Proof of Uncountable Domain for f:A-->B

[happyg1]

hi,
the question I am working on is this:
If f:A-->B and the range of f is uncountable, prove that the domain of f is uncountable.
Intuitively this seems to be true. If the range is uncountable, then function has to map an uncountable number of elements from the domain to the range. I don't know how to make this a rigorous proof. Does the function have to be one to one? Can I say that the inverse exists? I know that this is an easy problem, but I am stuck. Please point me down the right path.
Thanks,
CC

 

[neurocomp2003]

is f: said to be a bijection or 1-1 or onto?...cuz i can't see a inverse trig function satisfying that without more info.

You can't say the inverse exists without knowing 1-1 or onto or both...or can you...its been so long.

 

[fourier jr]

off the top of my head i would try to do it by contradiction. suppose the range of uncountable and the domain is countable. then f must map each element of A to more than one element of the range. contradiction since f is a function. (something like that anyway)

or like this, spose f:A-->B is a function, B is uncountable and A is countable. then make a list of all the elements of the range: f(a_1), f(a_2),... etc then all the elements of the range have been listed since f can't map one thing to more than one thing, so the range is countable. contradiction.

 

[matt grime]

Let f be a function from A to B, then if the domain of f is countable it follows that f is a surjection from a countable set onto its image, thus what can you say about the image? (this isn't a proof by contradiction but a proof by the contrpositive, ie rather than show A implies B we are proving the equivalent statement not(B) implies not(A))

 

[happyg1]

Hi guys,
That's all of the info that was given for the problem. My quandry was that I need it to be bijective and I don't know how I can say that it is. I'll try the contrapositive suggestion. I tried contradiction, but I'm not confident with my results. My teacher's hint was "read the book", which I've done...over and over.
Thanks for the input,
CC

 

[happyg1]

Hi all,
I have been working on this danged problem some more, and I am even more confused...here's why: f(x)=1. Isn't this a function that maps an uncountable domian into a countable range? If this is true then uncountable domain does not necessarily imply uncountable range. Then I am right back where I started...uncountable range, and I need to get at the domain. My head hurts.
CC

 

[HallsofIvy]

Hi all,
I have been working on this danged problem some more, and I am even more confused...here's why: f(x)=1. Isn't this a function that maps an uncountable domian into a countable range? If this is true then uncountable domain does not necessarily imply uncountable range. Then I am right back where I started...uncountable range, and I need to get at the domain. My head hurts.
CC

Yes, that's true but not relevant to the original question which was whether a countable domain could be mapped to an uncountable range. If f is not invertible, as f(x)= 1 is not, those are not the same question.

The definition of function requires that two different y values cannot be given by the same y value. Your reverse example has many different x values giving the same y.

Try this. Suppose the domain is countable. Then we can "list" all members of the domain:x₁, x₂, ..., x_i,...
For any y in the range, label it y_i where y= f(x_i).
What does that tell you?

 

[neurocomp2003]

does it say if the function is continous.

 

[happyg1]

Hi,
Thanks for the clarification. I think I have been looking at this particular problem so long that it has stopped making sense to me. I have done the exact thing that Hallsof Ivy suggested. I just kept thinking about 1-1 and onto...but that doesn't matter. I am unproving what I have written by throwing myself some red herrings.

No, it doesn't say that the function is continuous, just that it's a function.

I think I am convinced that this is true. I will write up my proof and go submit it to my professor, who will hopefully agree with this reasoning.

Thanks so much for your time and attention.
CC

 

[matt grime]

continuity has nothing to do with it, neurocomp.