If $X$ is compact, then every sequence in $X$ has a convergent subsequence. The following is the text given from a textbook:
"Statement: If $X$ is compact, then every sequence in $X$ has a convergent subsequence.
This statment is false. The counter example is given by the cofinite topology on $\mathbb N$, i.e. $$\tau=\{A\subseteq \mathbb N \vert A^{c} \;\text{is finite}\}\cup\{0\}.$$Then the sequence $\{n\}$ has no convergent subsequence as is if it has, then that eventually becomes constant. Further we know that every space with co-finite topology is compact."
My doubt is that I think this proof is wrong. Because the sequence $\{n\}$ is convergent to $\mathbb N$. Consider the subsequence $\{2n+1\}$, then this subsequence converges to say, $a$, for some $a \in \mathbb N$. Consider the open set $\mathbb R - \{a_0,a_1,\cdots a_n\}, a_i\neq a $. Then the sequence $\{2n+1\}$ converges to $a$, because we can find large enough $n$ such that some point of the sub-sequence lie in the open set.
I am not very clear why the author says "then that eventually becomes constant"? Can someone please explain what am I doing wrong here or Is my explanation correct?
 A: There appears to be some wires crossed in your textbook. The cofinite topology always has the property that every sequence contains a convergent subsequence. This is simply because every open set contains all but finitely many points. So either the sequence has a finite range, in which case it contains a constant subsequence. Or the sequence has an infinite range, in which case it has an injective subsequence, which converges to every point. Indeed suppose $(a_n)_{n=0}^\infty$ is an injective sequence and $a$ is a point. Then any open neighborhood of $a$ contains all but finitely many $a_n$.
The property that every sequence has a convergent subsequence is called sequential compactness. For examples of compact spaces that are not sequentially compact, see: compactness / sequentially compact
A: We can use the fact that a metrizable space is sequentially compact iff it is compact. Then, take $\beta \mathbb Z$, and suppose it is metrizable. Then, since $\mathbb Z$ is not compact, there is an $b\in \beta \mathbb Z\setminus \mathbb Z$ and since $\mathbb Z$ is dense in $\beta \mathbb Z$, there is a sequence of integers $X=\left \{ x_i \right \}_{i\in \mathbb N}$ that converges to $b$.
Now take any two disjoint subsequences $X_1$ and $X_2$ of $X$ and note that they are closed in $\mathbb Z$. As $\mathbb Z$ inherits the metric on $\beta \mathbb Z$, it is a normal space. Therefore, there is a continuous $f:\mathbb Z\to [0,1]$ s.t.$f(X_1)=0$ and $f(X_2)=1$, which extends continuously to a $g:\beta \mathbb Z\to [0,1]$.
But $X_1$ and $X_2$ both converge to $b$ and as $g$ is continuous we must have $g(b)=g(\lim X_1)=\lim g(X_1)=\lim f(X_1)=0$ and also $g(b)=g(\lim X_2)=\lim f(X_2)=1$, which is a contradiction.
For a more straightforward, example, with $I=[0,1]$, take $X=I^I,$ which is compact in the product topology (pointwise convergence), by Tychonoff's Theorem. Define $f_n(x)=$ the $n^{th}$ digit in the binary expansion of $x,$ having chosen the one that ends in an infinite string of zeros. If $(f_{n_k})$ is a convergent subsequence, then $f_{n_k}(x)$ converges for each $x\in I.$
On the other hand, if we define $x\in I$ by setting
$$
n^{th}\ \text{digit of}\  x=
\begin{cases}
1 \ \text{if}\  n = n_k\  \text{and}\  k\  \text{is even}\\
0\ \text{otherwise}\\
\end{cases}
$$
then $(f(x_{n_k}))$ does not converge.
