[SOLVED] Cardinality (easy)

• Apr 16th 2010, 11:13 PM
Bruno J.
[SOLVED] Cardinality (easy)
It's trivial that we can have a subset $\displaystyle S \subset \mathbb{R}$ such that any bounded neighbourhood of $\displaystyle 0$ contains $\displaystyle \aleph_0$ elements of $\displaystyle S$, and such that the complement of any bounded neighbourhood of $\displaystyle 0$ contains finitely many elements of $\displaystyle S$. Now is it possible to replace "$\displaystyle \aleph_0$" by "$\displaystyle \aleph_1$" and "finitely many" by "$\displaystyle \aleph_0$"?
• Apr 18th 2010, 04:04 PM
Drexel28
Quote:

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

I don't think so.

Let $\displaystyle E$ be the set in question. We may assume WLOG that $\displaystyle 0\notin E$ (since it won't affect the cardinality). $\displaystyle U_n=E\cap\left(\mathbb{R}-B_{\frac{1}{n}}(0)\right)$ (where $\displaystyle B_{\frac{1}{n}}(0)$ is the open ball of radius $\displaystyle \frac{1}{n}$ around $\displaystyle 0$). Then, I claim that $\displaystyle E=\bigcup_{n=1}^{\infty} U_n$. To see this let $\displaystyle x\in E$ then since $\displaystyle x\ne 0$ we see $\displaystyle d(x,0)>0$ and by the Archimedean principle there exists some $\displaystyle m\in\mathbb{N}$ such that $\displaystyle 0<\frac{1}{m}<d(x,0)$. Thus, $\displaystyle x\notin B_{\frac{1}{m}}\implies x\in U_m\subseteq\bigcup_{n=1}^{\infty}U_n$ and since it's trivial that $\displaystyle \bigcup_{n=1}^{\infty}U_n\subseteq E$ the conclusion follows. But, each $\displaystyle U_n$ is the intersection with $\displaystyle E$ of the complement of a bounded neighborhood of $\displaystyle 0$ and thus countable. So, $\displaystyle E=\bigcup_{n=1}^{\infty}U_n$ is the countable union of countable sets and thus countable. But, this contradicts the assumption that $\displaystyle E\cap [-1,1]$ is uncountable.
• Apr 18th 2010, 05:22 PM
Bruno J.
Perfect! (Rock)

My own problem, hope you liked it.
• Apr 18th 2010, 05:45 PM
Drexel28
Quote:

Originally Posted by Bruno J.
Perfect! (Rock)

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.
• Apr 19th 2010, 03:04 PM
Drexel28
Quote:

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

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

$\displaystyle 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)$$\displaystyle =E\cap\bigcup_{N\in\mathfrak{N}}N'=\bigcup_{N\in\m athfrak{N}}\left(E\cap N'\right)$

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