Show Question
Math and science::Analysis::Tao::05. The real numbers

# Existence of least upper bound, theorem

Let $$E$$ be a subset of $$\mathbb{R}$$. If $$E$$ has some upper bound, then it must have a least upper bound.

While it may seem banal at first, the result is quite amazing. To appreciate this, consider why a set of rationals does not necessarily have a rational least upper bound.

Proof outline
The proof of this is somewhat involved (but not difficult). The proof (p118) is a great example of the utilization of the properties of Cauchy sequences and equivalent sequences.

It starts by constructing two Cauchy sequences. One sequence, $$(a_n)_{n=1}^{\infty}$$, consists of only upper bounds, and the other, $$(b_n)_{n=1}^{\infty}$$ only of not upper bounds. There is a relationship between elements in the sequences: $$a_k$$ and $$b_k$$ differ by a rational $$\frac{1}{n}$$. As $$n$$ increases, it is shown that both sequences are equivalent and define the same real, $$S$$. Then, considering any other upper bound $$M$$, it must be greater or equal to all elements of $$(b_n)_{n=1}^{\infty}$$, otherwise it would not be an upper bound. Consequently, it must be greater to or equal to $$S$$. This makes $$S$$ less than or equal to all other upper bounds.

### Example

Upper bound def → least upper bound def→ uniqueness of least upper bound → existence of least upper bound → supremum def

Tao, Analysis I