Cluster Points and Subsets Let $A\subset S$. Show that if $c$ is a cluster point of $A$, then $c$ is a cluster point of $S.$ I wrote up a quick proof of this however, I feel I made it too simplistic. I do not know where to start and feel as though I am missing something in my proof.
 A: Hint
Here is the outline of the proof.
Let $A \subset S$ and assume that $c \in A$ is a cluster point of $A$. Then, [insert the conclusion from the definition of a cluster point here]
Now, since $A \subset S$, [extend (and justify) whatever conclusion about c,A you drew to c,S].
Thus, $c$ is a cluster point for $S$.
A: It’s not clear what you mean (in your comment) by using the boundaries of $A$. What if, for instance, $A=\Bbb Q$, the set of rational numbers? It’s also not true that if $c$ is a cluster point of $A$, then $c$ has a neighborhood $[c-d,c+d]$ contained in $A$. For example, $A$ might be $\left\{\frac1n:n\in\Bbb Z^+\right\}$: $c=0$ is a cluster point of $A$, but no interval around $0$ is contained in $A$. (In fact $A$ contains no non-degenerate intervals at all.)
You need to back up and focus on the definition of cluster point. (In fact it’s always a good idea when dealing with new ideas to go back to the definitions: a great many elementary facts, like this one, have proofs that are direct, straightforward applications of the relevant definitions.)
Assuming that you’re working just in $\Bbb R$, $c$ is a cluster point of $A$ if every open interval around $c$ contains a point of $A$ different from $c$. In symbols, $c$ is a cluster point of $A$ if $(u,v)\cap(A\setminus\{c\})\ne\varnothing$ whenever $u<c<v$. And we actually need only look at open intervals that are symmetric about $c$: $c$ is a cluster point of $A$ if $(c-\epsilon,c+\epsilon)\cap(A\setminus\{c\})\ne\varnothing$ for each $\epsilon>0$. From what you wrote in the comment, I suspect that you may be most familiar with this last version of the definition, so I’ll use it.
Suppose that $c$ is a cluster point of $A$. Then for each $\epsilon>0$ the set $(c-\epsilon,c+\epsilon)\cap A$ must contain at least one point different from $c$. What about the set $(c-\epsilon,c+\epsilon)\cap S$?
