# Show that sup B = inf A [duplicate]

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

What I gather from this is that $A=[b,\infty)$ and $B=(- \infty, b]$, since, as the problem states, $\inf A=\sup B$.

What I would like help with is how do you show that?

And part b: Use this to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness.

• How do you gather that $A=[b,\infty)$ and $B=(-\infty,b]$ ? What is $b$ here? How do you get that both $A$ and $B$ are intervals? They may not be connected.
– R_D
Jan 26, 2016 at 6:12
• Clearly, $\sup B \le \inf A$. [Later: Okay, maybe "clearly" isn't the right word; prove that you can't have $\sup B > \inf A$, using their definitions.] If they're distinct, then what can you say about their average? Jan 26, 2016 at 6:12
• @Jabernet, Carl's comment and the definitions of inf and sup is all you need. Can you try for yourself from that? Also take note that $A$ and $B$ need not be intervals.
– R_D
Jan 26, 2016 at 6:18
• – R_D
Jan 26, 2016 at 6:26
• @Jabernet, What you have said is correct. Now $A\cap B$ can at most have one point. Do you see the contradiction if it has say 2 points?
– R_D
Jan 26, 2016 at 6:46

You'll need some kind of statement that guarantees the completeness. It could be the axiom of greatest lower bound, but it could be expressed in another way.

Note that you can't conclude that $A=[b,\infty)$ form the prerequisites. There's nowhere stated that it should be an interval. On the other hand we can conclude that $B$ is an interval, that is either $B=(-\infty,b)$ or $B=(-\infty,b]$ for some $b$.

One should of course also note that both the existense of $\sup B$ and $\inf A$ depends on the axiom of greatest lowest bound or some other statement that guarantees completeness. If we're guaranteed their existence the proof would become trivial since $\sup B\le\inf A$ (because all elements in $B$ are less or equals elements of $A$) and $\sup B\ge\inf A$ (because that would mean there's an element in $B$ that is larger than some element in $A$).

Now to prove this we use that $A$ is bounded below, that is there's an lower bound $b_0\in B$. Also we have that $A$ is non-empty that is we have an $a_0\in A$. Now we have $b_0<a_0$.

We will now build sequences $a_j$ and $b_j$. Now consider $(a_j + b_j)/2$, either it's a lower bound to $A$ or it isn't. If it is we set $b_{j+1} = (a_j+b_j)/2$ and $a_{j+1} = a_j$, otherwise $a_{j+1} = (a_j+b_j)/2$ and $b_{j+1} = b_j$.

Now it's obvious that $b_j$ are all lower bounds to $A$, but no $a_j$ is (note that $a_j$ may not be an element of $A$, but that's not a problem). Since the difference $b_j-a_j$ halfes every step we have $a_j-b_j = (a_0-b_0)2^{-j}$.

Now we would use the statement that guarantees the completeness and conclude that there's precisely one $b$ such that $b_j<b<a_j$ for all $j$.

One statement that guarantees the completeness would of course be that any two sequences, a lower that is increasing and a upper that is decreasing with the property that the difference becoming sufficiently small given that the index is large enough. Then it becomes obvious.

Another is to use the statement that any Cauchy-sequence is convergent. We then form the alternating sequence $a_0$, $b_0$, $a_1$, $b_1$, etc and show that this is an Cauchy sequence. We then show that it's limit must be $b=\sup B=\inf A$.