# Proving that $\inf A=\sup B$

I have been trying to prove the following statement:

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

I know this question has been asked before, here and here. However, the first one uses a lemma that I would prefer not to use (my textbook does not mention that particular lemma, so I would like to be able to do the proof without reference to it), and the second one I did not understand. So I hope you'll forgive me for asking again.

This is what I've got so far:

First, assume that $A$ is a subset of $\mathbb{R}$ (this is implied, since the exercise is from a chapter on real numbers). The Since $A$ is non-empty and bounded below, then it can be shown from the Axiom of Completeness that $\inf A$ exists. (I realize I have to prove this, and I intend to do so, but for now just assume that $\inf A$ exists.) Let $s=\inf A$. Now, since $s$ is indeed a lower bound for $A$, $s\in B$ by definition of $B$, so $B$ is non-empty. Furthermore, for all $b\in B$, $b\le s$ (this follows from the definition of $s$, since $s$ is larger than any lower bound for $A$, and $B$ is the set of all lower bounds for $A$.) Then $B$ is upwardly bounded, and it follows from the Axiom of Completeness that $\sup B$ exists. Let $s'=\sup B$.

To show that $\inf A=\sup B$, it suffices to show that $s\le s'$ and $s'\le s$. To prove that $s\le s'$, assume (for contradiction) that $s>s'$. Then there is an element in $B$ which is greater than $s'$, namely $s$. However, this contradicts the assumption that $s'=\sup B$. Then it cannot be true that $s>s'$, and so it must be true that $s\le s'$.

This is where I get stuck. I tried assuming that $s<s'$ (again for contradiction), and then arguing that if this is so, then there exists a $b\in B$ which is larger than $s$, which would contradict the assumption that $s$ is a greatest lower bound for $A$. However, I don't think this is correct. If I could prove that $s'\in B$, then I think it would be correct, but despite my efforts, I haven't been able to.

I would very much like som pointers on how to go about proving that $s'\in B$, or any other hints or corrections that anyone might have to offer.

• $s' \in B$ is false in general. E.g., take $A = [0, \infty)$, then $B = (-\infty, 0)$ and $s' = 0 \not\in B$. But all you need is that if $s < s'$ then there exists a $b \in B$ with $s < b$ and this is immediate from your definition of $s'$ as $\sup\,B$. – Rob Arthan Jan 28 '17 at 20:31

It's truly a matter of definition and being as comfortable with them as you can be.

1) An upper (lower) bound is an element in the universe that is larger (smaller) or equal to any element in $A$. In math notation: if $x \ge a$ for all $a \in A$ then $x$ is an upper bound.

2) A sup (inf) is the least upper (greatest lower) bound, that is to say, it is an upper (lower) bound but it is the least (greatest) such bound. That means nothing smaller (larger) is an upper (lower) bound. Which means anything smaller (larger) then the sup (inf) will have an element in $A$ larger (smaller) than it. or in math notation: $\sup A \ge a$ for all $a \in A$ and for any $x < \sup A$ there is an $a \in A$ so that $x < a \le \sup A$.

Caveat: A $\sup$ of $A$ need not exist. But if $A$ is bounded and not empty and your universal set $U$ has "the least upper bound property" then $\sup A$ ($\inf A$) will exist for any non-trivial set that is bounded above (below). And $\mathbb R$ has the least upper bound property. A set $U$ has the least upper bound property if and only if it has the greatest lower bound property.

So if $A$ is bounded below and non empty, then $B = \{$ all the lower bounds of $A \}$ then $B$ is non empty and bounded above and $\sup B = \inf A$ is very straightforward to show.

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$\inf A$ exists: Because $A$ is bounded below and non-empty. As $\mathbb R$ has the least upper bound property, $\inf A$ exists.

$B$ is not empty: $\inf A$ is a lower bound of $A$. So $\inf A \in B$ so $B$ is not empty.

$B$ is bounded above by $\inf A$: Let $x > \inf A$. Then $x$ is not a lower bound of $A$ (by definition of $\inf$) So if $b \in B$ then $b$ is a lower bound of $A$ and $b \not > \inf A$ so $b \le \inf A$ for all $b \in B$. So $\inf A$ is an upperbound of $B$.

As $B \subset \mathbb R$ and $\mathbb R$ has least upper bound property then $\sup B$ exists and as $\inf A$ is an upperbound of $B$ while $\sup B$ is the least upper bound then $\inf A \ge \sup B$.

But $\sup B \ge b \in B$ and $\inf A \in B$ so $\sup B \ge \inf A$.

So $\sup B = \inf A$.

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The trick is simply not to have your eyes glaze over or to lose track of what refers to what. If you can keep track of what everything is, and what every term means, it is utterly impossible not to reach this conclusion. But I do empathize and understand how obtuse and off-putting the terminology can get.

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as per egreg's comment: If you prove the rather simple lemma:

Lemma: If $\max A$ ($\min A$) is defined to be the maximum (minimum) element in $A$, i.e. $\max A = a$ ($\min A = a$) so that $a \ge x$ ($a \le x$) for all $x \in A$. Then IF as set $A$ has a maximum (minimum) element then $A$ has a suprememum (infinum) and $\sup A = \max A$ ($\inf A = \min A$).

Then the proof is a matter of stating that as $\inf A$ is a lower bound of $A$ that $\inf A \in B$. And as $\inf A$ is the greatest lower bound and $x > \inf A \implies x \not \in B$ then $\inf A = \max B = \sup B$.

But you do have to prove the lemma. Which is basically the exact argument as in my above "lengthy" proof.

• Thank you, this was truly very helpful! Just a quick follow-up: Is it because of the Archimedean property that we are allowed to assume that there is a $y\in\mathbb{R}$ such that $x<y<\inf A$? Or does this follow from something else? – omvendtgribb Jan 29 '17 at 13:04
• In my second to last paragraph where I said "for any y so that $x < y < \inf A$, yes, that is due to the archideman principal (which to be honest, maybe I'm not allowed to assume at this point). In the third paragraph when I said "for any $x < \inf A$ there is an $a \in A$ so that $x < a \le \inf A$ is from the definition of $\inf$. – fleablood Jan 29 '17 at 17:11
• i can maybe redo to the second to last paragraph to not rely on the archimedean property. If $\inf A \ne \sup B$ then $\inf A < \sup B$ (because $\inf A$ is an lower bound so it is $\le \sup B$) and that would mean there is a $b: \inf A < b \le \sup B$ we $b$ is a lower bound but that would defie that $\inf A$ is the greatest lower bound. I'll think about that and reedit this this evening. – fleablood Jan 29 '17 at 17:23
• I don't understand the lengthy proof: if you define the infimum as the largest lower bound, then you have nothing at all to prove: $\inf A=\max B=\sup B$. – egreg Jan 30 '17 at 13:06
• $\max B$ need not exist. And why does max B = sup B if it does? Sup B need not be a lower bound. But if it's not a lower bound there is an element a of A less than it. Which means that a is larger or equal to all lower bounds meaning a is an upper bound of B that is smaller than Sup B which is a contradiction. The "lengthy" proof is simply going over the definitions and seeing what has to follow. I shorter proof is "it follows directly from definitions". But we can't assume that either inf A is in A. Or that sup B is in B. – fleablood Jan 30 '17 at 17:55

Since $s$ is the infimum of $A$, there exists $a\in A$ very close to $s$. Similarly, since $s'$ is the supremum of $B$, there exists $b\in B$ very close to $s'$. Intuitively, if $s<s'$, then we can take them so close, that $a<b$. But that would mean that $b$ is not a lower bound for $A$, contradicting the definition of $B$. To prove rigourously that we can find such $a$ and $b$, we can take $\varepsilon<<s'-s$, say, $\varepsilon=(s'-s)/100$ and use the definitions of supremum and infimum for this epsilon.

Here is a late and alternative proof, which still might help the OP or someone else.% \require{begingroup} \begingroup \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}{\\ #1 \quad & \quad \{\:} \newcommand{\hints}{\mbox{#1} \\ \quad & \quad \phantom{\{\:} } \newcommand{\hint}{\mbox{#1} \:\} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} %

Writing out your statement formally, expanding $\;B\;$ and the definition of lower bound, you want to prove $$\tag{0} \sup(\{b \mid \langle \forall x \in A :: b \le x \rangle\}) \;=\; \inf(A)$$ (Why did I expand the definitions of $\;B\;$ and of lower bound? To make more visible that this problem is essentially one of simplifying the left hand side to the right hand side.)

What you not mention explicitly, is your formal definitions of $\;\inf\;$ and $\;\sup\;$. So let me choose the simplest definitions I know: for any non-empty lower-bounded $\;A\;$ and non-empty upper-bounded $\;D\;$, for all $\;b,c\;$, \begin{align} \tag{1i} b \le \inf(A) \;\equiv\; \langle \forall x \in A :: b \le x \rangle \\ \tag{1s} \sup(D) \le c \;\equiv\; \langle \forall x \in D :: x \le c \rangle \\ \end{align} In this context, we have the following helpful related ordering rules: \begin{align} \tag{2i} \langle \forall b :: b \le v \equiv b \le w \rangle \;\equiv\; v = w \\ \tag{2s} \langle \forall c :: v \le c \equiv w \le c \rangle \;\equiv\; v = w \\ \end{align} or in words: equality is equivalent to having the same lower bounds (resp. upper bounds). Also, we have the 'single direction' equivalents of these: \begin{align} \tag{3i} \langle \forall b :: b \le v \then b \le w \rangle \;\equiv\; v \le w \\ \tag{3s} \langle \forall c :: v \le c \then w \le c \rangle \;\equiv\; w \le v \\ \end{align}

With this background the proof becomes straightforward.

We will try to simplify the left hand side of $\Ref{0}$ to reach the right hand side. And the shape of the LHS suggests to use definition $\Ref{1s}$, and therefore we investigate the upper bounds of the LHS. For any $\;c\;$, $$\calc \sup(\{b \mid \langle \forall x \in A :: b \le x \rangle\}) \;\le\; c \op\equiv\hints{definition \Ref{1i} -- this introduces our goal \;\inf(A)\;, and }\hint{allows us to prove non-emptyness and boundedness} \sup(\{b \mid b \le \inf(A)\}) \;\le\; c \op\equiv\hints{definition \Ref{1s}, allowed because \;\sup\;'s argument is non-empty}\hint{upper-bounded (witness: \;\inf(A)\;) -- as suggested above} \langle \forall b :: b \le \inf(A) \;\then\; b \le c \rangle \op\equiv\hint{property \Ref{3i}} \inf(A) \le c \endcalc$$

Therefore by $\Ref{2s}$ we have proved $\Ref{0}$.

Let $s=\inf A$ (we assume it exists, of course). Then $s$ is a lower bound for $A$ by definition. Hence $s\in B$.

If we prove that $s\ge b$, for every $b\in B$, we are done, because this proves $s$ is the maximum for $B$, so its supremum. Note the emphasis on if.

Let $b\in B$. Suppose $b>s$ and set $\varepsilon=b-s$. By definition of infimum, there is $a\in A$ such that $a<s+\varepsilon$. Therefore $a<s+(b-s)=b$, contradicting that $b$ is a lower bound for $A$. Hence $b\le s$.

To clarify, I assume that you define the infimum of $A$ to be a number $s$ satisfying the two properties

1. $s$ is a lower bound for $A$
2. no number $s'>s$ is a lower bound for $A$

Property 2 can be stated in terms of addition as “for every $\varepsilon>0$, there exists $a\in A$ with $a<s+\varepsilon$”.

Now look at the above proof. Property 1 is used to show that $s\in B$. Now we have to use also property 2. If $b\in B$, we cannot have $b>s$, otherwise this contradicts $b$ being a lower bound for $A$.

Thus $s$ is the maximum lower bound for $A$. The maximum of the set $B$ obviously coincides with its supremum.

• Thank you for answering my question! I do have some follow-up questions: 1) Where did I conclude too early? 2) How did you decide on setting $\varepsilon=b-s$? I understand why it works, I just wouldn't have thought of it myself. Is this a standard technique? – omvendtgribb Jan 28 '17 at 22:15
• @omvendtgribb When you say that $s\ge b$ for all $b\in B$, you are using what you want to prove, namely that the infimum of $A$ is the maximum lower bound. I assume your definition of infimum is that $s$ is a lower bound and that, for every $\varepsilon>0$ there is $a\in A$ with $a<s+\varepsilon$. That choice is indeed standard: the hypothesis $s<b$ is the same as $b-s>0$, and so we can apply the second property of the infimum. – egreg Jan 28 '17 at 22:22
• If someone is able to spot where the answer is incorrect, I'll be grateful. – egreg Jan 30 '17 at 11:21