0

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.

Novice
  • 1,127

3 Answers3

1

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.

José Carlos Santos
  • 427,504
  • I really appreciate your solution. Thanks for great help. – Novice Aug 31 '20 at 13:52
  • Anyway I have only one question. To show that $S$ is separable we have to check if the subset is countable and dense. I don't know from which fact or property we obtain the density. – Novice Aug 31 '20 at 14:15
  • A metric space $M$ is separable if it has a subset $D$ which is countable and dense. Asserting that $D$ is dense means that $\overline D=M$. So, if $M$ itself is countable, simply take $D=M$ and you're done. – José Carlos Santos Aug 31 '20 at 14:28
1
  1. The set containing exactly $2019$ ones is not closed. For, consider the sequence starting with $2018$ ones and then a $1$ in the $(2018+n)$th position.

$$x_n:=(1,\ldots,1,0,\ldots,0,1,0,\ldots)$$ Then $x_n\to x=(1,\ldots,1,0,\ldots)$ which has $2018$ ones, outside the set.

  1. The Cantor space $\{0,1\}^\mathbb{N}$ is separable, by taking the set of finite sequences, that is, the sequences that have a finite number of ones. This set is dense since, given any sequence $x$, one can take an initial segment as approximation. It is countable, e.g. using the binary numbers as index. $\{0,1\}^\mathbb{N}$ is compact by Tychonoff's theorem, since $\{0,1\}$ is compact.
Chrystomath
  • 10,798
1

For $n\in\Bbb N$ let $X_n=\{0,1\}$ with the discrete topology, and let $X=\prod_{n\in\Bbb N}X_n$ with the product topology; you can show without much trouble that the metric $d$ also generates the product topology, and once you’ve done that, you’ll be able to use whatever you know about product spaces. Depending on how much that is, you might be able to answer one or both parts of $(2)$ immediately.

It also helps with $(1)$. Let $D$ be the set of sequences in which $1$ occurs exactly $2019$ times. For each $x\in D$ let $F_x=\{n\in\Bbb N:x_n=1\}$, and let $$B_x=\{y\in X:y_n=1\text{ for each }n\in F\}\,;$$ $B_x$ is an open nbhd of $x$ such that $B_x\cap D=\{x\}$, so $D$ is a discrete set. It’s easy to see that $X$ has no isolated points, so $D$ cannot be open. And $D$ is not closed, either: any basic open nbhd of the zero sequence contains points of $D$. In fact you could try to prove that the closure of $D$ is the set of sequences that have at most $2019$ ones. And for separability you could consider the closure of the set of all sequences with only finitely many ones.

Brian M. Scott
  • 616,228