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Let $(M, D)$ be the metric space of real sequences with comparative metric ($D(x,y) =0$ if $x=y$ and $D(x,y) =1/n$ if $x \ne y$ where $x_1=y_1,…, x_{n-1} =y_{n-1},x_n \ne y_n$ so $n$ is the first place where the sequences $x$ and $y$ differ) and $E$ subset in $M$ of all real sequences that do not contain any number $1$. Is the set $E$ open or closed?

Attempt: $bd(E)=M$, because for each sphere with center in $x$ which is the element in $M$ and radius $r> 0$ has a non-empty cross section with $E$ and $M-E$. Therefore $E$ is not open and $E$ in not closed. Am I right?

General123
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  • One pedagogical approach would be to consider $E$ as the intersection $\cap_{i \ge 1} E_i$ where $E_n$ is the set of all real sequences that do not have a $1$ at the $n$-th place. Can you notice that $(M,d)$ is the product topology/distance of $\prod_{n \ge 1}(\mathbb{R}, \tau)$ where $\tau$ cotains all subsets of $\mathbb{R}$, i.e. it is the discrete topology? Now notice that $f_n:M \to (\mathbb{R},\tau)$ given by $f_n(x) = x_n$ is continuous (it is the n-th projection) and $E_n = f^{-1}(\mathbb{R}-{1})$ so $\mathbb{R}-{1}$ is closed in the product topology, i.e. in $M$. – Marko Karbevski Apr 18 '22 at 22:45
  • @MarkoKarbevski, and what's $int(E)$, $ext(E)$ and $bd(E)$? – General123 Apr 18 '22 at 22:47
  • For the interior: Notice that in the product topology, if $U$ is a non-empty open subset of $M$ then there exist finitely many open subsets $U_1, ..., U_n$ of $(\mathbb{R}, \tau)$ such that $(\prod_{n \ge i \ge 1} U_i) \times (\prod_{i \ge n}\mathbb{R}, \tau) \subset U$. Do you think that such a subset can be contained in $E$?

    For the $ext(E)$, if $E$ is closed, what can you say about $E^c$? You can conclude $int(E^c)$ from here.

     – Marko Karbevski Apr 18 '22 at 23:01
  • We have not yet had a topology jet ... – General123 Apr 18 '22 at 23:03
  • Oh, sorry. Basically $(\mathbb{R}, \tau)$ is the same as $(\mathbb{R}, d_{\text{discrete}})$ where $d_{\text{discrete}}(x,y) = 0$ if $x=y$ and $1$ otherwise. Have you studied the countable infinite products of metric spaces? – Marko Karbevski Apr 18 '22 at 23:10
  • Uff, also no :( – General123 Apr 18 '22 at 23:23

1 Answers1

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It's open if every point is an interior point, in other words, for every point, there is some small enough $\delta >0$ such that all points within $\delta$ are in your set. That's clearly false here, as given any $x\in M$ you can make a sequence of points $y_n\notin $M that get as close to $x$ as you want, by defining ${y_n}_n=1$ and ${y_n}_k=x_k$ for all $n \neq k$. Then $d(x,y_n)=\frac 1 n$, so it can't be open.

However, if you take a convergent sequence of points $x_n \in M$, the limit sequence can't have a $1$ anywhere in it. If it did, it would contradict the sequence converging meaning the distance goes to 0, and here they would be fixed at $\frac 1 n$ where $n$ is the first appearance of 1 in the limit sequence. Thus, since it contains all the limits of points from within, it is closed.

Alan
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