uncountable set: limit point

**particlejohn** · July 9th 2008, 08:07 PM

Prove that an uncountable set of $\bold{R}$ must have a limit point.

If a set $S \subset \bold{R}$ and is uncountable, then $\aleph_{0} < \text{card}(S)$ . Now a point $x_0$ is a limit point of $S \subset \bold{R}$ if for every $\varepsilon > 0$ , $(x_{0}- \varepsilon, \ x_{0}+ \varepsilon) \cap S \ \backslash \{x_0 \} \neq \emptyset$ .

Suppose for contradiction that $S$ did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?

**kalagota** · July 9th 2008, 11:19 PM

Originally Posted by particlejohn

Prove that an uncountable set of $\bold{R}$ must have a limit point.

If a set $S \subset \bold{R}$ and is uncountable, then $\aleph_{0} < \text{card}(S)$ . Now a point $x_0$ is a limit point of $S \subset \bold{R}$ if for every $\varepsilon > 0$ , $(x_{0}- \varepsilon, \ x_{0}+ \varepsilon) \cap S \ \backslash \{x_0 \} \neq \emptyset$ .

Suppose for contradiction that $S$ did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?

i'll try it this way..

if $S$ does not have a limit point, then every $s\in S$ , there is an epsilon neighborhood $V_{\varepsilon,s}$ of $s$ contains finitely many points in $S$ .

Then, $S \subseteq \bigcup_{s \in S} V_{\varepsilon,s}$ is still finite.

EDIT: my mistake, sorry john..

**particlejohn** · July 10th 2008, 12:15 AM

Originally Posted by kalagota

i'll try it this way..

if $S$ does not have a limit point, then every $s\in S$ , there is an epsilon neighborhood $V_{\varepsilon,s}$ of $s$ contains finitely many points in $S$ .

Then, $S \subseteq \bigcup_{s \in S} V_{\varepsilon,s}$ is still finite.

which is the contradiction I was aiming for?

**flyingsquirrel** · July 10th 2008, 12:16 AM

Hi

Originally Posted by kalagota

if $S$ does not have a limit point, then every $s\in S$ , there is an epsilon neighborhood $V_{\varepsilon,s}$ of $s$ contains finitely many points in $S$ .

Then, $S \subseteq \bigcup_{s \in S} V_{\varepsilon,s}$ is still finite.

I don't know how to answer the original question but this solution doesn't seem right to me : take $S=[0,1]$ , for $s\in S,\, \{s\}$ is a finite set but $\bigcup_{s\in S}\{s\}=S$ is uncountable. (a union of countable sets is not necessarily countable)

**kalagota** · July 10th 2008, 02:18 AM

oh yeah! it is the intersection of an arbitrary collection of finite sets is finite..

thanks for that and sorry john for the mistakes..

EDIT: I also jumbled the concepts of countability and finity..

**Plato** · July 10th 2008, 08:57 AM

I did not try an answer before because I was unsure of what you are working with,
But here are some observations.
If $S$ has no limits points, then every point of $S$ is not a limit point of $S$ .
If $t \in S$ there is an open interval such that $V_t = \left( {t - \delta ,t + \delta } \right)\,\& \,S \cap V_t = \left\{ t \right\}$ .
Because between any two numbers there is a rational number, $\left( {\exists x_t \in Q} \right)\left[ {t \in \left( {x_t - \varepsilon ,x_t + \varepsilon } \right) \subseteq \left( {t - \delta ,t + \delta } \right)} \right]$ .
Name $\left( {x_t - \varepsilon ,x_t + \varepsilon } \right) = U_{x_t }$ . Note that $U_{x_t }$ contains exactly one point of $S$ .

How many $U_{x_t }$ could there be? (This is a modification of a proof by John Kelly.)

**particlejohn** · July 10th 2008, 09:34 AM

Originally Posted by Plato

I did not try an answer before because I was unsure of what you are working with,
But here are some observations.
If $S$ has no limits points, then every point of $S$ is not a limit point of $S$ .
If $t \in S$ there is an open interval such that $V_t = \left( {t - \delta ,t + \delta } \right)\,\& \,S \cap V_t = \left\{ t \right\}$ .
Because between any two numbers there is a rational number, $\left( {\exists x_t \in Q} \right)\left[ {t \in \left( {x_t - \varepsilon ,x_t + \varepsilon } \right) \subseteq \left( {t - \delta ,t + \delta } \right)} \right]$ .
Name $\left( {x_t - \varepsilon ,x_t + \varepsilon } \right) = U_{x_t }$ . Note that $U_{x_t }$ contains exactly one point of $S$ .

How many $U_{x_t }$ could there be? (This is a modification of a proof by John Kelly.)

There can be an infinite number of $U_{x_t}$ .

**Plato** · July 10th 2008, 09:42 AM

Originally Posted by particlejohn

There can be an infinite number of $U_{x_t}$ .

Yes, but only countably many! The rationals are countable.
But you are given that $S$ is uncountable.
Is that a contradiction?

**particlejohn** · July 10th 2008, 10:25 AM

That is a contradiction. Thanks. Just a quick question: $\left( {\exists x_t \in Q} \right)\left[ {t \in \left( {x_t - \varepsilon ,x_t + \varepsilon } \right) \subseteq \left( {t - \delta ,t + \delta } \right)} \right]$ .

Is this basically saying that there is a rational number $x_t$ , such that the neighborhood of $x_t$ is a subset of the neighborhood of the real number?

**Plato** · July 10th 2008, 10:47 AM

Originally Posted by particlejohn

Is this basically saying that there is a rational number $x_t$ , such that the neighborhood of $x_t$ is a subset of the neighborhood of the real number?

What in the world does that mean: “subset of a number”?

It says that given any point $t \in S$ there is a neighborhood, V, of $t$ that contains no other point of $S$ .
Then there is neighborhood, U, with a rational center that contains $t$ and is a subset of neighborhood, V.

**particlejohn** · July 10th 2008, 10:58 AM

Originally Posted by Plato

What in the world does that mean: “subset of a number”?

It says that given any point $t \in S$ there is a neighborhood, V, of $t$ that contains no other point of $S$ .
Then there is neighborhood, U, with a rational center that contains $t$ and is a subset of neighborhood, V.

I said subset of neighborhood.

**awkward** · July 10th 2008, 02:57 PM

Originally Posted by particlejohn

Prove that an uncountable set of $\bold{R}$ must have a limit point.

If a set $S \subset \bold{R}$ and is uncountable, then $\aleph_{0} < \text{card}(S)$ . Now a point $x_0$ is a limit point of $S \subset \bold{R}$ if for every $\varepsilon > 0$ , $(x_{0}- \varepsilon, \ x_{0}+ \varepsilon) \cap S \ \backslash \{x_0 \} \neq \emptyset$ .

Suppose for contradiction that $S$ did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?

Break the real line into countably many disjoint, bounded intervals ..., [-2,1), [-1,0), [0, 1), [1, 2), ...

At least one of these intervals must contain infinitely many members of S-- otherwise S would be the union of countably many finite sets and would be countable. So...

**kalagota** · July 10th 2008, 06:49 PM

Originally Posted by awkward

Break the real line into countably many disjoint, bounded intervals ..., [-2,1), [-1,0), [0, 1), [1, 2), ...

At least one of these intervals must contain infinitely many members of S-- otherwise S would be the union of countably many finite sets and would be countable. So...

i think, we have to distinguish the definition of limit points in topology.
the points in any interval in R are limit points. so in fact, all of those intervals are set of limit points, even the end points.

**kalagota** · July 10th 2008, 06:53 PM

I then list some candidates for S.

1. a set that contains an interval.
2. a set that contains the irrationals.
3. (both of course)
4. (and of course, R)

what are the other sets that are uncountable

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