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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.

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Why don't you add your proof? After all, you say it's "quick". We can point you in the right direction once we see what you've written, and can't otherwise know what might be missing. – amWhy Jun 27 '13 at 18:31
I do not have it formally written. I just am using the ideas that if c is a cluster point of A then the neighborhood can be [c-d,c+d] within A. Since A is a subset of S, use the boundaries of A to be the neighborhood of the cluster point of S [c-d2,c+d2]. Since that is in A and A is a subset of S, then c is thus a cluster point of S – user72195 Jun 27 '13 at 18:38
I'd suggest you add exactly that, but add it within the posted question itself :-) – amWhy Jun 27 '13 at 18:43

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 $S$.

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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 $(c-\epsilon,c+\epsilon)\cap S$?

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