Let $x$ and $y$ be real numbers. Prove that if $x\le y + \epsilon$ for every positive real number $\epsilon$, then $x \leq y$.

Question

How do you prove that? I know that if $x\le y + \epsilon$ and $x = 5$ and $(y = 4) + (\epsilon = 1)...$ or if $x \le y + \epsilon$ and $x = -4$ and $y = -3 + (\epsilon = 1)$ then the statement holds true. What are the major oppositions to test against?

Answer 1

Assume the contrapositive: if $x>y$ then there exists some $\epsilon > 0$ such that x > y + $\epsilon$. Then let $\epsilon=\frac12(x-y)$.

Answer 2

Argue by contradiction. Assume that $x>y.$ Then $x-y>0.$ But for $\epsilon=\frac{x-y}{2}>0$ it is, by assumption,

$$x\le y +\epsilon = y+\frac{x-y}{2}=\frac{x+y}{2},$$ from where,

$x\le y.$ This contradicts the assumed fact $x>y.$

Answer 3

Proposition. If $x \le y + \epsilon$ for all real numbers $\epsilon > 0$ then $x \le y$.

Proof. Suppose for sake of contradition that $y < x$. Since $x - y > 0$, by the Archimedean property, we have there exists a positive integer $N > 0$ such that $(x-y)N > 1$, i.e., $x > y + 1/N$. If we define $\epsilon := 1/N$, then exists a real number $\epsilon$ which contradicts the hypothesis "$x \le y + \epsilon$ for all real numbers $\epsilon > 0$".