Showing that the the closure of a closure is just closure

[Mr Davis 97]

Homework Statement

Let $M$ be a metric space. Prove that $\overline{\overline{S}} = \overline{S}$ for $S\subseteq M$.

Homework Equations

The Attempt at a Solution

First we know that $\overline{S} \subseteq \overline{\overline{S}}$ is true (just take this for granted, since I know how to do it and it's the other direction that I am interested in).

Second, we want to show that $\overline{\overline{S}} \subseteq \overline{S}$. So let $x$ be a closure point for $\overline{S}$. So there exists a sequence $\{x_n\} \subseteq \overline{S}$ such that $\lim_{n\to\infty}x_n = x$. So by the definition of convergence there exists $N_1$ such that $d(x_{N_1},x) < \epsilon /2$, where $\epsilon$ is an arbitrary positive real number. Note that in particular that $x_{N_1}\in \overline{S}$, so there exists a sequence $\{y_m \}\subseteq S$ such that $\lim_{m\to\infty}y_m = x_{N_1}$. So by the definition of convergence there exists $N_2\in \mathbb{N}$ such that $m\ge N_2$ implies $d(y_m,x_{N_1}) < \epsilon /2$. So if $m \ge N_2$ then $d(y_m,x) \le d(y_m,x_{N_1}) + d(x_{N_1},x) < \epsilon /2 + \epsilon /2 = \epsilon$. Hence $\lim_{m\to\infty}y_m = x$, and so that $x$ is a limit point for $S$.

 

[RPinPA]

I can't see any flaw. Looks solid. Was that your question?

 

[fresh_42]

Homework Statement

Let $M$ be a metric space. Prove that $\overline{\overline{S}} = \overline{S}$ for $S\subseteq M$.

Homework Equations

The Attempt at a Solution

First we know that $\overline{S} \subseteq \overline{\overline{S}}$ is true (just take this for granted, since I know how to do it and it's the other direction that I am interested in).

Second, we want to show that $\overline{\overline{S}} \subseteq \overline{S}$. So let $x$ be a closure point for $\overline{S}$. So there exists a sequence $\{x_n\} \subseteq \overline{S}$ such that $\lim_{n\to\infty}x_n = x$. So by the definition of convergence there exists $N_1$ such that $d(x_{N_1},x) < \epsilon /2$, where $\epsilon$ is an arbitrary positive real number. Note that in particular that $x_{N_1}\in \overline{S}$, so there exists a sequence $\{y_m \}\subseteq S$ such that $\lim_{m\to\infty} x_{N_1}$.

I think you mean $\lim_{m\to\infty}y_m=x_{N_1}$.

So by the definition of convergence there exists $N_2\in \mathbb{N}$ such that $m\ge N_2$ implies $d(y_m,x_{N_1}) < \epsilon /2$. So if $m \ge N_2$ then $d(y_m,x) \le d(y_m,x_{N_1}) + d(x_{N_1},x) < \epsilon /2 + \epsilon /2 = \epsilon$

... for $m > \max\{\,N_1,N_2\,\}.$

Hence $\lim_{m\to\infty}y_m = x$, and so that $x$ is a limit point for $S$.

 

[Mr Davis 97]

I think you mean $\lim_{m\to\infty}y_m=x_{N_1}$.

... for $m > \max\{\,N_1,N_2\,\}.$

Why does $m > \operatorname{max}\{ N_1,N_2 \}$?

 

[member 587159]

While it may be correct, it is instructive to look for a proof where you don't use the metric: the statement is true in general topological spaces.

 

[fresh_42]

Why does $m > \operatorname{max}\{ N_1,N_2 \}$?

You need $m > N_1$ for $d(x,x_{N_1}) < \varepsilon /2$ and $m > N_2$ for $d(y_m,x_{N_1}) < \varepsilon /2\,.$

I'd like to explicitly support @Math_QED's suggestion to prove it without metric.

 

[Mr Davis 97]

You need $m > N_1$ for $d(x,x_{N_1}) < \varepsilon /2$ and $m > N_2$ for $d(y_m,x_{N_1}) < \varepsilon /2\,.$

I'd like to explicitly support @Math_QED's suggestion to prove it without metric.

I will consider that.

One more question though. Why does $m> N_1$ need to be the case for $d(x,x_{N_1}) < \epsilon /2$ if the latter statement does not depend on $m$ at all?

 

[fresh_42]

I will consider that.

One more question though. Why does $m> N_1$ need to be the case for $d(x,x_{N_1}) < \epsilon /2$ if the latter statement does not depend on $m$ at all?

I haven't analyzed whether it is actually needed or not. In case of doubt, simply increase $N$. We only consider elements which are surely close, so why bother any others? If $x_{N_1}$ is close to both, $y_m$ and $x$, I just thought "take no risk", because otherwise I would had to draw some pictures and analyze whether it comes automatically or not. And then again, why bother? We can chose $N$ as large as we like. But if you want to do the work ... go ahead; I don't think it's obvious, at least not to me.