Proof that '$\exists{y}>0$ s.t $\forall{x}>0,y<x$' is false.

Question

Does my proof of the false statement '$\exists{y}>0$ s.t $\forall{x}>0,y<x$' make logical sense using the contradiction method?

Suppose that there exists $y>0$ such that for all $x>0$ we have that $y<x$. Then take $y=x+1$ for any $x>0$ so that $y>0$. Thus $y>x$ which is a contradiction. Therefore the statement is false.

Answer 1

The easiest way of proving a statement to be false is often by counter example. Assuming we're working in the reals (or the rationals), consider $x = \frac{y}{2}$.

As for your proof, after supposing such a $y$ exists, it doesn't make much sense to then choose $y = x + 1$ for some arbitrary $x > 0$. You should be choosing your $x$ to show that such a $y$ cannot exist, not the other way around.

Answer 2

You proof is wrong. In the proof by contradiction you may only assume that there exists some $y>0$ such that for all $x>0$ we have $y<x$. This $y$ may be smaller than 1, so you can't just take $y=x+1>1$. Instead you have to use your $y$ to make an $x>0$ which is smaller as I did below:

Note that the negation of $\exists y>0\text{ such that }\forall x>0\text{ we have }y>x$ is given by the statement $$\forall y>0\exists x>0\text{ such that }y>x.$$ So let $y>0$ and let $x=\frac{1}{2}y$. Note that $0<x=\frac{1}{2}y<y$.

Answer 3

Your proof doesn't make logical sense. After you "Suppose that there exists $y>0$", y is fixed, you can't " take y=x+1 for any $x>0$". You should prove for this y, the statement "for all $x>0$ we have that $y<x$" is wrong, to show this, you can take $x=y/2$.