Show that the set is countable or finite.

[math25]

Hi, can someone please help me with this problem.

Let A be an open subset of the interval [0; 1].
1. Show that the set W = {C(x) : x is in A} is countable or finite.

This is what I have...

Suppose W is an infinite subset of N. Then we have f : W-> N, which
is one-to-one. By the fact that any infinite set contains a countable subset, then W has a countable subset E, since w is infinite. So we have a one-to-one map
g : N ~E -> W. By Cantor-Berstein-Schroeder's Th, W~N.

thanks

 

[Chaos2009]

I'm sorry, but what is $C \left( x \right)$?

 

[math25]

Sorry, this is what I have before problem:

Let A be an open subset of the interval [0; 1]. Our goal is to show that
A =the union In where for all n in N, In is a possibly empty open interval and for
all n and k in N, if n is not equal to k then In intersect Ik = empty set We begin by defining a relation ~ on A by the
formula
for all x; y in A; x ~ y if and only if every number between x and y is in A:
For each z in A, we let C(z) = {t in A : t ~ z}


#1 Show that the set W = {C(x) : x is in A} is countable or finite. (Hint: For this problem it's better to think of W as W = {U : there exist x in A such that U =
C(x)}. The reason is for every element U of W will be more than one number
x for which U = C(x).)

 

[Chaos2009]

Alright, well, I think it might be easier to define a mapping between these $C \left( x \right)$ and a subset of the rational numbers. Can you think of a way to assign a unique $r \in \mathbb{Q}$ to each $U \in W$?

 

[math25]

Well, the rational numbers are countable, so any open set in R is the countable union of components. But I am not sure how to write a proof...

 

[math25]

Would this proof work?

let r=1/n for n=1,2,3... and {In, n in N where In is the set corresponding to r=1/n, that is In is a maximal set having the property that is greater then or equal to 1/n for any a,b in In

Let W =Un In since W is the union of countably many countable sets, W is countable.

thanks

 

[Chaos2009]

Hmm, I'm afraid that doesn't quite work. Say $A = \left( \frac{1}{2}, \frac{3}{4} \right) \cup \left( \frac{3}{4}, 1 \right)$, then what would $I_{2}$ be?

I think it would be better to use the fact that the rational numbers are dense within the real numbers. You should be able to show that each $C \left( x \right)$ is non-empty and open and thus contains at least one rational number, and since they are disjoint, these rational numbers must be unique.