $I^I$ is not a compactification of $\mathbb{N}$ is indeed correct: suppose $e: \mathbb{N} \to I^I$ were an embedding, then $\{e(0)\}$ is open in $D:=e[\mathbb{N}]$, so that there is an open subset $O$ of $I^I$ such that $O \cap D = \{e(0)\}$ but then, as $I^I$ has no isolated points and is $T_1$, $O \setminus \{e(0)\}$ is an open non-empty set that misses $D$, so $D$ cannot be dense, and $e$ cannot be an embedding to make $I^I$ a compactification of $\mathbb{N}$. So the argument holds in all crowded $T_1$ spaces, in fact.