Is a set closed if it has no accumulation points? I was wondering 

if a set $A$ has no accumulation point, is this set $A$ closed?

I think this is true, but I'm not quite sure.
Here's my thinking:
By closed set definition: A set $A$ is closed if every accumulation point of $A$ is a point of $A$. 
Since $A$ has no accumulation points, it is closed. Am I saying this right?
 A: Well, we need to be a little careful with our wording.
Consider the set $S = \{\frac{1}{n} : n \in \mathbb{N} \}$.  Certainly $0$ is an accumulation point, but $0 \not\in S$.  Therefore, $S$ has no accumulation point within $S$, but $S$ is certainly not closed relative to the larger metric space $\mathbb{R}$.  However, as cesfat was saying below, if we look at $S$ in its own right with the subspace topology induced from $\mathbb{R}$, then $S$ is closed relative to itself (since it contains no accumulation points within itself).
Bottom line: when we talk about closure, we do so relative to a given metric space.
So to answer your question, if a set $A$ is embedded in a larger metric space $X$ and $A$ has no accumulation point anywhere in $X$, then it is vacuously true that $A$ is closed in $X$.
A: The set of all accumulation points of S is the empty set and every set has the empty set as a subset. Therefore, S contains all its accumulation pointz and thus S is closed. Very simple.
A: Yes it is just a consequence of logic and has nothing to do with the properties of closed sets, which may be why you find it strange. Anyway for instance an empty set is closed.
