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.
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.
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.
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?