Collection of continuum subsets

1. Jul 22, 2013

jostpuur

Let's define a set (collection) $\mathcal{C}$ by the following conditions.

$X\in\mathcal{C}$ iff all following conditions hold:

1: $X\subset [0,1]$.

2: $X$ is closed.

3: If $x\in X$ and $x<1$, then there exists $x'\in X$ such that $x<x'$.

4: For all $x\in X$ there exists a $\delta_x >0$ such that $]x,x+\delta_x[\;\cap\; X=\emptyset$. (So $x$ is not "cluster point from right".)

For example, if

$$X=\big\{1-\frac{1}{n}\;\big|\; n\in\{1,2,3,\ldots\}\big\}\;\cup\;\{1\}$$

then $X\in\mathcal{C}$.

It is easy to construct all kinds of members of $\mathcal{C}$, but they all seem to be countable. My question is that does there exist an uncountable member in $\mathcal{C}$?

Last edited: Jul 22, 2013
2. Jul 22, 2013

economicsnerd

No; every element of $\mathcal C$ is countable. Consider any $X\in\mathcal C$, and fix a function $\delta:X\to(0,\infty)$ as guaranteed by (4). For any $n \in \mathbb N,$ let
$$X_n:= \left\{x\in X: \enspace \delta_x \geq \frac 1 n\right\}.$$
As $(x, x+\delta_x)\cap(y,y+\delta_y)=\emptyset$ for any distinct $x,y\in X_n,$ it follows (from a pigeonhole argument) that $$1\geq \sum_{x\in X_n}\delta_x\geq\sum_{x\in X_n}\frac 1 n = \frac 1 n |X_n|,$$ i.e. $|X_n|\leq n.$ In particular, $X_n$ is finite. Then, as a countable union of finite sets, $$X = \bigcup_{n=1}^\infty X_n$$ is countable.

3. Jul 22, 2013

jostpuur

Nice proof. I was hoping that an uncoutable member would have existed, so that's why I didn't succeed in proving this myself...

4. Jul 22, 2013

jostpuur

Yes it is all clear. A collection of disjoint intervals is always countable. Now $X$ has the same cardinality as a collection of disjoint intervals. I must have been distracted by the other details.

5. Jul 24, 2013

jostpuur

I was hoping that I could define a mapping that maps all countable ordinal numbers into some set of $\mathcal{C}$. So the result was that this cannot be done. How unfortunate. But if you look at the first countable ordinal numbers, it would seem an intuitively reasonable feat.

6. Jul 24, 2013

economicsnerd

Yeah, it's weird. It seems there exists an element $X_\alpha \in \mathcal C$ which is order-isomphoric to $[0,\alpha]$ for any countable ordinal $\alpha,$ but not for any uncountable one (even the first uncountable one).

Last edited: Jul 24, 2013