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Let $A$ be a subset of a metric space $\Omega$. By definition, the closure of $A$ is the smallest closed set that contains $A$. How to prove that alternativelly, the closure is given by

(1): $\bar A = \{a_* \vert a_* = \lim_{n\rightarrow\infty} a_n; \forall a_n \in A\}$

i.e. that $\bar A$ is given by limits of all converging sequences in $A$.

I know how to prove that

(2): A set $F\subset\Omega$ is closed iff the limits of all convergengent sequences from $F$ are in $F$.

I feel that (1) and (2) should be related but how to prove (1)?

All the help appreciated.

EDIT: I was thinking about trying this, but it is too engineering-like in spirit, I am not sure whether it would work. Take the set of all possible converging sequences which are obtained by taking elements in $A$, and call it $S$.

$S = \{a_* \vert a_* \text{ as in Eq. (1)}\}$

Then there are two types of sequences, the ones that "saturate" i.e. where the same element starts repeating itself, call them $S_0$, and the ones that are "genuine" in the sense that all elements of the sequence are different, call it $S_*$. Thus

$S=S_0\cup S_*$

The saturating ones all represent $A$. Thus is should be that,

$S_0=A$

What is left, is the border $\partial A\equiv\bar A\setminus A$, and this border should be somehow made of genuine sequences, i.e. one should somehow prove that

$S_* = \partial A$

e.g. I presume by exploiting (2). Would a proof based on this strategy be possible?

zorank
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  • Your definition for $\partial A$ only holds iff $A\subset \Omega$ is open. – Dan Robertson May 28 '15 at 16:03
  • Thank you! This was an extremely helpful comment. Is it possible for you to say where my reasoning in the EDIT above breaks? For example, if $A$ were closed, where would I stumble? I know it might appear trivial to you, but I would be very grateful if you could point out the step at which the EDIT idea would break. – zorank May 29 '15 at 14:48
  • The definition of $\partial A$ is $\bar A\backslash \mathrm{Int}(A)$ – Dan Robertson May 29 '15 at 15:45
  • And that's it: if $Int(A)=A$ the my def is ok. I see it now. Thanks! – zorank Jun 01 '15 at 09:09

1 Answers1

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Consider the set $X=\{x\,|\,a_n\to x;\;\forall n\, a_n\in A\}$. Clearly $A\subset X$. You should be able to prove that $X\subset\Omega$ is closed. Now suppose that $Y\subset\Omega$ is closed, $Y\ne X$ and $A\subset Y\subset X$. Then there must be some $x\in X$ such that $x\ne Y$. Work from there to show that no such $Y$ exists and then you are done.

Dan Robertson
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  • $X\subset\Omega$ could be shown by using (2) in my post (I think). Then one would have to show that taking all possible sequences of elements that are already limits does not lead to anything new. Or? – zorank May 28 '15 at 15:50
  • You're exactly right about proving that $X\subset\Omega$ is closed (you only need the if and not the "only if"). For the rest you just suppose that you have $x$ as in my answer and from the definition of a closed set you should be able to construct a sequence that shows that $Y$ isn't closed. – Dan Robertson May 28 '15 at 16:01