real numbers

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Jul 30th 2011, 06:29 AM
thomasboateng

real numbers

I need help with this question,

Does any open interval in R have a maximum? Explain your answer.

Thomas
Jul 30th 2011, 06:34 AM
TKHunny

Re: real numbers

If it does, is it open? What say you?

There must be a defintion of "Open Interval" sitting about somewhere. Why not have a good, close look at it?
Jul 30th 2011, 06:48 AM
thomasboateng

Re: real numbers

i know it doesnt have any end points because you can always get a little bit more larger for example 0.1 then 0.11
but i dont know to explain correctly
Jul 30th 2011, 07:05 AM
Plato

Re: real numbers

Quote:

Originally Posted by thomasboateng

i know it doesnt have any end points because you can always get a little bit more larger for example 0.1 then 0.11 but i dont know to explain correctly

That may or may not be correct. It depends on which endpoint 0.1 is.
However, this is the essential point: between any two real numbers there is a third number.
If $x\in (a,b)$ then $a<x<b$ .
Therefore $\left( {\exists y \in (x,b)} \right)\left[ {x < y < b} \right]$ .

So can $(a,b)$ have a maximal element?
Jul 30th 2011, 07:25 AM
Also sprach Zarathustra

Re: real numbers

Quote:

Originally Posted by thomasboateng

I need help with this question,

Does any open interval in R have a maximum? Explain your answer.

Thomas

$J=(a,b)$

$\text{inf}\{ J \}=a$

$\text{sup}\{ J \}=b$

$a,b \not\in (a,b)$ , hence by definition:

$\text{min}\{ J \}\neq a$

$\text{max}\{ J \}\neq b$
Jul 31st 2011, 01:59 AM
HallsofIvy

Re: real numbers

Indirect proof: Suppose the open interval, (a, b), does have a maximum, M. Since M is in (a, b), a< M< b. Let $N= \frac{M+b}{2}$ .

Prove:
1) N< b.
2) a< M< N
so N is in (a, b)

3) M< N, contradicting the hypothesis.

Do you understand the difference between a "maximum" and a "supremum" (least upper bound)? That is crucial here.