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I seem to recall that you can say a set is closed if there exists a sequence that converges to a limit point of that set...obviously that is not correct but the idea is that you can deduce a set is closed because of the existence of some converging sequence, something along those lines. I think it was and "if and only if" theorem, so that the set being closed also gives you information about the sequences.

Does anyone know the proper theorem for this concept?

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up vote 1 down vote accepted

Let $X$ be a metric space and $A \subset X$. $A$ is closed in $X$ iff any sequence in $A$, which converges in $X$, converges in $A$.

In topological space, sequence has to be replaced by filter or net.

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Is that statement a collolary of this fact - "A point $x$ is said to be the limit point of a set $S$ if there exists a sequence $x_n$ which converges to $x$, where $x_n \in S$ and $x_n \neq x$"? – sonicboom Nov 2 '12 at 8:58
In a metric space (and more generally in every first-countable space), yes. – Seirios Nov 2 '12 at 9:19

In a general topological space $X$, a set $A$ is said to be closed if it contains all its limit points.

An equivalent and sometimes easier definition to check is the following: A set $A$ is closed in a topological space $X$ if $X \backslash A$ is open.

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I know those points, but I don't see what they have to do with sequences. It's the relationship between sequences and closed sets that I am interested in. – sonicboom Nov 2 '12 at 8:20
@sonicboom One way of defining closed sets is through sequences. Think about how to define a closed set? What exactly we want when we say a set if closed? – user17762 Nov 2 '12 at 8:22
@Marvis: In general you can’t define closed sets through sequences: having a topology determined by which sequences converge is a pretty strong property. – Brian M. Scott Nov 2 '12 at 8:24
@BrianM.Scott Ok. Is it because we need additional armor to even define what a limit point in a general topology is? (I know very little topology) I got sequences into picture since the OP wanted to understand the relationship between sequences and closed sets. – user17762 Nov 2 '12 at 8:38
No, the problem is that in general sequences don’t give enough information; you need to use nets or filters. Sequences do determine the topology in first countable spaces (spaces in which every point has a countable local base), for instance, and therefore in all metric spaces, but they’re not enough even for a reasonably nice space like $\{0,1\}^{\omega_1}$. – Brian M. Scott Nov 2 '12 at 8:42

In real analysis $A⊂R$ is closed iff $a_n∈A$ , $a_n$ goes to $a$ , $a∈A$.

= that means $x∈A^c$ (complement) iff there exist $a_n∈A : a_n$ goes to $x$.

also $A$ is closed iff $A= (A$ complement $)$

$x∈A^c$ iff there exist $a_n∈A$ : $a_n$ goes to infinity.

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Welcome to Math.SE! Please read the Help Center FAQ on how to participate in the community. Your answer is hard to follow. It might help to use $LaTeX$ markup in your post, which is enabled here by MathJax software, but my efforts to make these changes for you were frustrated by not quite understanding your meaning after the first line. – hardmath Dec 9 '14 at 21:47

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