# Terence Tao, Analysis I, Thm. 5.5.9, Least upper bound

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?

• What are $m_n$? Commented Aug 26, 2019 at 20:19
• $m_n$ is an integer, s.t. $L < m \le K$ where $K/n$ is an upper bound of $E$, and $L/n$ is not. With $\mathbb{Z} \ni n \ge 1$. Commented Aug 26, 2019 at 20:22
• 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. Commented Aug 26, 2019 at 21:56
• 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. Commented Aug 27, 2019 at 6:37

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.

• 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? Commented Aug 27, 2019 at 6:40
• The definition of an upper bound of $E$. In Tao's book that's definition 5.5.1 Commented Aug 27, 2019 at 8:25
• 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 $E$. Commented Aug 27, 2019 at 8:41
• @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. Commented Aug 27, 2019 at 9:17
• $y$ is upper bound $\Leftrightarrow$ $(\forall x \in E) x \le y$, whereas $y$ is no upper bound $\Leftrightarrow$ $(\exists x \in E) x > y$. Commented Aug 27, 2019 at 9:31