I am misunderstanding if one requires, for the pumping lemma proof by contradiction to apply, to prove only a single case of $y^i$, or a specific set. Namely, if I show that, at some point, provided $y^i$ reaches a certain $i$, that the lemma no longer holds, is that sufficient to prove that a language is not regular?
I have been given the following language:
$ L = \{a^n b^m c^k : n + m > k > 0\} $
I figured I would take the following $w = a^{m+1}b^{m}c^{2m}$, with the logic that, for $y^0$, so for $w = xz$, then $w_0 = a^{m}b^{m}c^{2m}$, in which case the condition no longer holds. This works, obviously, when $|y| = 1$. Is this case too "specific"?
Below is the entirety of my proof:
We assume the given language to be regular. If that is the case, then the pumping lemma must apply. Let $m$ be the pumping length.
I pick a string $w$ such that $w \in L$ and $|w| \geq m$:
I pick $w = a^{m+1}b^{m}c^{2m}$
From pumping lemma, we write $w = a^{m+1}b^{m}c^{2m} = xyz$
with $|xy| \leq m$ and $ |y| \geq 1$
such that:
Thus, $y = a^k$, $k \geq 1$
From pumping lemma, $w_i = xy^iz \in L$, $i = 0,1,2....$
We take $i = 0$: $w_0 = xz\in L$
Thus, $a^{m-k}b^{m}c^{2m} \in L$, $k \geq 1$
But clearly this is a contradiction, as we need $n+m > k$.
Thus, it must be that $L$ is not regular.
The pumping lemma for regular languages says that if the language $L$ is regular, from a sufficiently long word in $L$ you can generate an infinite set of words that are all in $L$.
"Sufficiently long" here means long enough that a run of this word in an automaton that accepted $L$ would have to visit at least one state twice. If we are trying to prove that a language is not regular, we certainly have no such automaton. Hence we say, "if there is such an automaton, it has a certain number of states. Just pick a word with at least those many letters." And so we work with an unspecified $m$---the pumping length---only assuming that it is finite.
If, for any pumping length $m$ you can find a word $w$ in $L$ of at least that length, and you find that even one word in the resulting set $xy^iz$ is not in $L$, the pumping lemma implies that $L$ is not regular.
Just as we cannot make assumptions about the number of states of the hypothetical automaton that accepts $L$, we cannot make assumptions about the division of $w$ into $x$, $y$, and $z$, except for $|xy| \leq m$ and $|y| \geq 1$.
That is, to prove that $L$ is not regular, we have to show that for every $m$ and for every possible division of $w$ into $x$, $y$, and $z$, there exists at least one value of $i$ for which $xy^iz$ is not in $L$.
As usual when multiple quantifiers are involved, order matters. In this case our proof obligation is: for every $m$ there exists a long-enough $w \in L$ such that, for every $x$, $y$, $z$ such that $w = xyz$, $|xy| \leq m$, and $|y| \geq 1$, there exists $i \geq 0$ such that $xy^iz$ is not in $L$.
This is what your proof shows: You have not imposed restrictions on $m$; have chosen a long-enough $w$ and shown that, no matter how you legally split it into $x$, $y$, and $z$, $xy^0z$ is not in $L$.