I want to prove that
$A$ = {ww: w $\in \sum^{*}$}
is not regular. This is what I have done so far:
Suppose A is regular. Let $p$ be the constant that must exist if $A$ is regular. Then, we can choose $ww \in A$ such that $|w| = p$. Because $|ww| \geq p$, we know that exists $x,y,z \in \sum^{*}$ such that:
(a) $ww = xyz$
(b) $|xy| \leq p$
(c) $xyyz \in A$
Since $|xy| \leq p$, we know that xy consist only in elements from the first $w$ of $ww$. Let's pick $y$ such that $|y|$ is odd, so $|xyyz|$ is odd as well. However, that is impossible as there is no string $w$ that $ww = xyyz$ because $|ww|$ is always even. Thus, $xyyz \notin A$.
Hence, we proved by contradiction that $A$ is not regular.
PS: If I wrote anything wrong feel free to change and tell me. English is not my first language.
Notice that $\Sigma$ must have more than one element, because $L = \{ww : w\in \{a\}^*\}$ is regular. So $\Sigma$ has at least two elements $a$ and $b$. You can't simply choose the substring $y$ that suits you. It is given by the lemma. We need to construct a different string, so consider $a^pb^pa^pb^p.$ Now no matter which $y$ the lemma gives us, we know it contains only $a$'s. Thus pumping up, we get more $a$'s in the first segment of $a$'s than in the second, a contradiction.