# Show that if a ∈ R, then: sup{r ∈ Q : r < a} = a .

• Sep 30th 2010, 09:14 AM
habsfan31
Show that if a ∈ R, then: sup{r ∈ Q : r < a} = a .
Show that if a ∈ R, then:
sup{r ∈ Q : r < a} = a .

Seems like a very simple question, but the r is throwing me off. Do you prove it the same way as a natural number using an epsilon? or how do you prove a supremum with a rational number.
• Sep 30th 2010, 09:28 AM
Plato
Quote:

Originally Posted by habsfan31
Show that if a ∈ R, then:
sup{r ∈ Q : r < a} = a .

We know that $\displaystyle \sigma = \sup \left\{ {r \in \mathbb{Q}:r < a}\right\}$ exists.
Also $\displaystyle \sigma \le a$ because $\displaystyle a$ is an upper bound for the set.
Suppose that $\displaystyle \sigma<a$ then $\displaystyle \sigma<\dfrac{a+\sigma}{2}<a$.
That will give a contradication.
• Sep 30th 2010, 09:41 AM
habsfan31
Thanks, but im confused. How is http://www.mathhelpforum.com/math-he...cee5e4cfae.png a contradiction?
• Sep 30th 2010, 09:51 AM
Plato
Do you realize that between any two numbers there is a rational number?
• Sep 30th 2010, 09:55 AM
habsfan31
Yes.
• Sep 30th 2010, 03:16 PM
habsfan31
K i received help from a friend who took the course, hes not sure if hes right though.

He says that since Q is dense in R, every nonempty interval of R contains a rational.

From the definition of the set S = sup{r ∈ Q : r < a} it follows a is an upper bound of the set S. For every ε >0, the density of the rationals in R shows there is a rational r in (a - ε, a). So, a - ε < r < a, which shows a = sup S is the least upper bound of S.

Is this approach correct?
• Sep 30th 2010, 03:19 PM
Plato
Well of course. That was the whole point.