Q:

Checking if the set is open, closed and if the space is separable in given metric space

On the set of $\{0,1\}^\mathbb{N}$ (zero-one sequences) we define a metric
$$ d(x,y)= \left\{ \begin{array}{ll}
0 & \textrm{when $x=y$}\\
\frac{1}{2^n}& \textrm{when $x\neq y$ and $n=\min\{k: x(k)\neq y(k)$}\}
\end{array} \right.$$
Having defined that I have two problems
$1.$ Is the set of sequences, which take in one exactly $2019$ times is closed/open?
$2.$ Is the space $\{0,1\}^\mathbb{N}$ separable? Is the space $\{0,1\}^\mathbb{N}$ compact?
My attempts
$1.$ For me condition of taking in one exactly $2019$ times is kind of enigmatic. I do not know if this set is open or close, but I know that to show that this set is not open we ought to find a sequence which doesn't belong to our space, but converges in metric to element from our space.
Simultaneously to show, that this set is not close we should find a sequence which belongs to our set, but its limit not. Unfortunately I don't see any examples of such sequences and limits here.
$2.$ To check if $\{0,1\}^\mathbb{N}$ is separable we should check if it contains a countable, dense subset. I think in this case such subset doesn't exist but I may be wrong. I also have to show the compactness but I think it has already been answered here Is $\{0,1\}^{\mathbb{N}}$ compact in $\mathbb{N}^{\mathbb{N}}$?.
Thanks.

A:

Let $S$ be your set. Then $S$ is not open, because if $(x_n)_{n\in\Bbb N}\in S$ and if $\varepsilon>0$, then the open ball $B_\varepsilon\bigl((x_n)_{n\in\Bbb N}\bigr)$ contains elements of $\{0,1\}^{\Bbb N}\setminus S$; for instance, take $m\in\Bbb N$ such that $2^{-m}<\varepsilon$ and than $x_m=0$ and let $(y_n)_{n\in\Bbb N}\in\{0,1\}^{\Bbb N}$ be such that$$y_n=\begin{cases}x_n&\text{ if }n\ne m\\1&\text{ if }m=n.\end{cases}$$Then$(y_n)_{n\in\Bbb N}\notin S$, and$$d\bigl((x_n)_{n\in\Bbb N},(y_n)_{n\in\Bbb N}\bigr)=2^{-m}<\varepsilon.$$
And $S$ is not closed. If it was closed, $\{0,1\}^{\Bbb N}\setminus S$ would be an open set. But the null sequence $N$ belongs to $\{0,1\}^{\Bbb N}\setminus S$ and if $\varepsilon>0$ and if $m\in\Bbb N$ is such that $2^{-m}<\varepsilon$, if you define $(x_n)_{n\in\Bbb N}$ as$$x_n=\begin{cases}1&\text{ if }n\in\{m,m+1,\ldots,m+2018\}\\0&\text{ otherwise,}\end{cases}$$then $(x_n)_{n\in\Bbb N}\in S$and $d\bigl(N,(x_n)_{n\in\Bbb N}\bigr)=2^{-m}<\varepsilon$.
Since $S$ is not a closed subset of $\{0,1\}^{\Bbb N}$ it is not compact. That other question that you have mentioned is about another distance.
Finally, since $S$ is countable, it is separable.

