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 $\forall \epsilon>0, E\cap B(p,\epsilon)$ 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!
 A: 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.
A: $[0,1]\times [0,1]$ can be covered with $4$ closed squares (from here on out all squares will be closed) with side length $1/2$ so surely there is a square $S_1$ with side length $1/2$ that contains uncountably many points in $E$. By the same argument, their is a square $S_2 \subset S_1$ with side length $1/4$ that contains uncountably many points in $E$. Proceed inductively to get a nested sequence of squares $S_1 \supset S_2 \supset S_3\ldots $ such that each $S_n$ has side length $2^{-n}$ and contains uncountably many points in $E$.
Letting $s_n$ be the center of $S_n$ we see that the sequence $\{s_n\}_n$ is eventually contained in each $S_k$ and thus converges to an $s$. For the same reason $s$ lies in all $S_n$ squares. Any $B(s; \epsilon)$ ball centered will of course contain an $S_k$ square; in particular contain uncountably many points in $E$.
A: Let $B_Q=\{B(p,q):p\in \Bbb Q^2\land q\in \Bbb Q^+\}.$ Then $B_Q$ is a countable base (basis) for the topology on $\Bbb R^2.$
Let $C$ be the set of all $b\in B_Q$ such that $E\cap b$ is countable. Then $C$ is countable (... because $C\subseteq B_Q$...). So $\cup_{b\in C}(E\cap b)=E\cap(\bigcup C)$ is countable. So $E\setminus (\bigcup C)$ is uncountable.
Every  $p'\in E\setminus (\bigcup C)$ is an accumulation point of $E.$
Proof: Suppose not. Take $p'\in E\setminus (\bigcup C)$ and an  open set $U\subset \Bbb R^2$ such that $p'\in U$ and such that $E\cap U$ is countable. Since $B_Q$ is a base, there exists $b\in B_Q$ such that $p'\in b\subseteq U.$ But then $E\cap b$ is countable (... because $E\cap b\subseteq E\cap U$ ...) so $b\in C$ so $$p'\in E\cap b\subseteq E\cap (\bigcup C)$$ contrary to $p'\in E\setminus (\bigcup C).$
Remarks: This works verbatim for any uncountable $E\subset \Bbb R^2.$ Note that the set of members of $E$ that are $not$ accumulation points of $E$, that is, $E\cap (\bigcup C),$ is only countable. And $\bigcup C$ is the $\subseteq $-largest open set to have countable intersection with $E.$
A: By contradiction, assume that for any $p \in [0,1] \times [0,1]$ there exists $\epsilon_p >  0$ 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.
