For an uncountable subset $E$ in $[0,1]\times[0,1]$, show there is a condensation point of $E$. I'm not sure how to approach this question:
Given $E\subseteq[0,1]\times[0,1]$ and $E$ is uncountable, show that there is some $p\in [0,1]\times[0,1]$ such that $\forall \epsilon&gt;0, E\cap B(p,\epsilon)$ is uncountable.
I tried using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. I defined a set $A$ to be the collection of all open balls centered at rational points with some $\epsilon$ such that the unit square is completely covered. Surely at least one of the balls (centered at say $x$) contains uncountably many points in $E$. But, I don't know how to make this work (or even if it could work). I'm concerned about the overlap of the balls and how to find some $x$ that doesn't move around as $\epsilon$ changes. This is my introduction to analysis, so I don't have a good intuition for much of these concepts yet. Any help would be greatly appreciated!
Hint:I would divide the square into four smaller $1/2\times 1/2$ (closed) squares: one of them must contain uncountably many elements of $E$. Then, I would divide that square further into four (closed) $1/4\times 1/4$ squares, etc.
We get a chain of ever smaller squares, which (due to compactness of the original square) must have nonempty intersection. Let $p$ be a point in all those squares: it should now be an easy check that $p$ satisfied the requirement from your question.
By contradiction, assume that for any $p \in [0,1] \times [0,1]$ there exists $\epsilon_p &gt;  0$ such that $E \cap B(p, \epsilon_p)$ is countable.
Consider the union $B = \bigcup B(p,\epsilon_p)$ over all rational points $p$ in $[0,1] \times [0,1]$ (that is, points where both components are rational). Then $E \subseteq B$ since $\mathbb{Q}$ is dense in $\mathbb{R}$. But $B$ is the countable union of countable sets, so $B$ is countable and thus $E$ is countable, a contradiction.