Let $A$ be nonempty and bounded below, and define $B=\{b\in\textbf{R}:b\text{ is a lower bound for }A\}$. Show that $\sup B=\inf A$.

Since $\sup B$ exists by the axiom of completeness, and since $\sup B$ is a lower bound for $A$ by defintion, $\sup B\in B$ and is a maximum of $B$.

Let $i$ be a lower bound for $A$. Then $i\in B$. Since $\sup B$ is a maximum of $B$, $\sup B\geq i$. We now have that $\sup B$ is a lower bound for $A$ and that $\sup B$ is greater than or equal to any other lower bound for $A$, so we may conclude that $\sup B=\inf A$.

Is this proof correct? It seems simpler than most other proofs of this, so I would like to make sure I'm not making a mistake anywhere.

  • $\begingroup$ Can you elaborate on "$\sup B$ is a lower bound for $A$ by definition"? $\endgroup$ – angryavian Jun 10 at 4:40
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    $\begingroup$ "Since supB exists by the axiom of completeness" How do you know $B$ is bounded above? "and since supB is a lower bound for A by defintion" No, it isn't. If $\sup B\not \in B$ then $\sup B$ is not a lower bound by definition and you have no reason to assume that $\sup B$ (if it exists) is an element of $B$.. $\endgroup$ – fleablood Jun 10 at 4:40
  • $\begingroup$ @angryavian I messed that up, I think. As fleablood commented that is not necessarily the case, which invalidates this proof $\endgroup$ – csch2 Jun 10 at 4:44
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    $\begingroup$ However proving $B$ is bounded above is easy (use definitions) and proving $\sup B$ is a lower bound of $A$ (use definitions but be very precise and very careful. Once you prove those (and you must prove them the rest of your proof is good. (Except you say $\sup B$ is a maximum of $B$. Sups are maxes unless they are elements and being a max isn't as relevant as being a $\sup$. $\endgroup$ – fleablood Jun 10 at 4:45

You assumed that $B$ is bounded above. You must prove that.

But that's easy with precise definitions. Is there a number that is as large or larger than all lower bounds of $A$? Well.... if you think this through....(Hint: what about actual elements of $A$? Are they at least as large as every lower bound of $A$?)

Second you assume that $\sup B$ is an lower bound of $A$ by definition. But it is not. $\sup B$ is the least upper bound of the lower bounds of $A$ but there is nothing in the definition that says it is an actual lower bound.

But if it isn't a lower bound of $A$ then there is an $a \in A$ so that $a < \sup B$... which says what about $a$, the set $B$ and upper bounds of $B$? Is $a$ an upper bound of $B$? Are there any elements of $B$ between $a$ and $\sup B$? What does that imply?

With those hints you can indeed prove that $B$ is bounded above and that $\sup B$ is an lower bound of $A$. Once you've done that your argument that that would mean $\sup B = \inf A$ is correct.



$A,B \subset \mathbb{R}$.

$A\not = \emptyset$, a bounded below.

$B=$ { $b| b$ is a lower bound for $A $}.

Show that $\sup B = \inf A$.

1) $B\not = \emptyset$ since $A$ is bounded below .

2) $B$ is bounded above since for $b \in B$: $b\le a$ , $a \in A(\not = \emptyset)$.

3) $A$ bounded below:

$\inf A $ exists, and $\inf A \le a$, $a \in A$.

4)Since $b \le a$, for $a \in A$:

$b \le \inf A$, for $b \in B$.

4) $\inf A$ is a lower bound for $A$, hence $\inf A \in B$.

5) $B$ is bounded above, $\sup B$ exists.

6) $\sup B \le \inf A$, refer to 4).

7) Since $\inf A \in B$ we have

$\inf A \le \sup B.$

8) Combining:

$\inf A \le \sup B \le \inf A.$


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