Define a bijection between (0,1) and [0,1].
I know that in order to prove this is to use a piecewise function. So this is what I have.
(This is a piecewise function just could not figure out how to put it on here)
This is what I did... Please help me if im wrong.
f(x) = x if for anyf(x) = if
I think what you were going for is something like
defined piecewise (I also haven't mastered the curly brace thing to make it look nice):
if there exists such that
if there does not exist such that
Now, can you prove that this is a bijection?
(A few edits.)
By definition, every element in the domain of a relation R must have an image in its codomain in order to be a function. In this case, we have no way of insuring every element in the domain to have an image since we don't even have the smallest element in it.
Would you mind re-reading my post and seeing if you still think it's problematic?
Edit: By the way, I wrote as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like .