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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?

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  • $\begingroup$ What are $m_n$? $\endgroup$
    – Berci
    Commented Aug 26, 2019 at 20:19
  • $\begingroup$ $m_n$ is an integer, s.t. $L < m \le K$ where $K/n$ is an upper bound of $E$, and $L/n$ is not. With $\mathbb{Z} \ni n \ge 1$. $\endgroup$ Commented Aug 26, 2019 at 20:22
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    $\begingroup$ Is your problem with the existence of the lub in the first place, or its unicity (that latter part is much easier)? Cannot really comment on the details of the book: I don't have it, so I don't know its definitions etc. $\endgroup$ Commented Aug 26, 2019 at 21:56
  • $\begingroup$ It's about existence and you are right that it's hard to make a statement without having the full context. Somebody posted the complete proof, but it still lacks the definitions and exercises the proof references. $\endgroup$ Commented Aug 27, 2019 at 6:37

1 Answer 1

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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

(...) if $a_n \le x$ for all $n \ge 1$, then $\text{LIM}_{n \rightarrow \infty} \; a_n \le x$

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

if $a_n < x$ for all $n \ge 1$, then $\text{LIM}_{n \rightarrow \infty} \; a_n < x$

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

And since $(m_n - 1)/n$ is not an upper bound, there are elements $x$ in $E$ such that $$ (m_n - 1)/n < x \le y.$$

is only true because $\mathbb{R}$ is totally ordered. In a general partially ordered set this is not nessecarily the case.

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  • $\begingroup$ Thanks a lot for this elaborate answer! Helped me a lot and it's clear and nicely put. You're mentioning a definition in the first sentence. Where can I find it? $\endgroup$ Commented Aug 27, 2019 at 6:40
  • $\begingroup$ The definition of an upper bound of $E$. In Tao's book that's definition 5.5.1 $\endgroup$ Commented Aug 27, 2019 at 8:25
  • $\begingroup$ The definition states that "We have that $M$ is an upper bound for $E$, iff we have $x \le M$ for every element $x$ in $E$". But $(m_n - 1)/n$ is not necessarily an element of $E$. $\endgroup$ Commented Aug 27, 2019 at 8:41
  • $\begingroup$ @MaxHerrmann I see, my mistake. Since $(m_n -1)/n$ is not an upper bound and $y$ is an upper bound, $(m_n-1)/n > y$ is false. Hence $(m_n-1)/n \leq y$ is true. To see this, assume $(m_n-1)/n > y$ to be true. Then $\forall e \in E: (m_n-1)/n > y \geq e$. Hence $(m_n-1)/n$ was an upper bound. Contradiction. $\endgroup$ Commented Aug 27, 2019 at 9:17
  • $\begingroup$ $y$ is upper bound $\Leftrightarrow$ $(\forall x \in E) x \le y$, whereas $y$ is no upper bound $\Leftrightarrow$ $(\exists x \in E) x > y$. $\endgroup$ Commented Aug 27, 2019 at 9:31

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