Cardinality 2

**Bruno J.** · April 19th 2010, 06:14 PM

Prove or disprove :

The set of limit points of an uncountable set of real numbers is uncountable .

**Drexel28** · April 19th 2010, 07:16 PM

Originally Posted by Bruno J.

Prove or disprove :

The set of limit points of an uncountable set of real numbers is uncountable .

This is easier said in more general terms.

Lemma: If $X$ is second countable and $E\subseteq X$ is covered by $\Omega$ then $\Omega$ omits a countable subcover $\Sigma$ .

Proof:Call the countable open base $\mathfrak{B}$ . For each $e\in E$ there exists some set $\omega\in\Omega$ such that $e\in\omega$ . But, we have that there exists some $B_e\in\mathfrak{B}$ such that $e\in B_e\subseteq\omega$ . Clearly then the collection $\left\{B_e\right\}_{e\in E}$ covers $E$ and is countable. Taking an $\omega$ which contains $B_e$ for each $e\in E$ finishes the argument. $\blacksquare$

Now we can prove the problem at hand.

Let $X$ be second countable and $E\subseteq X$ uncountable, then $E\cap D(E)$ is uncountable.

Assume not. Then, $D(E)$ is countable and so $E-D(E)$ is uncountable. But, since each point $e\in E-D(E)$ is in particular not a limit point of $E$ there exists a neighborhood $N_e$ such that $E\cap N_e=\{e\}$ . Clearly the class $\left\{N_e\right\}_{e\in E-D(E)}$ is an open cover for $E-D(E)$ and so by the lemma it must admit a finite subcover $\left\{N_{e_n}\right\}_{n\in\mathbb{N}}$ . But, $E-D(E)=\bigcup_{n\in\mathbb{N}}\left(E\cap N_{e_n}\right)$ but since each $E\cap N_{e_n}$ is a singleton it follows that $E-D(E)$ is countable Contradiction.

Your problem is a corollary since $\mathbb{Q}$ is a dense subset of $\mathbb{R}$ and every separable metric space is second countable.

**Bruno J.** · April 19th 2010, 07:40 PM

That doesn't work! (I thought you had it for a sec, I even wrote a post congratulating you

)

Removing the limit points from a set does not mean the resulting set has no limit points!

**Drexel28** · April 19th 2010, 08:04 PM

Originally Posted by Bruno J.

That doesn't work! (I thought you had it for a sec, I even wrote a post congratulating you

)

Removing the limit points from a set does not mean the resulting set has no limit points!

That wasn't the screw up, but my head was somewhere else. See edited post.

**Drexel28** · April 19th 2010, 08:11 PM

Originally Posted by Bruno J.

.

Derived set of $E$ . The set of limit points.

**Bruno J.** · April 19th 2010, 08:17 PM

Originally Posted by Drexel28

Derived set of $E$ . The set of limit points.

Figured it out just too late

Thanks for the proof, I'll give it the attention which it deserves when I'm not dead tired! Exams begin tomorrow.

(After all, what you had written before worked also. Why did you remove it? Every point of $E-D(E)$ is an isolated point, and so $E-D(E)$ has the cardinality of a collection of disjoint open sets, which is countable.)

**Drexel28** · April 19th 2010, 08:20 PM

Originally Posted by Bruno J.

Figured it out just too late

Thanks for the proof, I'll give it the attention which it deserves when I'm not dead tired! Exams begin tomorrow.

(After all, what you had written before worked also. Why did you remove it? Every point of $E-D(E)$ is an isolated point, and so $E-D(E)$ has the cardinality of a collection of disjoint open sets, which is countable.)

Because the needn't be disjoint. I was being stupid. Clearly they can intersect but they cannot contain each other. For example $(0,2)(1,3)$ could be two of the sets if [math[\frac{1}{2},2\in E[/tex] since they're intersection with $E$ could easily be a singleton but they aren't disjoint.

**Bruno J.** · April 19th 2010, 08:26 PM

Right - but taking them small enough?

Haha I really have to hit the bunk. 'Nite!

**Drexel28** · April 19th 2010, 09:06 PM

Originally Posted by Bruno J.

Right - but taking them small enough?

Haha I really have to hit the bunk. 'Nite!

I'm not sure right this second but I'm really tired. I'll look again tomorrow. It is immediate how to finish the argument assuming that $E-D(E)$ is closed.

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