# Let f:A->B be a bijection

1. Oct 1, 2013

### aaaa202

This is something I have been wondering about. Let f:A->B
be a bijection. If B is a subset of X. Can there still exist a bijection from A to X?

2. Oct 1, 2013

### UltrafastPED

But A is already fully mapped onto B, so which elements of A are left to map onto the X - B?

3. Oct 1, 2013

### aaaa202

Well the cardinality of the rationals, natural numbers and even natural numbers are the same.
So there exists a bijection between each of these sets even though, the even natural numbers are a subset of the natural numbers. So I think you're actually wrong?

4. Oct 1, 2013

### UltrafastPED

However it seems that you now know the answer: I think you are saying "for finite sets the answer is NO; for infinite sets the answer is YES". Next you should go ahead and construct an example using the set of natural numbers and subsets of odds and evens.

5. Oct 1, 2013

### HallsofIvy

Staff Emeritus
You might want to re-write the question to specify that B is a proper subset of X.

6. Oct 1, 2013

### jbunniii

It should be mentioned that it isn't always true for infinite sets. E.g., take $A = B = \mathbb{R}$ and $X = \mathbb{Z}$. In general it will be possible only if $X$ has the same cardinality as $B$ (and hence the same cardinality as $A$).