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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 $\le$).

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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 $L$.

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  • $\begingroup$ sup(L) = 3 = inf(B) its not 5. $\endgroup$
    – aram
    Commented Jan 3, 2018 at 14:18
  • $\begingroup$ @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. $\endgroup$ Commented Jan 3, 2018 at 14:23
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    $\begingroup$ But $3$ and $4$ are upper bounds of $L$, so $5$ doesn't fit the definition of the supremum. $\endgroup$ Commented Jan 3, 2018 at 14:37
  • $\begingroup$ @DanielFischer Ahh, I see it's saying. I was misreading the definition. Thank you for the clarification. $\endgroup$ Commented Jan 3, 2018 at 14:39
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    $\begingroup$ Your question's title is amusing! $\endgroup$
    – Paramanand Singh
    Commented Jan 4, 2018 at 6:56

1 Answer 1

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$\sup(L) \neq 5$: 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 $5$.

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