Limits in topological spaces I was trying to prove that the real numbers aren't compact in the co countable topology but I'm a bit confused about how sequences work in topological spaces.
I considered $ X_n = \mathbb{R} - \{n, n+1, ... \} $ and want to show $ \mathbb{R} $ is the union of all of these, but this relies on $ (x_n) := \{n : n \epsilon  \mathbb{N}\} $ going to infinity? Not sure how to show this for a topological space.
 A: There’s no need to consider sequences at all; you just need to show that $\mathscr{X}=\{X_n:n\in\Bbb N\}$ is an open cover of $\Bbb R$ that has no finite subcover. Clearly $\Bbb R\setminus\Bbb N\subseteq X_n$ for each $n\in\Bbb N$, and for $k,n\in\Bbb N$ we have $n\in X_k$ if and only if $k>n$, so in particular $n\in X_{n+1}$, and $\mathscr{X}$ is an open cover of $\Bbb R$. But if $\mathscr{R}$ is a finite subset of $\mathscr{X}$, let $m=\max\{n\in\Bbb N:X_n\in\mathscr{R}\}$; then $m\notin\bigcup\mathscr{R}$, so $\mathscr{X}$ has no finite subcover.
You can get an even more efficient example by letting $C_n=\Bbb N\setminus\{n\}$ for each $n\in\Bbb N$ and then setting $U_n=\Bbb R\setminus C_n$: $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is an open cover of $\Bbb R$, but for each $n\in\Bbb N$ the only member of $\mathscr{U}$ that contains $n$ is $U_n$, so clearly $\mathscr{U}$ has no proper subcover, let along a finite one.
By the way, the convergence of sequences in the co-countable topology on $\Bbb R$ is extremely simple: it’s a nice (and quite easy) exercise to show that a sequence $\langle x_n:n\in\Bbb N\rangle$ converges if and only if it is eventually constant, i.e., if and only if there is an $m\in\Bbb N$ such that $x_n=x_m$ for all $n\ge m$. Of course in this case the sequence converges to $x_m$.
