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To demonstrate that a set is open, you can show that any element of it is an interior point. That is, for any element $x$ of an open set $S$ there exists some ball of center $x$ and positive radius. Likewise, if you find that all elements of a set are interior points, you can figure that the set is open.

What property can be used to show that a set is closed? By a definition of closed sets you could show that its complement is open, but that method involves elements outside of the set.

Are there any properties specific to elements of closed sets?

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All convergent sequences in a closed set have their limits inside the closed set –  Amr Dec 29 '12 at 16:22
    
If the set contains all its limit points, then it is closed. –  Ram Dec 29 '12 at 16:22
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up vote 5 down vote accepted

You can’t limit yourself entirely to points of the closed set, because whether a set is closed depends on how it interacts with the surrounding space. For example, $\Bbb Q\times\{0\}$ is a closed subset of $\Bbb Q^2$ but not of $\Bbb R^2$. You don’t, however, have to look just at the complement.

Let $X$ be a topological space and $F\subseteq X$. Then the following are equivalent:

  • $F$ is closed in $X$.
  • If $\nu$ is a net in $F$ that converges to some $x\in X$, then $x\in F$.
  • If $\mathscr{F}$ is a filter on $F$ that converges to $x\in X$, then $x\in F$.

Note that both the net and the filter versions deal specifically with points of $F$, though they (necessarily) also refer to points that may be outside $F$, namely, the limit points of the nets or filters.

This PDF is a good introduction to nets and filters in topology.

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The explanation in PDF is very very clear!! –  007resu Dec 29 '12 at 19:25
    
@user1710036: Yes, I like that PDF quite a lot. –  Brian M. Scott Dec 29 '12 at 19:26
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Well, point-set topology is hardly my métier, but it seems to me that when one speaks of a set being closed, one always has in mind its setting within a larger space. One says (or ought to say), “$K$ is closed in $X$”, because the property of being closed is not intrinsic.

Once you understand that, you see that the property of being open is not intrinsic, either. Here's an example. In the space $X=[0,1]$, that is, $X$ is the closed unit interval, the subset $U=\langle1/2,0]$, that is the half-open interval from $1/2$ to $1$, is open. But it is not open in $\mathbb R$. About the point $1\in X$, the “$X$-ball” of radius $1/4$ is wholly contained in $U$, because the $X$-ball of radius $1/4$ contains no points to the right of $1$.

There is a far deeper result that says that an open subset of $\mathbb R^n$, when mapped continuously and in one-to-one fashion into another $\mathbb R^n$ (same dimension), has image that is necessarily open. This is “Invariance of Domain”, which you can google, and it does say that in this restricted sense, openness in $\mathbb R^n$ is intrinsic. But as I say, that is a deep theorem.

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There is no "intrinsic" property of closedness for sets or spaces $A$. Closedness is not a property of sets or of topological spaces, but a property of pairs $(A,X)$ where $X$ is a topological space and $A$ is an arbitrary subset of $X$. The set $A$ is closed in $X$ if the properties mentioned in Brian Scott's answer hold.

Given an arbitrary subset $A$ of some topological space $X$ this set may be closed in $X$ or not. Assume $A$ is not closed (and there are very strange subsets of $X$). Nevertheless the set $A$ inherits topological information from $X$, and you can consider $A$ as a topological subspace of $X$, equipped with the so-called relative topology. Looking at the pair $(A,A)$, the full relative space $A$ is trivially a closed set in $A$, even though $A$ may be the strangest subset of $X$ you can think of.

Contrasting the above, compactness is an "unary" property of topological spaces and of their subsets, the latter considered as spaces in their own right, using the relative topology. I won't go into this here.

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