Proving That T Has a Supremum: A Mathematical Exercise

[dargar]

Homework Statement

Let S be a set of positive real numbers with an infimum c > 0 and let the set T = {$$\frac{1}{t}$$ : t $$\in$$ S}.

Show that T has a supremum and what is it's value.

The attempt at a solution

Ok, so the value must be $$\frac{1}{c}$$.

But I'm unsure how to start proving that T must have a supremum. Any starting hints would be great :) thanks

 

[boombaby]

1/c is absolutely an upper bound, thus you need to prove it is the supremum, which means, for any e>0, you can find a 1/s such that 1/s > 1/c-e. find such s based on the fact that c is the infimum of S.

 

[dargar]

Okay I think I have something but I'm unsure whether it's right.

for $$\frac{1}{c}$$ = sup T it must meet the following two criteria.

1) $$\frac{1}{c}$$ is an upper bound such that $$\frac{1}{c}$$ $$\geq$$ $$\frac{1}{t}$$ $$\forall$$ t $$\in$$ S

2) $$\forall$$ e > 0, $$\exists$$ x $$\in$$ A with $$\frac{1}{c+e}$$< $$\frac{1}{t}$$ $$\leq$$ $$\frac{1}{c}$$

So my attempted proof follows that we can argue by contradiction.

supposing $$\frac{1}{c}$$ satisfies 1 and 2.

So 1) $$\Rightarrow$$ $$\frac{1}{c}$$ is an upper bound

Assume that $$\frac{1}{c'}$$ = sup T so $$\frac{1}{c'}$$ < $$\frac{1}{c}$$.

From this I get for some e > 0 $$\frac{1}{c'}$$ = $$\frac{1}{c+e}$$.

2) $$\Rightarrow$$ $$\exists$$ $$\frac{1}{t}$$ $$\in$$ T with $$\frac{1}{c+e}$$< $$\frac{1}{t}$$ $$\leq$$ $$\frac{1}{c}$$ $$\Rightarrow$$ $$\frac{1}{c'}$$< $$\frac{1}{t}$$ $$\leq$$ $$\frac{1}{c}$$ which contadicts the fact that c' = sup T

Think I've made a bit of a mess of it as I'm trying to base it off an example that's kind of similar in my notes

 

[boombaby]

what is the x in your 2) criteria?
What is the A in your 2) criteria?
By the definition, you 2) criteria should be:
for any e>0, there exists a t in S such that 1/t > 1/c - e.
If you think your criteria is equavalent to this one, you have to show us. ( I havn't checked it)

If you want to argue by contradiction, you made something wrong.
You already assume that 1/c satisfies 1 and 2, which implies that 1/c is the least upper bound. you assume again that 1/d should be the least upper bound, which is less than 1/c. From these two assupmtions alone, you already get a contradiction