Math Proof: Uncountable binary sequence and a bijection from R to R-{0}

[iceblits]

Homework Statement

Question 1:
Prove that the cardinality of R (the set of all real numbers)is the same as the cardinality of R-{0} by constructing a bijective function from R to R-{0}

Question 2: Let A be the infinite sequence of binary numbers as follows:
A={(a1,a2,a3...)|ai= 0or 1 for all i in the natural numbers}

Show that A is uncountable

Homework Equations

The Attempt at a Solution

For question 2 I think I have to use a proof similar to Cantor's diagonalization argument for proving that the set of real numbers is uncountable. I think I have to use contradiction and assume that the set is countable.

 

[micromass]

Yes, for (2) you need to do something very similar to Cantor diagonalization.

For one, you'll want to find a bijection $f:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}$.

Hint, if $x\notin \mathbb{N}$, define f(x)=x.

 

[iceblits]

I think Imay have misunderstood the second part. Isnt f(x) still f(x)=x?

 

[micromass]

I think Imay have misunderstood the second part. Isnt f(x) still f(x)=x?

You define f(x)=x in (1) for all x not in $\mathbb{N}$.

 

[iceblits]

so {(x| x =/= 1,2,3,4...)} ?

 

[micromass]

so {(x| x =/= 1,2,3,4...)} ?

Uuh, what do you mean with that??

Also note that I consider 0 to be in $\mathbb{N}$.

 

[iceblits]

gahh I am so sorry I don't understand..am I looking for a function that hits all numbers except for 0,1,2,3...?

 

[micromass]

gahh I am so sorry I don't understand..am I looking for a function that hits all numbers except for 0,1,2,3...?

No, that's not what I stated. I just said you had to define f(x)=x for $x\notin\{0,1,2,3,...\}$. You still need to define f(0), f(1), f(2), ...

But you have to end up with a bijection $f:\mathbb{R}\rightarrow \mathbb{R}\setminus \{0\}$.

 

[iceblits]

so how about f(x)={x+1 for the natural numbers and x otherwise) would that make it so that x is not in the natural numbers but f(x) exists for the natural numbers?

 

[micromass]

so how about f(x)={x+1 for the natural numbers and x otherwise) would that make it so that x is not in the natural numbers but f(x) exists for the natural numbers?

That's a nice proposal!

 

[iceblits]

yay...i can't believe it took me that long to understand what you were trying to say..its obvious now though :)

 

[BahBahSheep]

Just wondering if under that function, the preimage of 1.