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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.


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


Tao, Analysis I