# Showing that the the closure of a closure is just closure

## Homework Statement

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

## 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##.

Last edited:

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

fresh_42
Mentor
2021 Award

## Homework Statement

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

## 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##.

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.

Mr Davis 97
fresh_42
Mentor
2021 Award
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
Mentor
2021 Award
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.