Collection of continuum subsets

[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

<br /> X=\big\{1-\frac{1}{n}\;\big|\; n\in\{1,2,3,\ldots\}\big\}\;\cup\;\{1\}<br />

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}?

 

[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.

 

[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...

 

[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.

 

[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.

 

[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).