How to prove existence of a supremum or infimum – "Math for Non-Geeks"

General procedure

How do we find the supremum/infimum of a set? There are some strategies:

1. Visualize the set: What does the set look like? Try to draw it. You may use a paper, a computer or your imagination!
2. Make a hypothesis: Is the set bounded from above? If yes, what is a suitable candidate for a supremum? And why? If not, why is it unbounded? And what about the supremum?
3. Find a proof for the supremum/ infimum: Now, how can you show that the suspected supremum/ infimum is indeed one? Or that there is no supremum or infimum due to unboundedness? Take a piece of paper and try to construct a proof. Perhaps, within the proof, you may notice that your first step was wrong for some reason. This is a good sign: detecting your own wrong thoughts takes you closer to the actual truth.
4. Write down the proof: If you have a plan how to make a proof, formulate it in a short and understandable way - such that anyone who reads it and doesn't know the solution gets the answer as quickly and nicely as possible.

General structure of proof

Proofs that a set has a supremum/ infimum generally follow some patterns, which we want to list here:

Supremum: structure of proof

In order to show that a number $s$ is the supremum of a set $M$ , you may proceed as follows:

1. Prove that $s$ is an upper bound of $M$ : This is done by showing $y\le s$ for all $y\in M$.
2. Prove that no number $x<s$ is an upper bound of $M$: Take any $x<s$ and show that there is a $y\in M$ with $y>x$.

Infimum: structure of proof

A proof that $\tilde{s}$ is the infimum of a set $M$ could look as follows:

1. Prove that $\tilde{s}$ is a lower bound of $M$ : This is done by showing $y\ge \tilde{s}$ for all $y\in M$.
2. Prove that no number $x>\tilde{s}$ is a lower bound of $M$ : Take any $x>\tilde{s}$ and show that there is a $y\in M$ with $y<x$.

Maximum: structure of proof

The proof makes use of the definition of the maximum:

1. Prove that $m$ is an upper bound of $M$: This is done by showing $y\le m$ for all $y\in M$.
2. Prove that $m\in M$.

Minimum: structure of proof

In order to show that $\tilde{m}$ is the minimum of $M$ , you may proceed as for the maximum:

1. Prove that $\tilde{m}$ is a lower bound of $M$: This is done by showing $y\ge \tilde{m}$ for all $y\in M$.
2. Prove that $\tilde{m}\in M$.

Examples for determining supremum/infimum

Finite sets

With finite nonempty sets of real numbers, determining the infimum and supremum is simple. These sets always have a maximum (greatest number) and a minimum (smallest number). The maximum of the set is automatically its supremum, and the minimum is automatically its infimum.

Example (supremum and infimum of a finite set)

Consider the set $M=\{23,42,-1,5\}$. Its maximum is $42$ and its minimum $-1$. The element $42$ is included within the set and no element is greater than $42$. So$42$ is the maximum. Analogously, the minimum is $-1$.

As $42$ is a maximum, it is also a supremum of $M$ . And analogously, $-1$ is an infimum.

Exercise (for understanding): Determine the supremum and the infimum of the following sets:

1. $M_1=\{4,72,-5,99,42\}$
2. $M_2=\{(-1{)}^k\cdot 2^{-k}:1\le k\le 5\wedge k\in \mathbb{N}\}$
3. $M_3=\{\pi ,\sqrt{2},1\}\cap \{\sqrt{2},-1,4\}$

Solution:

1. The supremum of $M_1$ is its biggest element $99$ . The infimum is the smallest element $-5$.
2. There is $\{(-1{)}^k\cdot 2^{-k}:1\le k\le 5\wedge k\in \mathbb{N}\}=\{-\frac{1}{2},\frac{1}{4},-\frac{1}{8},\frac{1}{16},-\frac{1}{32}\}$. So the supremum is $\frac{1}{4}$ and the infimum is $-\frac{1}{2}$.
3. There is $\{\pi ,\sqrt{2},1\}\cap \{\sqrt{2},-1,4\}=\{\sqrt{2}\}$. So supremum and infimum of $M_3$ are both given by $\sqrt{2}$.

Intervals

For nonempty bounded intervals, the lower endpoint is the infimum and the upper endpoint is the supremum:

Theorem (supremum and infimum of intervals)

Let $a,b\in ℝ$ with $a<b$, and let $I$ be one of the following four intervals:

1. $I=[a,b]$
2. $I=(a,b]$
3. $I=[a,b)$
4. $I=(a,b)$

In that case, $a$ is the infimum and $b$ the supremum of the interval.

How to get to the proof? (supremum and infimum of intervals)

The above intervals differ in whether the endpoints $a$, $b$ are included or not. In any case, we know that for each $x\in I$ the following holds: $a\le x\le b$. So we know that $a$ is a lower bound and $b$ is an upper bound. That means, $a\le infI$, $b\ge supI$ . We still have to show that $a$ is indeed the greatest lower bound and $b$ the smallest upper bound.

Let us assume that there is a $\tilde{a}>a$, so that $\tilde{a}$ is also a lower bound. We can show that this leads to a contradiction:

To show that $\tilde{a}$ cannot be a lower bound, we need to find an $x\in I$, so that $x<\tilde{a}$. To construct such a $x$, we take the mean value between $a$ and $\tilde{a}$, which by definition is greater than $a$. It could happen that $\tilde{a}$ is so large that the mean value becomes larger than $b$. In this case, $\tilde{a}$ is even larger than $b$, so any point of the interval is below $\tilde{a}$ and $\tilde{a}$ is not a lower bound. In any other case, the mean value between $a$ and $\tilde{a}$ is an element of the interval and it is smaller than $\tilde{a}$. So $\tilde{a}$ is not a lower bound in this case either. This is the contradiction, we have been looking for: we just showed that in any case, $\tilde{a}$ cannot be a lower bound. So $a$ is the greatest lower bound, i.e. the infimum. Showing that $b$ is a supremum works by analogous arguments.

Proof (supremum and infimum of intervals)

Let $I$ be an interval. It does not matter, here, whether the boundary points are included or not. For each $x\in I$, we have $a\le x$, so $a$ is a lower bound. We show that there is no greater lower bound than $a$ by contradiction:

Assume that $\tilde{a}>a$ also was a lower bound of $I$. Then there is no $x\in I$, such that $x<\tilde{a}$. We distinguish two cases:

Case 1: $\tilde{a}>b$ (small interval)

Let $x:=\frac{a+b}{2}$. We can estimate $\begin{array}{cc}& a=\frac{a+a}{2}\\ & \downarrow a<b\\ <& \frac{a+b}{2}=x\\ & \downarrow a<b\\ <& \frac{b+b}{2}=b\end{array}.$ So $a<x<b$ and hence $x\in I$. By $\tilde{a}>b$ we get $x=\frac{a+b}{2}<b<\tilde{a}.$ This contradicts the assumption of $\tilde{a}$ being a lower bound of $I$.

Case 2: $\tilde{a}\le b$ (large interval)

Now, take $x:=\frac{a+\tilde{a}}{2}$. By our assumptions, $\begin{array}{cc}& a=\frac{a+a}{2}\\ & \downarrow a<\tilde{a}\\ <& \frac{a+\tilde{a}}{2}=x\\ & \downarrow a<\tilde{a}\le b\\ <& \frac{b+b}{2}=b\end{array}.$ So $a<x<b$ and we have $x\in I$.

But since $a<\tilde{a}$ we have $x=\frac{a+\tilde{a}}{2}<\frac{\tilde{a}+\tilde{a}}{2}=\tilde{a}$. Again, this contradicts $\tilde{a}$ being a lower bound of $I$.

In both cases, we get a contradiction to $\tilde{a}$ being a further lower bound of $I$ . So $a$ is indeed the infimum of $I$.

The proof that $b$ is the supremum is done analogously. We assume that $\tilde{b}$ was an upper bound of $I$, which is lower than $b$ . If we consider the number $x:=\frac{a+b}{2}$ for $\tilde{b}<a$ and $x:=\frac{b+\tilde{b}}{2}$ for $\tilde{b}\ge a$, we will get a violation of this bound.

Alternatively, one may use the properties of suprema and infima and apply the following trick: We use $sup(I)=-inf(-I)$. $-b$ is an infimum of $-I$ by the above arguments, so $b$ is a supremum of $I$.

Exercise (for understanding): Determine the supremum and the infimum of the following sets:

1. $M_1=(-1,3]$
2. $M_2=\{x\in ℝ:2\le 2x\le 8\}$
3. $M_3=\{x\in ℝ:|x-1|<4\}$
4. $M_4=\{\sin (x):x\in ℝ\}$ (here you may use, what you have learned in school about the sine function )

1. The supremum is $3$ and the infimum is $-1$.
2. Since $2\le 2x\le 8⟺1\le x\le 4$ , we have that $M_2=[1,4]$. This is an interval with infimum 1 and supremum 4.
3. $M_3$ is the set of all $x$ with distance to $1$ being less than $4$ . That is an open interval $(1-4,1+4)=(-3,5)$. Its infimum is $-3$ and the supremum is $5$.
4. $M_4$ is the set of all real numbers being covered by the sine function. The sine oscillates between $-1$ and $1$. So we just have an interval $M_4=[-1,1]$. Hence, the infimum of $M_4$ is $-1$ and the supremum is $1$.

Intervals of integers

Exercise (for understanding): Now, let us consider integer boundaries $a,b\in \mathbb{Z}$ with $a<b$. And we only pick out the integer numbers of the intervals:

1. $[a,b]\cap \mathbb{Z}$
2. $(a,b]\cap \mathbb{Z}$
3. $[a,b)\cap \mathbb{Z}$
4. $(a,b)\cap \mathbb{Z}$

What are supremum and infimum now? In which case do they even exist?

It helps a lot to know that the sets are all finite:

1. $[a,b]\cap \mathbb{Z}=\{x\in \mathbb{Z}:a\le x\le b\}$
2. $(a,b]\cap \mathbb{Z}=\{x\in \mathbb{Z}:a<x\le b\}$
3. $[a,b)\cap \mathbb{Z}=\{x\in \mathbb{Z}:a\le x<b\}$
4. $(a,b)\cap \mathbb{Z}=\{x\in \mathbb{Z}:a<x<b\}$

(Be careful with open boundaries! They remove one entire integer element.) In the first 3 cases, there is $a<b$ and the set is non-empty.

1. $sup([a,b]\cap \mathbb{Z})=b$ and $inf([a,b]\cap \mathbb{Z})=a$
2. $sup((a,b]\cap \mathbb{Z})=b$ and $inf((a,b]\cap \mathbb{Z})=a+1$
3. $sup([a,b)\cap \mathbb{Z})=b-1$ and $inf([a,b)\cap \mathbb{Z})=a$

So in each of these three cases, both a supremum and an infimum exist. However, in the last case $(a,b)\cap \mathbb{Z}$ can be empty. This happens exactly if $a+1=b$. In that case, there is neither a supremum nor an infimum. We only have the improper supremum $sup\mathrm{\varnothing }=-\mathrm{\infty }$ and the improper infimum $inf\mathrm{\varnothing }=\mathrm{\infty }$. However, those are no real numbers, i.e. no proper suprema/ infima. For $a+1\ne b$, the set is non-empty, and we have supremum $b-1$ and infimum $a+1$.

Sequences

The set $M=\{5+\frac{2}{n}:n\in \mathbb{N}\}$.

Suprema and infima of sequence elements are often required in mathematics. These elements form a set with at most countably many elements. When that set is bounded above or below, it has a real-valued supremum or infimum, respectively. Let's start with an example:

Exercise (Sets of sequence elements)

What are the supremum and the infimum of the set $M=\{5+\frac{2}{n}:n\in \mathbb{N}\}$? Are they also a maximum or a minimum? Prove all your assumptions!

How to get to the proof? (Sets of sequence elements)

We follow a step-by-step approach:

Proof step: Visualize the set $M$.

The first elements of $M$ are:

1. $5+\frac{2}{1}=7⟹7\in M$
2. $5+\frac{2}{2}=6⟹6\in M$
3. $5+\frac{2}{3}=5\frac{2}{3}⟹5\frac{2}{3}\in M$
4. $5+\frac{2}{4}=5\frac{1}{2}⟹5\frac{1}{2}\in M$
5. ?

So the set $M$ is of the form $M=\{7,6,5\frac{2}{3},5\frac{1}{2},5\frac{2}{5},\dots \}$, with all further elements approaching $5$.

Proof step: Give a hypothesis what may be the infimum or the supremum.

We see that the set is bounded from above by $7$. At the same time $7$ is an element of the set, so $7$ must be the maximum of the set. Furthermore, the set is bounded from below by $5$. Since the elements of the set are getting closer and closer to $5$, there can be no lower bound greater than $5$. It follows that $5$ is probably the infimum of the set. Note that we are only making assumptions here because we are arguing intuitively. We still need a solid proof.

Proof step: Find a proof for the supremum / maximum.

We have already established that $7$ is probably the maximum. So we have to show two things:

• $7\in M$
• $7\ge x$ for all $x\in M$

The number $7$ is definitely an element of $M$ , namely for $n=1$ there is $5+\frac{2}{n}=7$. In order to prove that $7$ is an upper bound of $M$ , we need to show $7\ge 5+\frac{2}{n}$. This can be achieved by equivalent transformations:

$\begin{array}{cc}7& \ge 5+\frac{2}{n}\\ 2& \ge \frac{2}{n}\\ 2n& \ge 2\\ n& \ge 1\end{array}$

Now $n\ge 1$ is an obvious statement for natural numbers. But in the proof we must go the opposite way: Since we want to show the inequality $7\ge 5+\frac{2}{n}$, we have to start at $n\ge 1$ and gradually transform this inequality into $7\ge 5+\frac{2}{n}$. This is feasible because we have only used equivalent transformations above.

In the last chapter, we have seen that every maximum of a set is automatically the supremum of the set (only the other way round it is not always the case). From this, it follows that $7$ is the supremum of $M$.

Proof step: Find a proof of the infimum / minimum.

To prove that $5$ is an infimum, we have to show

• $5\le x$ for all $x\in M$
• For all $y>5$ there is an $x\in M$ with $x<y$

In order to show that in addition, $5$ is not a minimum, we also have to prove that $5\notin M$. First, we have to find a proof to $5\le 5+\frac{2}{n}$ for all $n\in \mathbb{N}$:

$\begin{array}{cc}5& \le 5+\frac{2}{n}\\ 0& \le \frac{2}{n}\end{array}$

Now $0\le \frac{2}{n}$ is an obviously true statement, since $\frac{2}{n}$ is positive. So in the later proof we can get to the inequality $5\le 5+\frac{2}{n}$ starting from $0\le \frac{2}{n}$ by performing the above equivalent transformation backwards (i.e. adding $5$ on both sides).

Let us now continue with an arbitrary (possibly better bound) $y>5$. To rule out that this is a better bound, we have to find an $x$ with $x=5+\frac{2}{N}$ and $N\in \mathbb{N}$ such that $y>x=5+\frac{2}{N}$. We choose here the variable $N$ and not $n$, because we want to find a concrete element of the set $M$ (in mathematics often $N$ is used, if we are looking for a concrete $n$). Let us reformulate this inequality to $N$ to find a matching $N\in \mathbb{N}$:

$\begin{array}{cc}y& >5+\frac{2}{N}\\ y-5& >\frac{2}{N}\\ \frac{y-5}{2}& >\frac{1}{N}\end{array}$

Because of $y>5$, we have $y-5>0$ and $\frac{y-5}{2}>0$. The Archimedean axiom now guarantees us that we can find a suitable $N$, because according to the Archimedean axiom the fraction $\frac{1}{N}$ becomes smaller than any positive real number.

Finally, we need to show that $5\notin M$. Or put in different words, $5\ne 5+\frac{2}{n}$ is valid for all $n\in \mathbb{N}$. But because $\frac{2}{n}>0$ there is $5+\frac{2}{n}>5$ and therefore $5\ne 5+\frac{2}{n}$. So this is rather simple. Now, we are ready to write down the proof:

Proof (Sets of sequence elements)

We prove that $7$ is a maximum (and therefore supremum) of the set $M$ and $5$ is an infimum but not a minimum of the set $M$.

Proof step: $7$ is a maximum of $M$

Proof step: $7$ is an element of $M$

For $n=1$ there is $5+\frac{2}{n}=7$. So $7\in M$.

Proof step: $7$ is an upper bound of $M$

For all $n\in \mathbb{N}$ there is

$\begin{array}{cccc}& & n& \ge 1\\ & ⟹& 2n& \ge 2\\ & ⟹& 2& \ge \frac{2}{n}\\ & ⟹& 7& \ge 5+\frac{2}{n}\end{array}$ So $7$ is greater or equal to any element of $M$.

Proof step: $5$ is the infimum of $M$

Proof step: $5$ is a lower bound of $M$

For all $n\in \mathbb{N}$ there is

$\begin{array}{cccc}& & 0& \le \frac{2}{n}\\ & ⟹& 5& \le 5+\frac{2}{n}\end{array}$

So $5$ is less than or equal to any element of $M$.

Proof step: No other number greater than $5$ can be a lower bound of $M$

Let $y>5$ be arbitrary. Then, $\frac{y-5}{2}>0$ and by the Archimedean axiom, there is an $N\in \mathbb{N}$ with $\frac{y-5}{2}>\frac{1}{N}$. There is

$\begin{array}{cccc}& & \frac{y-5}{2}& >\frac{1}{N}\\ & ⟹& y-5& >\frac{2}{N}\\ & ⟹& y& >5+\frac{2}{N}\end{array}$

Since $5+\frac{2}{N}\in M$ there is an element of $M$ which is smaller than $y$ . So $y$ is no lower bound of $M$.

Proof step: $5$ is not a minimum of $M$

There is $\frac{2}{n}>0$ and hence $\frac{2}{n}+5>5$. So $5$ is not an element of $M$ and hence not a minimum.

Sets of function values

Exercise

Determine the supremum and the infimum of the set

$M=\{\frac{1}{1+x^2}:x\in ℝ\}$

How to get to the proof?

Let $f:ℝ\to ℝ:x\mapsto \frac{1}{1+x^2}$. We proceed as follows:

Step 1: Visualize the set $M$.

The function $f(x)=\frac{1}{1+x^2}$ drawn in a diagram looks as follows:

The graph of the function f(x)=1/(1+x²)

The set $M$ now consists of all values being hit by the function $f$ , which is just the image of $f$.

Step 2: Make a hypothesis which numbers are the supremum and the infimum of the set.

We can assume that $1$ is the supremum of $M$. Because $f(0)=\frac{1}{1+0^2}=1$, $1$ is also hit by the function $f$. So this number is in $M$ and should therefore be the maximum of this set.

Furthermore, the assumption that $0$ is the infimum of $M$ is obvious. The function always seems to be positive, i.e. greater than zero. The greater the absolute value of $x$, the closer the function values are to zero (at least that's how it looks at first glance). So $0$ should be the infimum of $M$, whereas it is not directly in $M$ and therefore it should not be a minimum.

Step 3: Find a proof for the supremum/ maximum.

We assert that $1$ is the maximum of the set $M$. Because $\frac{1}{1+0^2}=1$, we can directly see that $1\in M$. Now all that is missing is the proof that $1$ is an upper bound of the set $M$. For this, we must prove that for all real numbers $x$ we have the following inequality:

$\frac{1}{1+x^2}\le 1$

Let us transform this inequality by means of equivalent transformations:

$\begin{array}{ccc}& \frac{1}{1+x^2}& \le 1\\ ⟺& 1& \le 1+x^2\\ ⟺& 0& \le x^2\end{array}$

We already know that the last inequality is fulfilled for all $x\in ℝ$ . Since we have only used equivalent transformations, we can later recover the inequality $\frac{1}{1+x^2}\le 1$.

Step 4: Find a proof of the infimum / minimum.

Here, we must first show that all elements of $M$ are greater than or equal to zero. However, $\frac{1}{1+x^2}$ is the quotient of two positive numbers, which must be positive again. All elements from $M$ are therefore positive, and hence greater than or equal to $0$.

Now, we need to show that $0$ is also the largest lower bound of $M$. For this, let $y>0$ be arbitrary. We have to find an element from $M$ which is smaller than $y$. So there must be a $x\in ℝ$ with

$\frac{1}{1+x^2}<y$

Let us transform this inequality by means of equivalent transformations:

$\begin{array}{ccc}⟹& \frac{1}{1+x^2}& <y\\ ⟺& 1& <y\cdot (1+x^2)\\ ⟺& 1& <y+y\cdot x^2\\ ⟺& 1-y& <y\cdot x^2\\ & & \downarrow y\text{ is positive}\\ ⟺& \frac{1-y}{y}& <x^2\end{array}$

In order to take the square root, $\frac{1-y}{y}\ge 0$ is required, i.e. $y\le 1$. But for the further proof it is no problem to assume $0<y\le 1$, because for $y>1$ the last inequality is always fulfilled. The square number $x^2$ is then always greater than the negative number $\frac{1-y}{y}$.

So consider $0<y\le 1$. Then:

$\begin{array}{ccc}& \frac{1-y}{y}& <x^2\\ ⟺& \sqrt{\frac{1-y}{y}}& <|x|\end{array}$

For $0<y\le 1$ we need to choose an $x\in ℝ$ with $|x|>\sqrt{\frac{1-y}{y}}$. This $x$ automatically fulfils

$\frac{1}{1+x^2}<y$

hence $y$ cannot be a lower bound of $M$.

Proof

We prove that $1$ is a maximum and hence a supremum of $M$ . Further, we prove that $0$ is an infimum but not a minimum of $M$.

Proof step: $1$ is the maximum of $M$

Proof step: $1$ is an element of $M$

For $x=0$ there is $\frac{1}{1+x^2}=1$. So $1\in M$.

Proof step: $1$ is an upper bound of $M$

There is

$\begin{array}{ccc}& 0& \le x^2\\ ⟹& 1& \le 1+x^2\\ ⟹& \frac{1}{1+x^2}& \le 1\end{array}$

So $1$ is an upper bound of $M$.

Proof step: $0$ is an infimum of $M$

Proof step: $0$ is a lower bound of $M$

There is for all $x\in ℝ$:

$\begin{array}{ccc}& x^2& \ge 0\\ ⟹& 1+x^2& >0\\ ⟹& \frac{1}{1+x^2}& >0\end{array}$

So 0 is a lower bound of $M$.

Proof step: No number greater than $0$ is a lower bound of $M$

Let $y>0$ be arbitrary. For $y>1$ there is

$\frac{1-y}{y}<x^2$

for any real $x$, since $1-y$ is negative, then. For $0<y\le 1$ choose $x$ such that $|x|>\sqrt{\frac{1-y}{y}}$ . Then the following inequality is satisfied.

For any $y>0$ there is at least one real number $x$ with $x^2>\frac{1-y}{y}$. For real numbers of this kind, we have

$\begin{array}{ccc}& \frac{1-y}{y}& <x^2\\ ⟹& 1-y& <y\cdot x^2\\ ⟹& 1& <y+y\cdot x^2\\ ⟹& 1& <y\cdot (1+x^2)\\ ⟹& \frac{1}{1+x^2}& <y\end{array}$

So $y$ cannot be a lower bound of $M$, which proves that $0$ is the largest lower bound of $M$.