What is the negation of the given statement in metric space

Question

Let $(X,d)$ be a metric space and $T:X\to X$ be a mapping.

$\bf{P:}$ "There exists a map $\phi:\Bbb R^+ \to \Bbb R$ such that $\displaystyle d(Tx,Ty)\le \phi(d(x,y))$ for all $x,y\in X$."

What is the negation of the statement $\bf P$ ?

I think it will be : "There exists $x,y\in X$ such that for all map $\phi:\Bbb R^+ \to \Bbb R$, we have $\displaystyle d(Tx,Ty)> \phi(d(x,y))$".

Am I correct ? If not then what it will be?

Answer 1

The negation is

For any map $\phi:\mathbb R^{+} \to \mathbb R$ there exist points $x$ and $y$ with $d(Tx,Ty) >\phi (d(x,y))$.

Answer 2

"For any map $\phi:\Bbb R^+ \to \Bbb R$, there exists $x,y\in X$ with $\displaystyle d(Tx,Ty) > \phi(d(x,y))$. "

Notice that the $x, y$ for each map can be different.

Answer 3

It can help to abstract from details in cases like this. P has the form

$\exists \phi\forall x \forall y\ A \leq B$

So its negation is

$\neg\exists \phi\forall x \forall y\ A \leq B$

Now apply brute force! You know that pushing a negation sign past a quantifier flips the quantifier into its dual: so the negation of P is equivalent to

$\forall \phi\exists x \exists y\ \neg(A \leq B)$

i.e.

$\forall \phi\exists x \exists y\ A > B$

Exactly as @KaveRamaMurthy said!