# Subsets of $\mathbb{R^2}$ using the discrete metric are closed, why?

I understand why all subsets of say, $\mathbb{R^2}$ are open with respect to the discrete metric - Let $U$ be a subset of $\mathbb{R^2}$. For all $x \in U$ we can choose an 0 > r > 1 such that we will have an open ball, $B$ that only contains a single point, x itself. And this open ball will be contained entirely in $U$ as $x \in U$ so $U$ is open.

But as we are talking about the discrete metric $U$ is also closed. Closed afaik means that the set contains it's limit points. So is this correct - The limit point(s) of $B$, which only contains a single point $x$, is just that $x$. Hence it's closed?

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In a space with the discrete topology no set has any limit points, so every set vacuously contains all of its limit points. Remember, $p$ is a limit point of a set $A$ if and only if every open neighborhood of $p$ contains a point of $A$ different from $p$. But in a discrete space $\{p\}$ is an open neighborhood of $p$, and obviously it doesn’t contain any point of $A$ different from $p$ no matter what set $A$ is.
To put it a little differently, let $A$ be any set in a discrete space. In order for $A$ not to be closed, $A$ would have to have a limit point $p$ such that $p\notin A$. But $A$ has no limit points, so it certainly has no limit points that aren’t in it!