# Proof verification from “Understanding Analysis”: $\sup B=\inf A$

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.

• Can you elaborate on "$\sup B$ is a lower bound for $A$ by definition"? – angryavian Jun 10 at 4:40
• "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 $B$.. – fleablood Jun 10 at 4:40
• @angryavian I messed that up, I think. As fleablood commented that is not necessarily the case, which invalidates this proof – csch2 Jun 10 at 4:44
• 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 $\sup$. – 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.

Attempt:

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

$$A\not = \emptyset$$, 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:

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