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.
Best Answer
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.