Let S ⊆ R, and recall M = supS is the least upper bound of S. Suppose that S is bounded, so supS ∈ R. (If S was not bounded, then we set supS = ∞). Show that for every ε > 0, there exists x ∈ S such that |x−M| < ε and that there exists a sequence (x_{n}) ^{∞}_{n=1} from S such that for each n ∈ N,
|x_{n }− M| < 1/2^{n}.
Define M to be the supremum of S. R and N are the real numbers and natural numbers, respectively.
Let S ⊆ R and M = sup S which is the least upper bound of S. We are given that S is bounded, so sup S ∈ R.
(i) We have to show that for every ε > 0, there exists x ∈ S such that |x − M| < ε .
Proof of (i)
If M is a least upper bound then for every ε > 0 there exist x ∈ S such that M – ε < x, if not, M – ε will be the least upper bound of the set S by its definition. Also, x ≤ M by the definition of least upper bound.
So, it is trivial that x ≤ M+ ε.
Hence,
M – ε < x ≤ M+ ε
or |x − M| < ε
Again,
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