Bijection?

I know there must exist a bijection between X = (0,1) and Y = X U {1}, because they have the same cardinality, but I just can't find one.  Can anyone show me an example?

1 to 1/2

1/2 to 1/4

1/4 to 1/8

and so on

1/2^n to 1/2^(n+1)

And the rest of Y maps to the rest of X

Or a function from X to Y:

f(x)=

(if log2(1/x) is an integer) then f(x)=2x

else f(x)=x

Yes. The idea perfectly applied by pawn_slayer666 forms part of a "paradox" about a hotel with an infinite number of rooms where someone arrives, so all the guests are moved along to make room for the new arrival.

Alternatively, choose:

the obvious injection f : (0,1) -> (0,1) U {1} = (0,1]

the injection g : (0,1] -> (0,1) defined by g(x) = x/2

and then apply the Schroeder-Cantor-Berstein theorem (which does actually give a precise definition for a bijection, which can be unravelled).

Thanks.  So what you are doing is basically choosing an N-sized subset of X and using that to make the one-point translation.

And if we wanted to map X with Z = X U {any number of points c < N}, we could make more room in the chosen subset:

For x | log2(1/x) = integer > c, f(x) = (2^c)x

For x | log2(1/x) =/= integer, f(x) = x

And there are c points leftover to map to the introduced set.

If we let c = alef-0, we could do a translation using the sets

{x | log3(1/x) is int.},

{x | log5(1/x) is int.},

and so on for all primes.  There are c of such sets, so each one could take a finite number of points.

Elroch, I already knew that the two sets were the same, what I was looking for was an explicit example.

What I meant was that, as well as the solution given by pawnslayer666, the proof given here (as "another proof") can be used to generate another explicit solution, as it can for any two sets and a pair of injective functions between them. Obviously, the bijection depends on which functions g and f you use.