Solution is nonsense: Prove that if $||x-y|| \geq k||T(x)-T(y)||$ then $T$ need not be invertible

Question

For all vectors $x, y \in \mathbb{R}^n$, $T: \mathbb{R}^n \longrightarrow \mathbb{R}^n$ and $k > 0$, if $||x-y|| \geq k||T(x)-T(y)||$ then $T$ not be invertible.

The solution doesn't make sense. They first choose a $T$ that is non invertible and show that it satisfies the equality shown above.

This doesn't prove the fact that if the equality is true, then $T$ might not be invertible. It just shows that if $T$ is not invertible, then the inequality is true, the other way. How can they do this?

Answer 1

Let be $P(T)$ the property $$\|x-y\|\ge k\|T(x)-T(y)\|$$ and $I(T)$ "$T$ is invertible". The title says $$\neg(\forall T: P(T)\implies I(T)),$$ or equivalently $$\exists T: P(T)\land\neg I(T),$$ an this can be proved showing one noninvertible $T$ what has the property $P$.

About

They first choose a $T$ that is non invertible and show that it satisfies the equality shown above.

this is exactly $\exists T: P(T)\land\neg I(T)$.

This doesn't prove the fact that if the [in]equality is true, then $T$ might not be invertible. It just shows that if $T$ is not invertible, then the inequality is true, the other way.

False. Showing one $T$ that verifies $P(T)$ and $\neg I(T)$ proves again $\exists T: P(T)\land\neg I(T)$, never $\forall\cdots$