# Openness and Closedness in Metric Spaces

Let $X$ be a metric space. Furthermore, let $E$ be an open subset of $X$. Then, the complement of $E$, or all members of $X$ that are not in $E$, is closed, or contains all of its limit points. I understand this to be true locally around $E$.

However, why is this true when taking into account $X$ entirely. For instance, could there not exist a limit point of $X$ which is not a limit point of $E$? What if there is a point not in $E$ "distant" from $E$ which is a limit point of $X$ but not in $X$? Then, $E$ would still be open, but its complement would not be closed.

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In order to be closed, the complement $E^c$ needs to contain all limits of convergent sequences in $E^c$, not all limits of convergent sequences in $E$. Does that help? (Your remarks about points outside of $X$ puzzle me: openness and closedness are relative to $X$; even if there are any points outside of $X$ (there need not be in your setup), then you should ignore them in this context.) –  Pete L. Clark Jan 1 '12 at 16:45
"a limit point of $X$ but not in $X$" - if $X$ is the metric space you are working in, it does not make sense to talk about things "outside" of $X$, $X$ is all you have. –  Santiago Canez Jan 1 '12 at 16:45
@Santiago: I think the question should be rephrased to take what you point out into account! –  Jon Bannon Jan 1 '12 at 20:49

Subsets being open or closed would very much depend on the space. The real numbers are open, and closed, in the space $\mathbb R$; in the space $\mathbb C$, however, it is a closed set which is not open.

So the a set $E$ which is open in $X$ is such set that $X$ can recognize all those not in $E$ as a closed set.

Consider the set $(0,1)$ which is open in $\mathbb R$, and also in $\mathbb Q$. However $\dfrac{1}{\sqrt 2}$ is not a rational number, so it is a limit point of $(0,1)\cap\mathbb Q$ which $\mathbb Q$ does not know about, and therefore does not care for.

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What do you mean by the real numbers are open and closed? Because I am aware that $\mathbb R$ is connected. –  Hassan Muhammad Jan 1 '12 at 18:04
@Hassan: If $X$ is a topological space then in itself $X$ is both open and closed. In particular, the topological space of the real numbers thinks of itself as both an open set as well as a closed set. The complex numbers, however, do not agree that it is open but only that it is closed. –  Asaf Karagila Jan 1 '12 at 19:33

The complement of any open subset of $X$ is a closed subset of $X$. This is often taken to be the definition of a closed subset. Any limit point of $X$ is a point of $X$ (since $X$ is the essentially the entire universe you are working in), so if a limit point is not in $E$ then it must be in its complement.

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