# Stone-Čech compactifications and limits of sequences

I've been working on some old prelims from my university when they used to just be on point-set topology. We don't cover a couple of the topics so I've been teaching myself some of the material, one of the topics no longer covered is Stone-Čech Compactification. Which I have a somewhat tenuous understanding of, at any rate one of the questions reads:

Let $X$ be a completely regular topological space and let $\beta(X)$ denote the Stone-Čech compactification of $X$. Show that every $y \in \beta(X) \setminus X$ is a limit point of $X,$ but is not the limit of a sequence of points in $X$.

It's clear to me how to go about the first part, $X$ if considered as a subset of $\beta(X)$ is dense in $\beta(X)$. Then it follows that for every point $y \in \beta(X) \setminus X$ that every neighborhood of $y, U$ in $\beta(X)$ will touch $X$.

For the second part of the question, I must say, sadly, that I'm at a loss in general. Presumably we need to assume that we have some convergent sequence $\{x_i\}_{i \in \mathbb{N}}$ that converges to a point $y \in \beta(X) \setminus X$ and show that this is a contradiction. But, probably due to my weak understanding of the Stone-Čech Compactification I am unsure of how to go about this.

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How did you define the S-C compactification? –  Asaf Karagila Apr 27 '11 at 4:49
I'm going off Munkres' definition who defines it via closure of an embedding of $X$ into some product of $[0,1]$. Essentially taking all continuous $f_\lambda: X \rightarrow [0,1]$ and defining a function $h: X \rightarrow \prod_{\lambda \in \Lambda} [0,1]_\lambda$ by $h(x)=(f_\lambda(x))_{\lambda \in \Lambda}$ so that $\beta(x)$ is the closure of $h(X)$. –  JSchlather Apr 27 '11 at 5:06
I'm not sure how to answer this exactly (nor I have the time at the moment), the smidgen of intuition I can give you is that $\beta(X)$ is much too big for sequences, for example $|\mathbb N|=\aleph_0$ and $|\beta(\mathbb N)|=2^{2^{\aleph_0}}$. –  Asaf Karagila Apr 27 '11 at 8:11
For tha case that X is discrete countable space, it is Corollary 3.6.15 in Engelking's Genral topology. (In case you have the access to this book it might help.) See also here: thales.doa.fmph.uniba.sk/sleziak/texty/rozne/engel/engel.pdf The same result is shown here books.google.com/… –  Martin Sleziak Apr 27 '11 at 10:57
The result about $\beta\omega$ is also shown in this answer on MO. –  Martin Sleziak Jul 30 '13 at 12:06

EDIT: As pointed by Stefan H. in his comment, the solution I have suggested only works if $X$ is normal, since I am using Tietze extension theorem.

Perhaps I have overlooked something and I will be blushing, but I will give it a try. (This is my solution, I did not check the books I mentioned in the above comment. Perhaps the proofs from those books can give you a hint for a different proof.)

Let $x_n\in X$ be a sequence which converges to $x\in\beta X\setminus X$. We can assume that $x_n$'s are distinct. I will show bellow that $\{x_n; n\in\mathbb N\}$ is closed discrete subspace of $X$. But first I will show how to use this fact.

For any choice of $y_n\in[0,1]$, $n\in\mathbb N$, we can define $f(x_n)=y_n$ and extend it continuously (by Tietze's theorem) to the whole $X$. Now there exists a continuous extension $\overline f : \beta X \to [0,1]$. By continuity, the sequence $y_n=\overline f(x_n)$ converges to $\overline f(x)$. We have shown that every sequence in $[0,1]$ is convergent, a contradiction.

Now to the proof that $\{x_n; n\in\mathbb N\}$ is closed and discrete.

Since $\{x_n; n\in\mathbb N\}\cup\{x\}$ is a compact subset of $\beta X$, it is closed in $\beta X$. The intersection with $X$ is $\{x_n; n\in\mathbb N\}$ and it must be closed in $X$.

Now we consider $\{x_n; n\in\mathbb N\}$ as a subspace of $X$ and we want show that this subspace is discrete. Choose some $x_n$. By Hausdorffness, it can be separated from $x$, i.e. there exists a neighborhood $U\ni x$ such that $x_n\notin U$ and a neighborhood $V\ni x_n$ with $V\cap U=\emptyset$.

Now by convergence $U$ contains all but finitely many $x_n$'s, hence using Hausdorfness we can separate $x_n$ from the (finitely many) remaining ones.

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I have a question: How can you extend the function from the sequence to the whole $X$ if $X$ is not normal? –  Stefan Hamcke Sep 21 '13 at 10:44
@StefanH. Good point. I've edit my answer to point out that my answer works only for normal space (at least in the way it is formulated at this moment). I hope I will get back to this problem and think a little more about it when I have more time. –  Martin Sleziak Sep 21 '13 at 12:32