The following theorem in Rudin says:
If p is a limit points of a set E, then every neighborhood of p contains infinitely many points of E.
My question is, can we go as far as to say that the neighborhood contains an uncountably infinite number of points?
Defining x to be a limit point of the set E.
Take the open ball, B = { p | d(x,p) < 1 }
Then we have our starting distance value of 1. We then take each value in the interval [0,1] and take a smaller open ball around x that must intersect E (so 1, .99999, … 0 ). Each new ball we pick a unique unchosen point in the open ball, using the fact that x is a limit point to guarantee we can find a point in E. Hence each point from E is associated with a point in the uncountable interval in [0,1] (namely the distance value that produced our choice of that point).
I believe the above to be wrong, even if a better "starting value" instead of 1 is chosen (for instance a starting value that doesn't completely cover the set E). I'm thinking this is due to this mapping i've defined is not how uncountable sets are defined.
I'm hoping someone can point out why the above is incorrect.
Thanks!
Best Answer
"Each new ball we pick a unique unchosen point"
Yes, the problem is that no such "unchosen" point may exist! For example, considering the set $E = \{\frac 1n : n \in \mathbb N\} \cup \{0\}$. Pick an uncountable set $S \subset [0,1]$. Suppose that for each $ \epsilon \in S$, there is a unique point $x_\epsilon \in E$ such that $x_\epsilon$ belongs to $B(0,\epsilon)$ but not to any $B(0,\delta)$ for $\delta < \epsilon, \delta \in S$. Then, the above association provides a $1-1$ map from $S \to E$, given by mapping $\epsilon$ to the point $x_\epsilon$.This contradicts the fact that $E$ is countable.
Therefore, at least existentially, we can show that this map is not injective. In other words, after some time, you will definitely run out of points to choose for a given $\epsilon$, and will have to choose another point.
(Please use whatever definition of limit point you have to verify that $0$ is a limit point of $E$ as well)
Also, try to think about conditions which you can impose so that indeed any such neighbourhood will contain uncountably many points. Do you think an interior point (which is also a limit point) will have this property (in $\mathbb R$)?