# Mathematical Analysis by Walter Rudin, Theorem 1.11: Upper/Lower Bounds and Supremum/Infimum.

Theorem 1.11

Suppose $$S$$ is an ordered set with the least-upper-bound property, $$B \subset S$$, $$B$$ is not empty, and $$B$$ is bounded below. Let $$L$$ be the set of all lower bounds of $$B$$. Then

$$\alpha = \sup(L)$$

exists in $$S$$, and $$\alpha = \inf(B)$$.

In particular, $$\inf(B)$$ exists in $$S$$.

Rudin's definitions for upper/lower bounds and supremum/infimum are given as follows, respectively:

1.7 Definition

Suppose $$S$$ is an ordered set, $$E \subset S$$. If there exists a $$\beta \in S$$ such that $$x \le \beta$$ for every $$x \in E$$, we say that $$E$$ is bounded above, and call $$\beta$$ an upper bound of E.

Lower bounds are defined in the same way (with $$\ge$$ in place of $$\le$$).



1.8 Definition Suppose $$S$$ is an ordered set, $$E \subset S$$, and $$E$$ is bounded above.

Suppose there exists an $$\alpha \in S$$ with the following properties:

(i) $$\alpha$$ is an upper bound of $$E$$.

(ii) If $$\gamma < \alpha$$ then $$\gamma$$ is not an upper bound of $$E$$.

Then $$\alpha$$ is called the least upper bound of $$E$$ [that there is at most one such $$\alpha$$ is clear from (ii)] or the supremum of $$E$$, and we write

$$\alpha = \sup(E)$$.

The greatest lower bound, or infimum, of a set $$E$$ which is bounded below is defined in the same manner: The statement

$$\alpha = \inf(E)$$

means that $$\alpha$$ is a lower bound of $$E$$ and that no $$\beta$$ with $$\beta > \alpha$$ is a lower bound of $$E$$.

Let $$S = \{1, 2, 3, 4, 5 \}$$ and $$B = \{ 3, 4\}$$.

Using the two definitions above, we get that

$$B$$ is

bounded below by $$\{ 1, 2, 3\}$$ and

bounded above by $$\{4, 5\}$$.

Let $$L$$ be the set of all lower bounds of $$B$$: $$L = \{ 1, 2, 3\}$$.

L is

bounded below by $$\{ 1 \}$$ and

bounded above by $$\{3, 4, 5\}$$.

Since $$\gamma, \alpha \in S$$, we get that $$\sup(L) = 5$$ and $$\inf(B) = 1$$.

I've gone over this many times and tried to followed Rudin's instructions and definitions precisely. Even so, something is going wrong, and, as I said, I'm trying to follow his text precisely.

I would greatly appreciate it if people could please take the time to clarify this.

EDIT:

For $$\sup(L)$$, definition 1.8 gives us the following.

(i) $$\alpha$$ is one of $$\{3, 4, 5\}$$.

(ii) $$\gamma = 3, 4 < \alpha = 5$$. Therefore, $$\gamma = 3, 4$$ is not an upper bound of $$L$$.

Therefore, we get that $$\alpha = 5$$ is the least upper bound of $$L$$.

• sup(L) = 3 = inf(B) its not 5.
– aram
Commented Jan 3, 2018 at 14:18
• @Aram I know, but if you apply definition 1.8 to $L$, you get that (i) $\alpha = \{ 3, 4, 5 \}$ and (ii) $\alpha = 5 > 3, 4$. So my confusion pertains specifically to Rudin's definitions. Commented Jan 3, 2018 at 14:23
• But $3$ and $4$ are upper bounds of $L$, so $5$ doesn't fit the definition of the supremum. Commented Jan 3, 2018 at 14:37
• @DanielFischer Ahh, I see it's saying. I was misreading the definition. Thank you for the clarification. Commented Jan 3, 2018 at 14:39
• Your question's title is amusing! Commented Jan 4, 2018 at 6:56

$\sup(L) \neq 5$: this contradicts Definition 1.8.

Let $\gamma = 4$. We have that $\gamma < 5$, but $\gamma$ is an upper bound of $L$ (as you noted, $4$ is an upper bound of $L$). By definition, $\sup(L)$ must satisfy that whenever $\gamma < \sup(L)$, $\gamma$ is not an upper bound of $L$. $5$ doesn't satisfy this, so $\sup(L)$ can't be $5$.