Any uncountable subset $S$ of $\mathbb{C}$ must have a limit point .

Question

I am trying to prove this result by contradiction , if possible no point of $\mathbb{C}$ is a limit point of uncountable set $S$ then for every point of set $\mathbb{C}$ there exists a $r>0$ such that open ball centered at that point has no member of set $S$ other than that point , so , that $r$ will depend on the point taken from $\mathbb{C}$ ,i.e $B(a , r_a)$ contains no point of $S$ other than $a$ if $a$ is a point from set $S$ , and if $a$ does not belongs to set $S$ then it will contains no points from $S$, so , $\bigcup B(a ,r_a) = \mathbb{C}$ and $S$ is contained in $\bigcup B(a , r_a)$ put each ball either contain one point of $S$ or contain no point of $S$, that means $S$ is countable( how plz expain if am correct, I cannot fully sure about the reason) ????? which is a contradiction am I right , plz discuss , and plz discuss any other alternative approaches , how to think about this problem .

Answer 1

You have an uncountable number of balls $B(a,r_a)$, each of them containing at most one point of $S$. How do you know that countably many of them already cover ${\mathbb C}$?

A hint: There are countably many closed grid squares $Q_{jk}:=[j-1,j]\times[k-1,k]$. For at least one of them $S\cap Q_{jk}$ is uncountable. Now split this $Q_{jk}$ into four subsquares, etc.

Answer 2

Actually the limit point of $S$ must belong to $S$.

Suppose like you said that every point of $\mathbb{C}$ is not a limit point of $S$ , in particular any point $s$ of $S$ will not be a limit point.

Then for every $s \in S$ there is $r_s>0$ such that $B(s,r_s)\cap S =\{s\}.$

Now we take a countable subset of $\mathbb{C}$ which is dense , for example take

the $\mathbb{Q}\times\mathbb{Q}$. Then for every $s\in S$ there is $q_s \in \mathbb{Q}\times\mathbb{Q} $ such that $ q_s \in B(s,\dfrac{r_s}{2})$.

Now consider the correspondence $T:S \to \mathbb{Q}\times\mathbb{Q}$ , where $T(s) = q_s.$

If $T$ is not one to one then there exist two points $s_1 \neq s_2$ in $S$ such that $T(s_1)=T(s_2)$ then $q_{s_1}=q_{s_2}$. So ,$\ $ $q_{s_2} \in B(s_1,\dfrac{r_{s_1}}{2})\cap B(s_2,\dfrac{r_{s_2}}{2})$.

Now by applying the triangle inequality we get $|s_1-s_2| \leq |s_1-q_{s_2}| + |s_2 - q_{s_2}| < \dfrac{r_{s_1}}{2} +\dfrac{r_{s_2}}{2}$. (*)

And now you have two cases :

1)$ \ $ $r_{s_1}> r_{s_2}$ where from (*) you get that $|s_1-s_2| < r_{s_1}$ which is equivilent to $B(s_1,r_{s_1})\cap S \supseteq \{s_1,s_2\}.$

2) $\ $ $r_{s_2}> r_{s_1}$ which gives you $B(s_2,r_{s_2})\cap S \supseteq \{s_1,s_2\}.$

In any case we have a contradiction. So $T$ must be one to one which means that $S$ must be countable which gives us another contradiction . So $S$ must have at least one limit point !

Answer 3

With a bit more care you can prove the following: there exists a countable subset of $S$ such that its complement $S'$ in $S$ has the following property: every neighborhood of a point in $S'$ contains uncountably many points of $S'$.

Indeed, consider the subset $T$ of $S$ defined as follows:

$$T= \{t \in S \ | \text{ there exists neighborhood } U_t \text{ of } t, U_t \cap S \text{ countable} \}$$

Note that above we can choose $U_t$ to be an open box with rational vertices. One can see that $T$ equals $$\bigcup_{B \text{ rational},\ S\cap B \text{countable}} (S \cap B)$$

So $T$ is a countable union of countable sets, thus countable. Its complement $S'$ in $S$ has the property: every neighborhood $U$ of a point in $S'$ contains uncountably many points of $S$. Since $T= S\backslash S'$ is countable we can say "uncountably many points of $S'$ ".

Answer 4

There is a simple proof for this question. Reference to The one dimension case

As for the complex plane $\mathbb{C}$, we notice that the $\mathbb{C}$ acts like $\mathbb{R}^2$, A set in $\mathbb{C}$ is compact iff it is bounded and closed.

Use this fact and the proof for the one dimension, you can work out the proof for $\mathbb{C}$.