# Closed set in a metric space

Does a closed set in a metric space contain only it's limit points?

My thinking: A set has interior and boundary points. Interior points will be limit points as every neighbourhood around them will contain at least one point of the set. Boundary points of a closed set are its limit points. A closed set contains all its boundary points. Hence all points in a closed set are limit points.

Is this argument incorrect in any way?

• An interior point need not be a limit point, it can very well be an isolated point. Commented Sep 15, 2016 at 9:32

Let $\langle X,d\rangle$ be a metric space, and let $C$ be a closed set in $X$. Then $X\setminus C$ is an open set, so for each $x\in X\setminus C$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap C=\varnothing$. Thus, no point of $X\setminus C$ is a limit point of $C$. This shows that $C$ contains all of its limit points.

Conversely, suppose that $A\subseteq X$, and $A$ contains all of its limit points. Then if $x\in X\setminus A$, $x$ is not a limit point of $A$, so there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap A=\varnothing$. Let $U=X\setminus A$. What we just showed can be rephrased as follows: if $x\in U$, there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\subseteq U$. By definition, then, $U$ is an open set, and $A=X\setminus U$ must be closed.

In short, we’re proved that a subset of $X$ is closed if and only if it contains all of its limit points.

However, a subset of $X$ need not have any limit points, so you can’t make the stronger statement that a subset of $X$ is closed if and only if it is the set of its limit points. In $\Bbb R$, for example, every finite set is closed, but a finite set has no limit points. For a more interesting example, the set $\Bbb Z$ of integers is closed and has no limit points.

• Can you tell me where I'm going wrong if I try to prove that the complement of an open set is closed: Consider an open set $X$. Let it's complement be $Y$. For all points in $X$, there exists an open ball of a certain radius so that it's fully in $X$ (as it's open). Thus $Y$ contains all such points which don't satisfy the above i.e. all open balls around them contain at least one element not in $X$ i.e. in $Y$. Thus all elements of $Y$ become limit points (as all open balls contain at least one element in $Y$). But it's not universally true that all points in a closed set are limit points Commented Sep 17, 2016 at 11:57
• @Anant: A point of $Y$ might be an isolated point of $Y$, in which case it would be the only point of $Y$ is some open ball around it. For instance, suppose that your space is $[0,1]$, and let $X=(0,1)$. Then $Y=\{0,1\}$, and the open ball of radius $1$ centred at $0$ is $\{t\in[0,1]:|t-0|<1\}=[0,1)$, whose intersection with $Y$ is $\{0\}$. Commented Sep 17, 2016 at 18:39
• Ok understood it. Commented Sep 17, 2016 at 19:55

No this is not correct. The set $\{0\}\subset \mathbb{R}$ does not contain a limit point, but is closed.

• A closed set contains all limit points right? Then how is $\{0\}$ closed? Commented Sep 15, 2016 at 10:43
• $\{0\}$ does contain all its limit points; there just are none. So this is just the statement that $\emptyset\subset \{0\}$ Commented Sep 15, 2016 at 10:44

If $X$ is equipped with the (metrizable) discrete topology then limitpoints do not exist in $X$.