Here is the question: Let $X$ be a set with the discrete topology and $\beta X$ be its Stone-Čech compactification.

Let $\{{x_n}\}$ be a sequence in $X$ and suppose it converges in $\beta X$, show that $\{{x_n}\}$ is eventually constant.

My attempt: I know that if $\{{x_n}\}$ converges in $X$, then it must be eventually constant since $X$ has the discrete topology. However I am unsure in the case where the sequence in convergent in $\beta X$ instead.

Let $(\beta X,j)$ be the compactification of $X$. Then the sequence $\{j(x_n)\}$ converges in $\beta X$. Let $b$ be one such limit point.

I thought of using the extension property of the stone-cech compactification how have very little idea on how to begin. Any hints would be greatly appreciated.

  • $\begingroup$ I don't know if this is true. What if $X = \mathbb{N}$ and $x_n = n$? $\endgroup$ Mar 20 '18 at 2:49
  • $\begingroup$ The question requires that the sequence to be convergent in its Stone-Čech compactification. In this case the sequence would not be convergent in $\beta \mathbb{N}$ (correct me if im wrong) via the injective map $j$. $\endgroup$
    – Soby
    Mar 20 '18 at 2:52
  • $\begingroup$ Sorry I was thinking of the one point compactification? Do these two coincide for $\mathbb{N}$? $\endgroup$ Mar 20 '18 at 2:55
  • $\begingroup$ The thing you're trying to prove is definitely false whenever $\beta X$ is metrizable, since every sequence has a convergent subsequence. $\endgroup$ Mar 20 '18 at 3:56
  • $\begingroup$ I have managed to show that $\beta X$ is indeed not metrizable using the metrization lemma in a subsequent question. $\endgroup$
    – Soby
    Mar 20 '18 at 4:07

Suppose $x_n$ is a sequence in $X$ (seen as a sequence in $\beta X$; I'll ignore the $j$ embedding map and identify $x \in X$ with its copy in $\beta X$) that converges to some $p \in \beta X$ where $p \notin X$.

We could have $x_n =x_m$ for $n \neq m$, but we can assume that all terms in the sequence are distinct (by going to a subsequence, which will have the same limit), and there cannot be finitely many values $x_n$ as this would mean the sequence is eventually constant and can only converge to that constant (in $X$).

Now define a function $f: X \to [0,1]$ by: $f(x) = 0$ if $x = x_{2m}$ for some $m$, and $f(x)$ = 1 if $x = x_{2m+1}$ for some $m$, and $f(x)= 0$ (or any value you like) if $x$ is not a value of the sequence, so not of the form $x_n$ for some $n$. This $f$ is continuous, because all functions defined on a discrete space $X$ are continuous. As such, the universal property of the Cech-Stone tells us there is a (unique) continuous extension $\beta f: \beta X \to [0,1]$.

But now note that by continuity $$\beta f(p) = \beta f(\lim_n x_n) = \lim_n \beta f(x_n)$$ and $$\lim_n \beta f(x_{2n}) = \lim_n f(x_{2n}) = 0$$ while $$\lim_n \beta f(x_{2n+1}) = \lim_n f(x_{2n+1}) = 1$$ showing that $\beta f(p)$ should be both equal to $0$ and $1$, a clear contradiction.

So no such sequence can exist.

  • $\begingroup$ My answer is using a corollary of the properties of extensions of functions. As such, mine is not as "basic". $\endgroup$ Mar 20 '18 at 22:58

In General Topology by R. Engelking, in the chapter "Compact Spaces" there are many useful results about compactifications, and much detail about $\beta \Bbb N$ with the discrete topology on $\Bbb N.$

Observe that the 1-point compactification $\alpha Y$ of an infinite discrete space $Y$ is NOT $\beta Y,$ because in the partial order of compactifications of $Y$ it ranks strictly below any 2-point compactification.

Supposing WLOG that $id_X:X\to \beta X$ is the maximal compactification of $X,$ we have (Engelking 3.6.8): If $X$ is a normal space and $Y$ is a closed subset of $X$ then $id_Y:Y\to Cl_{\beta X}(Y)$ is the maximal compactification of $Y.$

So if $X$ is an infinite discrete space and $Y=\{x_n:n\in \Bbb N\}$ is an infinite subset of $X$ then $Cl_{\beta X}Y =\beta Y,$ which cannot be equal to $Y\cup Z$ for any $Z\subset X$ (because $Y\cup Z$ is not compact), and cannot be equal to $Y\cup \{p\}$ for any $p\in \beta X\setminus X,$ otherwise it would be equal to $\alpha Y.$

  • $\begingroup$ I found the section in Engelking on the Wallman Extension $wY$ to be very helpful for understanding $\beta Y$ for any discrete $Y,$ especially when $Y$ is countable. If $Y$ is a normal space then $wY=\beta Y.$ $\endgroup$ Mar 20 '18 at 22:50

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