Prove: If x ge 0 and x le (alpha) for all (alpha) 0, then x=0.

Any hints on how to start this or what the proof should include?

Question

Prove: If x ge 0 and x le (alpha) for all (alpha) > 0, then x=0.

Any hints on how to start this or what the proof should include?

anonymous · Answer

If $$x \ge 0$$ and $$x < (\alpha)$$ for all $$(\alpha) > 0$$ then x=0

anonymous · Answer

might want to start by assuming 
$x\neq 0$ and arrive at a contradiction

anonymous · Answer

since $x\geq 0$ if $x\neq 0$ then $x=\beta$ and $\beta>0$
now consider $\frac{\beta}{2}$ and get a contradiction

anonymous · Answer

ahhh i like that, thanks!

anonymous · Answer

Ug, my 2nd line should be $$x \le (\alpha)$$ not strictly less than. I think that changes things.

anonymous · Answer

It seems to me that because the inequalities include "or equal to" that x > 0 wouldnt provide the contradiction I am looking for.