Is $\{0,1\}^{\mathbb{N}} \setminus \{(x_n)_n \in \{0,1\}^{\mathbb{N}} \mid x_n=1\: \text{for almost all} \: n \in \mathbb{N}\}$ compact?

Question

Is $\{0,1\}^{\mathbb{N}} \setminus \{(x_n)_n \in \{0,1\}^{\mathbb{N}} \mid x_n=1\: \text{for almost all} \: n \in \mathbb{N}\}$ compact in $\{0,1\}^{\mathbb{N}}$?

Answer 1

The set you describe is dense so certainly not closed and hence not compact. Any basic open subset only depends on finitely many coordinates and so contains points in your set: pick alternating zeroes and ones on the remaining ones.

Or: define $a_n$ to be the point that is $1$ on the first $n$ coordinates and alternating $0$ and $1$ (or just all $0$) thereafter. This sequence lies in your set and converges in the product to the all $1$ sequence which is not.

Answer 2

By Tychonoff's Theorem $\{0,1\}^{\mathbb{N}}$ is compact in the product topology. So $$\{0,1\}^{\mathbb{N}} \setminus \{(x_n)_n \in \{0,1\}^{\mathbb{N}} \mid x_n=1\: \text{for almost all} \: n \in \mathbb{N}\}$$ is compact if and only if it is closed, which it is if and only if $$X=\{(x_n)_n \in \{0,1\}^{\mathbb{N}} \mid x_n=1\: \text{for almost all} \: n \in \mathbb{N}\}$$ is open.

$X$ is open if and only if it is the union of intersections of finitely many $p_{i}^{-1}(U_{i})$ where $p_{i}$ is the projection map of the $i$th coordinate and $U_{i}$ is some open set in $\{0,1\}$. Let $i_{1},...,i_{n}\in\mathbb{N}$ be a finite set of indice and let $U_{k}\subset\{0,1\}$. Note that $\bigcap_{k=1}^{n}p_{i_{k}}^{-1}(U_{k})$ is either empty if $U_{k}=\emptyset$ for some $1\leq k\leq n$ or contains $x\in\{0,1\}^{\mathbb{N}}$ which is given by $$x(n)=\begin{cases}0 &\text{ if }n\neq i_{k}\text{ for some }k.\\ \min U_{k}&\text{ if }n=i_{k}.\end{cases}$$ Since $x\not\in X$ it follows that $X$ is not open, so $\{0,1\}^{\mathbb{N}} \setminus X$ is not closed and thus not compact.