Hi

1) , . Note that both unions are in fact disjoint unions.

Because , there's a bijection . Now we're ready to construct a bijection like this:

Verify it is really a bijection!

2) means there's a bijection

means there's a bijection

Define as follows:

Verify it is a bijection.

3) Conditions say that and exist.

Is an upper bound of ? Yes: let , so for some , .

But and , so .

Is the least upper bound of ? Suppose it is not, then there exist such that is an upper bound of .

Because is a supremum, there must exist some such that .

Because is an infimum, there must exist some such that .

We get which contradicts the assumption that is an upper bound of .

We conclude that .

4) Because x->2x is a bijection from [0,1) to [0,2), it will suffice to construct a bijection between [0,1] and [0,1) and composition of the two bijections will be a bijection from [0,1] to [0,2).

If you know Cantor-Bernstein Theorem, it is easy, because there obviously exist an injective map from [0,1) to [0,1] (the identity map) and an injective map from [0,1] to [0,1) (x -> x/2).

If you don't know the theorem, consider set and set . You can construct bijection by setting (verify it is a bijection).

Next, we see that , so the identity on is a bijection from to .

Now we're ready to construct a bijection as follows: