Is it possible for an uncountable set to have at most a finite number of limit points? Let A be an uncountable set of real numbers.
a) Show that A has a (finite i.e. a real number) limit point.
b) Is it possible for A to have a finite set of limit points?
Prove what you state
I have proven A has a limit point and know that it is not possible for A to have at most a finite number of limit points since it is uncountable but the question says prove what you state and I am unable to come up with a rigorous explanation or proof (using introductory real analysis).
 A: Let us assume that $A$ has only finitely many limit point, $a_1, \dotsc, a_n$.
For every integer $k$, we form a set $X_k = \bigcup\limits_{i = 1}^n[a_i - 1/k, a_i + 1/k]$.
It is a simple exercise to check that the intersection $\bigcap\limits_{k = 1}^\infty X_k$ is just the set $\{a_1, \dotsc, a_n\}$.
In particular, the set $B = A \backslash \bigcap\limits_{k = 1}^\infty X_k$ is still uncountable.
But we can rewrite: $$ B = A \cap \left(\bigcap_{k = 1}^\infty X_k\right)^c = A \cap \left(\bigcup_{k = 1}^\infty X_k^c\right) = \bigcup_{k = 1}^\infty (A \cap X_k^c) = \bigcup_{k = 1}^\infty (A\backslash X_k).$$
Let $B_k$ denote the set $A\backslash X_k$. Since $B$ is the countable union of all $B_k$, the uncountability of $B$ implies that at least one $B_k$ is uncountable.
However, if some $B_k$ is uncountable, then by applying part a) of the question to $B_k$, we know that it must have a limit point. This limit point cannot be any of the $a_i$, as an interval around each of them is taken away.
Furthermore, since $B_k$ is contained in $A$, a limit point of $B_k$ is also a limit point of $A$.
This contradicts the original hypothesis that the $a_i$'s are all the limit points of $A$.
A: Let $A_n=A\cap [-n, n] $ so that $A=\bigcup\limits_{n=1}^{\infty} A_n$. If every $A_n$ is countable then so is $A$. Hence some $A_n$ is uncountable. Thus $A$ has an uncountable and bounded subset. Therefore we can assume without any loss of generality that $A$ itself is bounded. Now by Bolzano-Weierstrass theorem $A$ has a limit point.
Suppose there are only a finite number of limit points, say $c_1,c_2,\dots,c_k$ of $A$. And then for each $n$ consider the set $$B_n=A\setminus\bigcup_{i=1}^{k}A\cap(c_i-1/n,c_i+1/n)$$ Each $B_n$ is a finite set (otherwise we get another limit point). And so the union of these sets $B_n$ is countable. But by De Morgan law $$\bigcup_{n=1}^{\infty} B_n=A\setminus \bigcap_{n=1}^{\infty} \bigcup_{i=1}^{k}A\cap(c_i-1/n,c_i+1/n)=A-\{c_1,c_2,\dots, c_k\} $$ and we get an obvious contradiction as left hand side is countable and right hand side is uncountable. 
