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.
Best Answer
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.