How to prove comparability property & writing its proof

Question

I am given the relation in $\mathbb{R}$: $xRy$ if $x\le 2^y$. I want to prove this has the comparability property, so I know I start with let $x,y\in \mathbb{R}$. Then I need to show either $xRy$ or $yRx$. I am not too sure how to go about it.

Answer 1

Suppose for contradictory purposes that $R$ does not have the comparability property.

Then there must be some pair of $x$ and $y$ such that $x\not R y$ and $y\not R x$. In other words, there must be some pair of $x$ and $y$ such that $x>2^y$ and $y>2^x$.

Since $y>2^x$ implies that $2^y>2^{2^x}$ we have $x>2^y>2^{2^x}$ and in particular we have $x>2^{2^x}$.

Now, one of three things will be true about $x$. Either $x\leq 0$ or $0<x< 2$ or $x\geq 2$.

• Case 1: $x\leq 0$: In the case that $x\leq 0$ we have $0\geq x>2^{2^x}$ however since $2^a>0$ for all real $a$ we have $0\geq x>2^{2^x}\geq 2^0=1$ but since $0\not\geq 1$ this is a contradiction.

• Case 2: $0<x<2$:

  In the case that $0<x<2$ we have $2>x>2^{2^x}$ however since $2^a>1$ for all $a>0$ we have $2>x>2^{2^x}>2^1=2$ but since $2\not > 2$ this is a contradiction.

• Case 3: $2\leq x$: (left as excercise to reader) Use what you know about the comparison of $x$ vs $2^x$ when $x\geq 2$.

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Since in all three cases we arrive at a contradiction, we see that there cannot be any such $x$ and thus there cannot be any pair $x,y$ such that neither of $xRy$ nor $yRx$ is true, therefore at least one of $xRy$ or $yRx$ must be true and the relation is indeed comparable.