Applying the Pumping Lemma

Here's a more formal definition of the pumping lemma:

If L is an infinite regular language, then there exists some positive integer m such that any string w is a member of L whose length is m or greater can be decomposed into three parts, xyz, where

|xy| is less than or equal to m,
|y| > 0,
w_i = xyⁱz is also in L for all i = 0, 1, 2, 3, ....

Here's what it all means:

m is a (finite) number chosen so that strings of length m or greater must contain a cycle. Hence, m must be equal to or greater than the number of states in the dfa. Remember that we don't know the dfa, so we can't actually choose m; we just know that such an m must exist.
Since string w has length greater than or equal to m, we can break it into two parts, xy and z, such that xy must contain a cycle. We don't know the dfa, so we don't know exactly where to make this break, but we know that |xy| can be less than or equal to m.
We let x be the part before the cycle, y be the cycle, and z the part after the cycle. (It is possible that x and z contain cycles, but we don't care about that.) Again, we don't know exactly where to make this break.
Since y is the cycle we are interested in, we must have |y| > 0, otherwise it isn't a cycle.
By repeating y an arbitrary number of times, xy*z, we must get other strings in L.
If, despite all the above uncertainties, we can show that the dfa has to accept some string that we know is not in the language, then we can conclude that the language is not regular.

To use this lemma, we need to show:

For any choice of m,
for some w L that we get to choose (and we will choose one of length at least m),
for any way of decomposing w into xyz, so long as |xy| isn't greater than m and y isn't ,
we can choose an i such that xyⁱz is not in L.

We can view this as a game wherein our opponent makes moves 1 and 3 (choosing m and choosing xyz) and we make moves 2 and 4 (choosing w and choosing i). Our goal is to show that we can always beat our opponent. If we can show this, we have proved that L is not regular.