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

 

[pka]

Hi, can someone please help me with this problem.
1. Show that the set W = {C(x) : x is in A} is countable or finite.

There is no question there.
You must tell us what \(\displaystyle A\) is and what \(\displaystyle C(x)\) means.

 

[pka]

I you edited the OP to say what A is.
But we have no way of knowing what \(\displaystyle C(x), x\in A\) means.

 

[math25]

Sorry

is open, and we want to show that

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 defining 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). We'll define a one to one function from W to
Q. Once we've done that we'll know whether or W is countable or
finite.)
thanks

 

[pka]

Sorry

is open, and we want to show that

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 defining 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}

A component is a maximally connect subset of a top-space.
So all you needed to say is that \(\displaystyle C(z)\) is the component determined by \(\displaystyle z\).
Because the rational numbers are countable, any open set in \(\displaystyle \mathbb{R}\) is the countable union of components.