Proving x is the Infimum of a Non-Empty Group with a Given Low Bound

[transgalactic]

its given that "x" is a low bound of non empty group A.
prove that x=inf A for e>0 there is "y" which x<=y<=x+e
??

i can't understand this question

i know how to prove that a certain number is inf or sup using only the formula of the group

but here i don't have the formula of the geoup
and i have this 'y' i don't know why its for??

 

[VeeEight]

Perhaps you need to rewrite the question. x is a lower bound for nonempty set A? y is an element of A? For all e or a fixed e?

 

[transgalactic]

its given that "x" is a lower bound for non empty set A.
prove that x=inf A for e>0 there is "y" y is element of A which x<=y<=x+e
e is fixed
??

 

[transgalactic]

i know that inf is the biggest integer under the lower bound
and if we add a certain "e" then it will cease to be inf
and it will become inside the group

 

[VeeEight]

inf A is the greatest lower bound for the set A.
So inf A + e will not be a lower bound for the set A for some positive real e.

What does this statement mean: "for e>0 there is "y" y is element of A which x<=y<=x+e"?

 

[transgalactic]

that there is a fixed "e" for which x<=y<=x+e

y is element of A

 

[VeeEight]

It is stated that A is bounded below, x being a lower bound, and there is a specific e such that all elements of A fall in the interval [x, x + e]. So A is a bounded set. Perhaps you need to reread the question (or your notes that pertain to the question).

 

[transgalactic]

i think i got it
A bounded from the bottom
x is a low bound of A

i need to prove that x=inf Athis expression is just the definition of inf
x<=y<=x+eso i tried to solve it as if i got a formula
I presume that "x" is not inf
then our inf is x+e
now i proove that x+e is not infA
x+e>y

but here i got stuck because in this step i would do a formula operation.

??