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?
Best Answer
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 $X$.