I have a proof of a statement, and somehow a proof of its negation as well. I guess that the proof of its negation has mistakes?
Statement $\forall x \in \mathbb{R} \forall y \in \mathbb{R} \exists z \in \mathbb{R} |x-z| < |x-y|$.
Proof Let $x,y \in \mathbb{R}$. Choose $z = \frac{x+y}{2}$.
Negation $ \exists x \in \mathbb{R} \exists y\in \mathbb{R} \forall z\in \mathbb{R} |x-z| \geq |x-y|$
Proof Let $x \in \mathbb{R}$. Choose $x = y$. Let $z \in \mathbb{R}$.
$|x-y| = 0 \leq |x-z|$. (This is actually a proof for the negation with $ \forall x…$, but for sure would show the actual negation as well.)
Best Answer
The original statement is false: what happens when $x=1=y ?$
For the negation, when adapting the proof of $\forall x$ to $\exists x,$ remember that $x$ is no longer arbitrary, so you do need to specify a particular choice of $x$ (remember, this value is fixed while the arbitrary variables that come after it are varying).
So, a correct proof might instead go:
Choose $(x,y)=(1,1)$ and let $z \in \mathbb{R}.$ Then $|x-y| = 0 \leq |z-1|=|x-z|.$