# Thread: uncountable set: limit point

1. ## uncountable set: limit point

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?

2. 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..

3. 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?

4. 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)

5. 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..

6. 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.)

7. 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 $U_{x_t}$.

8. 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.

9. 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?

10. 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.

11. 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.

12. 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...

13. 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.

14. 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

,

,

,

,

,

,

,

,

### every countable subset of r has a limit point

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