Question on countable sets A countable set contained in $[0,1]$ with no limit points. The question asks if such a request is possible. I think it is. Consider $\{1/n\}$. This is a countable set and it's a subset of $[0,1]$. If I wanted a countable set in [0,1] with no isolated points, this would be impossible I believe.
 A: No, this is not possible. Since $[0,1]$ is compact, by Bolzano-Weierstrass theorem, countable set of points in $[0,1]$ must have a limit point. 
A: Mimic the proof of the
Bolzano-Weierstrass theorem.
Divide the segment into two parts.
One of them must have
an infinite number of points.
Divide that segment into two parts.
One of those must have an infinite number of points.
Keep on doing this,
each time keeping the segment
with an infinite number of points.
After $n$ times,
there is a segment
of length $1/2^n$
with an infinite number of points.
The limit of this segment
is a point in the set which is 
a limit of points in the set.
A: I'll appeal to the fact that $[0,1]$ is compact; that is, every open cover has a finite subcover.
Consider the open cover
$$U=\{G:G\text{ is an open set$\\$containing at most one$\\$point of }A\}$$
If $U$ covers $[0,1]$, then it has a finite subcover $U'$, each containing at most one point of $A$. This would imply that $A$ is finite, contradicting the hypothesis that $A$ is infinite.
Can you show that $U$ covers $[0,1]$, if $A$ has no limit point? (Hint: if $A$ equals $\{1,\frac12,\frac13,\dots\}$ — which does have a limit point — then no set in $U$ would contain $0$, and $U$ wouldn't cover $[0,1]$.)
