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&#039;} = sup T so \frac{1}{c&#039;} < \frac{1}{c}.

From this I get for some e > 0 \frac{1}{c&#039;} = \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&#039;}< \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