# How to turn a $p \Rightarrow q$-type statement into a $\neg q \Rightarrow \neg p$-type one?

I am working with the supposition that $$\lim\limits_{x\to \infty}{f(x)}=\infty$$. I have rewritten it as $$\forall R>0\;\exists r>0\;\text{ such that }\;|x|>r\Rightarrow |f(x)|>R.$$ I want to turn this "$$p\Rightarrow q$$"-type supposition into a $$\neg q\Rightarrow \neg p$$ one, but I am not sure how to negate it. Is $$\forall R>0\;\exists r>0\;\text{ such that }\;|f(x)|\leq R\Rightarrow |x|\leq r$$ correct?

• The function is $f: \mathbb{R}^m \to \mathbb{R}^n$ Sep 21 at 12:34

You have hidden quantifiers that aren't being accounted for. Let us take a closer look at "$$\lim_{x\to\infty}f(x)\neq\infty$$".

$$(\forall R>0)(\exists r>0)(|x|>r\implies|f(x)|>R)$$

While this seems sensible enough, the variable $$x$$ here is free. This is a question about a single value of $$x$$. If I were to plug in $$x=0$$ then we would have that this always holds (for every function $$f$$). Instead, what is really happening is

$$(\forall R>0)(\exists r>0)(\forall x)(|x|>r\implies|f(x)|>R)$$

Now you can use modus tollens to obtain the form that you seek.

$$(\forall R>0)(\exists r>0)(\forall x)(|f(x)|\leq R\implies |x|\leq r)$$

I think that you may want to do away with your absolute values, though. $$\lim_{x\to\infty}f(x)=\infty$$ is that in the positive $$x$$ direction $$f$$ eventually becomes arbitrarily large. Therefore, it would look more like this:

$$(\forall R>0)(\exists r>0)(\forall x)(f(x)\leq R\implies x\leq r)$$