Question

Prove: if x>=0 and x<=epsilon for all epsilon>0, then x=0.

Answer 1

@wio would you be able to help with this?

Answer 2

Hmmm, well what are you supposed to use to do this proof?

Answer 3

Looks similar to squeeze theorem.

Answer 4

im not sure what the squeeze theorem is. But I thought maybe you would just use contradiction and assume that either x>0 and x<0, and do both case so then by contradiction x>=0 and x<=0 so that means x=0. but Im really not sure.

Answer 5

Hmm

Answer 6

How about something like this... assume x>0. and we must show epsilson>0 such x>epsilon. Let epsilon=x/2. Since x>0, epsilson>0. then
epsilson = x/2 = x/2< (x+x)/2 = x.
so by contradiction x<=0.

or something like that.

Answer 7

are you still need help?

Answer 8

yeah, i really dont know what I am doing for if I am even on the right track

Answer 9

ok,maybe i can help you;)

Answer 10

Seems like we need the completeness axiom

Answer 11

A = {x|x >= 0}
B = {x|x <= (epsilon for all epsilon>0)}

\[A \lceil \rceil B = {0}\]
so,
x = 0

Answer 12

ok?

Answer 13

are A and B like two different cases or are A and B like just different sets.

Answer 14

I can prove it in this way too:
A = {x|x >= 0 , x <= (epsilon for all epsilon>0)}
A = {0}
so,
x = 0

Answer 15

But if you are given that x>=0 and e>0. then all you are saying is that A=0 so x must equal 0. that is all you have to say??

Answer 16

yes ;)

Answer 17

easy enough. thank you.

Answer 18

your welcome :)

Answer 19

density of rational/irrationals..

Answer 20

assume

\[x\ge0,x\le\epsilon\text{ for all }\epsilon>0\\\text{assume by contradiction }x\ne0\\\text{by density we have that there exists }\epsilon \le x\\a \space contradiction...\]