[SOLVED] Cardinality (easy)

**Bruno J.** · April 16th 2010, 11:13 PM

It's trivial that we can have a subset $S \subset \mathbb{R}$ such that any bounded neighbourhood of $0$ contains $\aleph_0$ elements of $S$ , and such that the complement of any bounded neighbourhood of $0$ contains finitely many elements of $S$ . Now is it possible to replace " $\aleph_0$ " by " $\aleph_1$ " and "finitely many" by " $\aleph_0$ "?

**Drexel28** · April 18th 2010, 04:04 PM

Originally Posted by Bruno J.

It's trivial that we can have a subset $S \subset \mathbb{R}$ such that any bounded neighbourhood of $0$ contains $\aleph_0$ elements of $S$ , and such that the complement of any bounded neighbourhood of $0$ contains finitely many elements of $S$ . Now is it possible to replace " $\aleph_0$ " by " $\aleph_1$ " and "finitely many" by " $\aleph_0$ "?

I don't think so.

Let $E$ be the set in question. We may assume WLOG that $0\notin E$ (since it won't affect the cardinality). $U_n=E\cap\left(\mathbb{R}-B_{\frac{1}{n}}(0)\right)$ (where $B_{\frac{1}{n}}(0)$ is the open ball of radius $\frac{1}{n}$ around $0$ ). Then, I claim that $E=\bigcup_{n=1}^{\infty} U_n$ . To see this let $x\in E$ then since $x\ne 0$ we see $d(x,0)>0$ and by the Archimedean principle there exists some $m\in\mathbb{N}$ such that $0<\frac{1}{m}<d(x,0)$ . Thus, $x\notin B_{\frac{1}{m}}\implies x\in U_m\subseteq\bigcup_{n=1}^{\infty}U_n$ and since it's trivial that $\bigcup_{n=1}^{\infty}U_n\subseteq E$ the conclusion follows. But, each $U_n$ is the intersection with $E$ of the complement of a bounded neighborhood of $0$ and thus countable. So, $E=\bigcup_{n=1}^{\infty}U_n$ is the countable union of countable sets and thus countable. But, this contradicts the assumption that $E\cap [-1,1]$ is uncountable.

**Bruno J.** · April 18th 2010, 05:22 PM

Perfect!

My own problem, hope you liked it.

**Drexel28** · April 18th 2010, 05:45 PM

Originally Posted by Bruno J.

Perfect!

My own problem, hope you liked it.

Nice problem!

P.S. I think the result (with a slight tweaking) can be extended to any first-countable Hausdorff space.

**Drexel28** · April 19th 2010, 03:04 PM

Originally Posted by Drexel28

Nice problem!

P.S. I think the result (with a slight tweaking) can be extended to any first-countable Hausdorff space.

Just an update, it does.

Theorem: Let $X$ be a $T_1$ topological space and $x\in X$ be such that $\{x\}$ has a countable neighborhood base. Then, if $E$ is a set such that $E\cap N'$ is countable for each neighborhood $N$ of $x$ then $E$ is countable.

Proof: We may assume WLOG again that $x\notin E$ . Then, if $\mathfrak{N}$ is the countable neighborhood base at $x$ then

$E=E\cap X=E\cap\left(\varnothing\right)'=E\left(E\cap\bigc ap_{N\in\mathfrak{N}}N\right)'=E\cap\left(E'\cup \bigcup_{N\in\mathfrak{N}}N'\right)$ $=E\cap\bigcup_{N\in\mathfrak{N}}N'=\bigcup_{N\in\m athfrak{N}}\left(E\cap N'\right)$

And since each $E\cap N'$ is countable and $\mathfrak{N}$ countable it follows that $E$ is the countable union of countable sets and thus countable. The conclusion follows.

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