Proving sets have the same cardinality

[Baron]

Prove that |(0,1)| = | N U (0,1) |
where N is the set of natural numbers and U stands for union

Let A = (0,1) and B = N U (0,1) = N U A

So to prove sets have the same cardinality, I have to show there exists a bijection from A to B. I can't think of a function and it seems difficult to think of one as the function has to be piecewise. (I'm thinking it might be easier to construct a bijection from B -> A)

My second approach was to use theorems I already know to prove the sets have the same cardinality. I know N is countable and A= (0,1) is uncountable. I also know the union of two countable sets are countable but I don't see how that is helpful.

So how do I prove this?

 

[lookagain]

Prove that |(0,1)| = | N U (0,1) |
where N is the set of natural numbers and U stands for union

Let A = (0,1) and B = N U (0,1) = N U A

So to prove sets have the same cardinality, I have to show there exists a bijection from A to B.
I can't think of a function and it seems difficult to think of one as the function has to be piecewise.
(I'm thinking it might be easier to construct a bijection from B -> A)

My second approach was to use theorems I already know to prove the sets have the same cardinality.
I know N is countable and A= (0,1) is uncountable. I also know the union of two countable sets are
countable but I don't see how that is helpful.

So how do I prove this?

Baron, you may find some partial help here:

http://www.people.vcu.edu/~rhammack/BookOfProof/Cardinality.pdf

bottom of page 218 with figure 13.1 and all (or almost all of) page 219

 

[Baron]

Let A = (0,1) and B = N U (0,1)
(R stands for real numbers throughout this post)

So I basically have to show there is a bijection from A -> R and from B -> R?

Let h: A -> R and h(x) = tan(pi *(x-1/2)).
h(x) is a bijection from A -> R

But how do I show there is a bijection from B -> R? I tried defining a piecewise function that defines N and (0,1) separately but any function I define is either not injective or surjective. I can't even sketch how the function is suppose to look like.

 

[pka]

Prove that |(0,1)| = | N U (0,1) |
where N is the set of natural numbers and U stands for union
Let A = (0,1) and B = N U (0,1) = N U A

We know that \(\displaystyle A\subset B\) therefore \(\displaystyle |A|\le |B|\).

Now define \(\displaystyle f:B \to A\) in the following way:
If \(\displaystyle x\in N\) then \(\displaystyle f(x)=(2x+1)^{-1}\); if \(\displaystyle x=n^{-1}\) for some \(\displaystyle n\in N\) then \(\displaystyle f(x)=(2n)^{-1}\); elsewise \(\displaystyle f(x)=x\)

Once you show that \(\displaystyle f\) is an injection you know \(\displaystyle |B|\le |A|\).

BTW. that trick is a shift of countable sets.

 

[Baron]

How is f(x) a function?

x = 1 is a natural so f(1) = 1/(2x+1) = 1/3
x = 1/n where n = 1 and 1 is a natural so f(1) = 1/(2*1/n) = 1/2

As f(x) maps 1 to two different points how can f(x) be a function?

 

[pka]

How is f(x) a function?
x = 1 is a natural so f(1) = 1/(2x+1) = 1/3
x = 1/n where n = 1 and 1 is a natural so f(1) = 1/(2*1/n) = 1/2
As f(x) maps 1 to two different points how can f(x) be a function?

The standard understanding is that if we have already defined the function for 1, we would not do it again.
If you are uncomfortable with that, then add n>1 to the second condition.

 

[Baron]

So I got f(x) is injective (which was what we wanted to prove), but for curiosity, f(x) is not surjective right?

As f(x) never maps any number to 1/2.

 

[pka]

So I got f(x) is injective (which was what we wanted to prove), but for curiosity, f(x) is not surjective right? As f(x) never maps any number to 1/2.

You are correct it is not surjective. That is not necessary.

Use \(\displaystyle \|A\|\) to stand for the cardinally of set \(\displaystyle A\).

Then \(\displaystyle \left\| A \right\| \le \left\| B \right\|\) if and only if there is an injection \(\displaystyle f: A\to B\).