Non-trivial convergent sequence in Stone-Čech compactification of $\mathbb{N}$ Why are there only trivial convergent sequences in the Stone-Čech compactification of $\mathbb{N}$?
 A: Let me give a fairly direct proof. Assume $R=(\mathcal{U}_{n})_{n}$ is a sequence of distinct ultrafilters on some set $X$. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence $(\mathcal{V}_{n})_{n}$ of $(\mathcal{U}_{n})_{n}$ such that $\{\mathcal{V}_{n}|n\in\mathbb{N}\}$ is a discrete subspace of $\beta X$. In particular, there is a sequence $(A_{n})_{n}$ of sets with $A_{n}\in\mathcal{V}_{m}$ if and only if $m=n$. If we let $B_{n}=A_{n}\setminus(A_{0}\cup...\cup A_{n-1})$, then the sequence $(B_{n})_{n}$ is pairwise disjoint and $B_{n}\in\mathcal{V}_{m}$ iff $m=n$. If we let $B=\bigcup_{n}B_{2n}$, then $B\in\mathcal{V}_{m}$ if and only if $m$ is even. In other words, if $\mathcal{B}=\{\mathcal{V}\in\beta X|B\in\mathcal{V}\}$, then $\mathcal{B}$ is a clopen set with $\mathcal{V}_{m}\in\mathcal{B}$ iff $m$ is even. Therefore, we conclude that the sequence $(\mathcal{V}_{m})_{m}$ cannot converge to any point, so the sequence $(\mathcal{U}_{n})_{n}$ cannot converge to any point either.
A: Here's an answer for sequences of arbitrary ultrafilters (I replace $\mathbf{N}$ by an arbitrary set $X$ since it holds in general). Let $\mathcal{U}_n$ be a sequence of ultrafilters converging to an ultrafilter $\mathcal{U}$. Define $A_n\subset 2^X$ as the set of subsets $B\subset X$ such that the sequence $(\mathbf{1}_{\mathcal{U}_k}(B))_{k\ge n}$ is constant. Then $A_n$ is a Boolean subalgebra of $2^X$, there are inclusions $A_n\subset A_{n+1}$ for all $n$ and $\bigcup A_n=2^X$. A result of Koppelberg and Tits ("Une propriété des
produits directs infinis de groupes finis isomorphes". CR Math. Acad. Sci. Paris, Sér. A 279 : 583-585, 1974) then implies that $A_n=2^X$ for some $n$. This means that $\mathcal{U}_k=\mathcal{U}$ for all $k\ge n$.
The Koppelberg-Tits result is now stated as "2^X" has cofinality $\neq\omega$ as a Boolean algebra. The above reference is in French; a simple generalization of the Koppelberg-Tits argument is given in Proposition 4.4 here, but some better references are hopefully available.
A: This result follows from much stronger results from general topology. These results can be found in [1].

$\mathbf{Theorem}$ Each non-discrete closed subset of $\beta X\setminus\upsilon X$ 
  contains a copy of $\beta\mathbb{N}$ (and in
  particular, its cardinality is at least $2^{2^{\aleph_{0}}}$). 

$\phantom{Here is a secret message.}$

$\mathbf{Corollary}$ If $X$ is locally compact and realcompact, then every infinite
  closed set in $\beta X\setminus X$ contains a copy of $\beta\mathbb{N}$ (thus, its cardinality is at least $2^{2^{\aleph_{0}}}$).

As a result of the above corollary, since every discrete space of non-measurable cardinality is realcompact, if $A$ is a discrete space of non-measurable cardinality, then every closed set in $\beta A\setminus A$ contains a copy of $\beta\mathbb{N}$. 
Now assume that $A$ is a set of non-measurable cardinality and $(x_{n})_{n}$ is a convergent sequence in $\beta A$ that converges to some point $x\in\beta A$. Take a subsequence $(y_{n})_{n}$ such that $\{y_{n}|n\in\mathbb{N}\}\subseteq A$ or $\{y_{n}|n\in\mathbb{N}\}\cap A=\emptyset$. If $\{y_{n}|n\in\mathbb{N}\}\subseteq A$, and $\{y_{n}|n\in\mathbb{N}\}$ takes infinitely many values, then take a subsequence $(z_{n})_{n}$ where each $z_{n}$ is distinct. Let $f:A\rightarrow[0,1]$ be a function where $z_{n}=0$ whenever $n$ is even and $z_{n}=1$ whenever $n$ is odd. Then $f$ extends to a continuous $\overline{f}:A\rightarrow[0,1]$. In particular, $\overline{f}(z_{n})\rightarrow f(x)$. This is a contradiction since $\overline{f}(z_{n})$ oscillates between $0$ and $1$ endlessly. Therefore $\{y_{n}|n\in\mathbb{N}\}$ can only take finitely many values, so $(y_{n})_{n}$ is eventually constant.
On the other hand, if $(y_{n})_{n}\subseteq\beta A\setminus A$, then since $\beta A\setminus A$ is closed, we have $x\in\beta A\setminus A$ as well. Therefore $\{y_{n}|n\in\mathbb{N}\}\cup\{x\}$ is a closed subset of $A$ of cardinality less than $2^{2^{\aleph_{0}}}$. Therefore, by the above corollary, the set $\{y_{n}|n\in\mathbb{N}\}\cup\{x\}$ is finite, so the sequence $(y_{n})_{n}$ must be eventually constant. QED
$\textbf{Added later}$ Using the notion of an $F$-space, we have other results that prove that $\beta\mathbb{N}$ has no non-trivial convergent sequence. If $X$ is a completely regular space, then a subset $Z\subseteq X$ is said to be a cozero set if there is a function $f:X\rightarrow[0,1]$ with $Z=f^{-1}(0,1]$. A subset $Z$ of a completely regular space $X$ is said to be $C^{*}$-embedded if every bounded continuous map $f:Z\rightarrow\mathbb{R}$ extends to a bounded continuous function $g:X\rightarrow\mathbb{R}$. A completely regular space is said to be an $F$-space if every cozero set is $C^{*}$-embedded. There are many characterizations of $F$-spaces and these characterizations can be found in [1]. The following result lists some of these characterizations

$\mathbf{Theorem}$. Let $X$ be a completely regular space. Then the
  following are equivalent.
  
  
*
  
*$X$ is an $F$-space.
  
*If $U,V$ are disjoint cozero sets in $X$, then there is a continuous function $f:X\rightarrow[0,1]$ with $U\subseteq f^{-1}[\{0\}]$ and $V\subseteq f^{-1}[\{1\}]$.
  
*$\beta X$ is an $F$-space.

From the above result, we conclude that if $D$ is a discrete space, then $\beta D$ is an $F$-space. The following result can be found in [2]

$\mathbf{Theorem}$ Every countable subspace of an $F$-space is
  $C^{*}$-embedded and hence, every infinite compact $F$-space contains
  a copy of $\beta\mathbb{N}$.

From the above result, we conclude that if $X$ is an $F$-space, then every sequence in $X$ is eventually constant. If $(x_{n})_{n}$ is not eventually constant, then take a subsequence $(y_{n})_{n}$ where each $y_{n}$ is distinct and where the set $Y:=\{y_{n}|n\in\mathbb{N}\}$ is discrete. Let $f:Y\rightarrow [0,1]$ be the map where $y_{n}=0$ whenever $n$ is even and $y_{n}=1$ whenever $n$ is odd. Then $f$ extends to continuous map $g:X\rightarrow[0,1]$. However, the sequence $g(y_{n})$ cannot converge to anything, so $(x_{n})_{n}$ is not convergent.  In particular, since $\beta D$ is an $F$-space for discrete $D$, there are no non-trivial convergent sequences in $\beta D$.
[1] Gillman, Leonard, and Meyer Jerison. Rings of Continuous Functions,. Princeton, NJ: Van Nostrand, 1960. 
[2] Walker, Russell C. The Stone-Cech Compactification. Berlin: Springer-Verlag, 1974.
A: The Stone-Čech compactification is compatible with finite disjoint unions: $\beta(A\sqcup B)=\beta A\sqcup \beta B$.
If $X=(x_n)_{n\in\mathbb N}$ were a convergent sequence with limit $x\in\beta\mathbb N$, then for any partition of $\mathbb N=A\sqcup B$ such that both $A\cap X$ and $B\cap X$ are infinite, we would have $x\in \beta A$ and $x\in \beta B$, a contradiction.
