Show that supr Q r a a for each a RSolutionSolution Let A

Show that sup{r Q: r < a} = a for each a R.

Solution

Solution :

Let A = {r in Q | r < a} I will prove this by showing that sup A < a and sup A > a cannot happen.

Note that a is an upper bound for A by definition of {r in Q | r < a}. Thus sup A > a cannot occur. Suppose by contradiction that sup A < a. Then we know that sup A belongs to the set A because sup A < a. But we can find a number r between sup A and a, thus having sup A < r < a but that is a contradiction because r < sup A for all r in A.

Thus, sup {r in Q | r < a} = a as needed to show.

OR

let a be in R
let S = {q in Q: q < a}
let s - sup(S)
if s < a, since there is a rational between any two distinct reals, there exists q in Q with s < q < a
but the q is in S, q<a contradicts that s = Sup(S)
if s > a, since there is a rational between any two distinct reals, there exists r in Q with a < r < s
therefore r > q for all q in S
but s, by definition of supremum, is the least real greater than or equal to all members of S.
r < s contradicts that s = sup(S)
therefore s = a