*I know why cases 1 and 2 are wrong because our language can have different numbers of 0's and 1's. But I'm not sure how case 3 can be proved wrong for our language.*

Exercise 1.30:
Describe the error in the following “proof” that $0^{∗}1^{∗}$ is not a regular language. (An error must exist because $0^{∗}1^{∗}$ is regular.)

> 
> The proof is by contradiction. Assume
> that $0^{∗}1^{∗}$ is regular. Let p be the pumping length for $0^{∗}1^{∗}$ given by the pumping
> lemma. Choose s to be the string $0^{p}1^{p}$. You know that s is a member of $0^{∗}1^{∗}$, but
> Example 1.73 shows that s cannot be pumped. Thus you have a contradiction. So
> $0^{∗}1^{∗}$ is not regular.
> 
> 
> 

Example 1.73:

> 
> Let B be the language $\{0^{n}1^{n}|n ≥ 0\}$. We use the pumping lemma to prove that
> B is not regular. The proof is by contradiction.
> Assume to the contrary that B is regular. Let p be the pumping length given
> by the pumping lemma. Choose s to be the string $0^{p}1^{p}$. Because s is a member
> of B and s has length more than p, the pumping lemma guarantees that s can be
> split into three pieces, s = xyz, where for any i ≥ 0 the string $xy^{i}z$ is in B. We
> consider three cases to show that this result is impossible.
> 
> 
> 1. The string y consists only of 0s. In this case, the string xyyz has more 0s
> than 1s and so is not a member of B, violating condition 1 of the pumping
> lemma. This case is a contradiction.
> 2. The string y consists only of 1s. This case also gives a contradiction.
> 3. The string y consists of both 0s and 1s. In this case, the string xyyz may
> have the same number of 0s and 1s, but they will be out of order with some
> 1s before 0s. Hence it is not a member of B, which is a contradiction.
> 
> 
> 

> 
> Thus a contradiction is unavoidable if we make the assumption that B is regular,
> so B is not regular. Note that we can simplify this argument by applying
> condition 3 of the pumping lemma to eliminate cases 2 and 3.
> In this example, finding the string s was easy because any string in B of
> length p or more would work.
> 
> 
> 

*This question is from the book 'Introduction to the Theory of Computations' by Michael Sipser, exercise 1.30*