Terence Tao, Analysis I, 3e, Thm. 5.5.9:
Theorem 5.5.9 (Existence of least upper bound). Let $E$ be a non-empty
subset of $\mathbb{R}$. If $E$ has an upper bound, (i.e., $E$ has some
upper bound $M$), then it must have exactly one least upper bound.
In the proof it says that
(…) Now we show it [$S := \text{LIM}_{n \rightarrow \infty} (m_n – 1)/n$] is a least upper bound. Suppose $y$ is an
upper bound for $E$. Since $(m_n – 1)/n$ is not an upper bound, we
conclude that $y \ge (m_n – 1)/n$ for all $n \ge 1$.
But from my understanding, if $y$ is an upper bound for $E$, then all elements in $E$ are smaller than or equal to $y$. And since $(m_n – 1)/n$ is not an upper bound, there are elements $x$ in $E$ such that
$$
(m_n – 1)/n < x \le y.
$$
If $y > (m_n – 1)/n$, then $y \ge (m_n – 1)/n$.
But why is equality used here? Is it some sort of trick to make use of
(…) if $a_n \le x$ for all $n \ge 1$, then $\text{LIM}_{n
\rightarrow \infty} \; a_n \le x$
in a following step of the proof?
Best Answer
All that can be said is that $(m_n -1)/n \leq y$; that is by definition. Given the fact that $\forall n \in \mathbb{N}$ the element $(m_n -1)/n$ is not an upper bound we can deduce that there is strict inequality.
In the following sentence he uses Exercise 5.4.8, which essentially says
Here the result is true regardless of whether $a_n \le x$ or $a_n < x$ for all $n \ge 1$. In short, it does not matter whether to write strict or non-strict inequality; both are valid and the conclusion is the same.
By the way
is NOT true. Consider the sequence $(\frac{-1}{n})_{n \in \mathbb{N}}$. Here each element is strictly smaller than $0$, but the limit equals $0$.
Also, the statement
is only true because $\mathbb{R}$ is totally ordered. In a general partially ordered set this is not nessecarily the case.