Properties of supremum and infimum – Serlo

Since the Supremum is applied to sets, a very obvious question is: What happens if we change the set? If we intersect it with another set, for example, or take the union with another set, or if we make it larger or smaller? Here we will learn some rules that will help you to work with the Supremum in such cases.

Overview about rules for supremum and infimum

First, let us introduce the following abbreviations:

Definition

For all sets $A,B\subseteq \mathbb {R}$ and all $\lambda \in \mathbb {R}$ we define:

• $-A:=\{-x:x\in A\}$
• $\lambda A:=\{\lambda x:x\in A\}$
• $A+B:=\{a+b:a\in A,b\in B\}$
• $A\cdot B:=\{a\cdot b:a\in A,b\in B\}$

Now, for the supremum and infimum the following rules apply (where $A,B,D\subseteq \mathbb {R}$ and $f,g:D\rightarrow \mathbb {R}$ and $\lambda \in \mathbb {R}$). In the following it is always assumed that the supremum or the infimum exists.

Rules for the supremum

• $\sup A\geq \inf A$
• $A\subseteq B\Rightarrow \sup A\leq \sup B$

• $\sup(A\cup B)=\max\{\sup A,\sup B\}$
• $\sup(A\cap B)\leq \min\{\sup A,\sup B\}$
• $\sup(-A)=\sup(\{-x:x\in A\})=-\inf(A)$
• $\sup(\{x^{-1}:x\in A\})=(\inf(A))^{-1}$, if $\inf(A)>0$ holds.
• $\sup(\lambda A)=\sup(\{\lambda x:x\in A\})=\lambda \cdot \sup(A)$ for $\lambda \geq 0$
• $\sup(A+B)=\sup(\{x+y:x\in A\land y\in B\})=\sup(A)+\sup(B)$
• $\sup(A\cdot B)=\sup(\{x\cdot y:x\in A\land y\in B\})=\sup(A)\cdot \sup(B)$, if $A$ and $B$ contain only non-negative elements.
• $\sup(f+g)(D)\leq \sup f(D)+\sup g(D)$
• There is a sequence $(a_{n})_{n\in \mathbb {N} }$ in $A$ with $\lim _{n\rightarrow \infty }a_{n}=\sup A$.

Question: Why does $\sup A\leq \sup B\Rightarrow A\subseteq B$ not hold? Find a counterexample!

A counterexample is $A:=\{1\},B:=\{2\}$.

Question: Why does $\sup(A\cap B)=\min\{\sup A,\sup B\}$ not hold? Find a counterexample!

Let $A=\{1,2\}$ and $B=\{1,3\}$. Then $A\cap B=\{1\}$ and hence $\sup A\cap B=1$, but $\sup A=2$ and $\sup B=3$, so $\min\{\sup A,\sup B\}=2$.

The supremum of a sum of two functions might be smaller than the sum of its suprema.

Question: Why does $\sup(f+g)(D)=\sup f(D)+\sup g(D)$ not hold? Find a counterexample!

We set $D:=[0,1]$. As a function, we choose $f:D\to \mathbb {R} ,x\mapsto x$ and $g:D\to \mathbb {R} ,x\mapsto 1-x$. Then $(f+g):D\to \mathbb {R} ,x\mapsto 1$, so

$\sup(f+g)(D)=1<1+1=\sup f(D)+\sup g(D)$

Rules for the infimum

• $\inf A\leq \sup A$
• $A\subseteq B\Rightarrow \inf A\geq \inf B$
• $\inf(A\cup B)=\min\{\inf A,\inf B\}$
• $\inf(A\cap B)\geq \max\{\inf A,\inf B\}$
• $\inf(-A)=\inf(\{-x:x\in A\})=-\sup(A)$
• $\inf(\lambda A)=\inf(\{\lambda x:x\in A\})=\lambda \cdot \inf(A)$ for $\lambda \geq 0$
• $\inf(A+B)=\inf(\{x+y:x\in A\land y\in B\})=\inf(A)+\inf(B)$
• $\inf(A\cdot B)=\inf(\{x\cdot y:x\in A\land y\in B\})=\inf(A)\cdot \inf(B)$, if $A$ and $B$ contain only non-negative elements.
• $\inf(f+g)(D)\geq \inf f(D)+\inf g(D)$
• There is a sequence $(a_{n})_{n\in \mathbb {N} }$ in $A$ with $\lim _{n\rightarrow \infty }a_{n}=\inf A$.

Proof of the rules

In the following sections we will prove the above properties only for the Supremum. The infimum is treated analogously.

Supremum is greater/equal to the infimum

Theorem

Let $A$ be a non-empty, bounded set. Then, there is $\sup A\geq \inf A$.

How to get to the proof?

This is quite an intuitive statement: The supremum is an upper bound for all elements and the infimum a lower bound. So the supremum should be above the infimum. The mathematical proof is simple, as well: upper bounds are above and lower bounds below any element. So we just pick one element and have that supremum is above and the infimum below it (and hence below the supremum)

Proof

The set $A$ is non-empty, so it must contain an element, say $x$. Since $\sup A$ is an upper bound, there is $x\leq \sup A$. Analogously $\inf A\leq x$. That means $\inf A\leq x\leq \sup A$ and hence $\inf A\leq \sup A$.

Estimating the supremum of subsets

Theorem

In case $A\subseteq B$, there is $\sup A\leq \sup B$.

Proof

By definition of the supremum, $\sup B$ must be in upper bound of $B$. Since $A\subseteq B$ , it is also an upper bound of $A$, i.e. for all $x\in A$ there is $x\leq \sup B$. The supremum of $A$ is now the exactly the smallest upper bound of $A$. So it must be lower or equal to $\sup B$ .

Supremum and union

Theorem

There is

$\sup(A\cup B)=\max\{\sup A,\sup B\}$

Proof

If $x\in A\cup B$, then there is $x\in A$ or $x\in B$. By definition of the supremum $x\leq \sup A$ or $x\leq \sup B$. In particular, that means $x\leq \max\{\sup A,\sup B\}$. Therefore $\max\{\sup A,\sup B\}$ is an upper bound of $A\cup B$.

But is it also the "smallest" upper bound? We always have $\sup A\leq \sup B$ or $\sup B\leq \sup A$. In the firts case, for all $x\in A$ we have $x\leq \sup A$ (the supremum is an upper bound), and since $\sup A\leq \sup B$ also $x\leq \sup B$. So $\sup B$ is both an upper bound for $A$ and $B$, or in other words, it is an upper bound of $A\cup B$. It must be the smallest upper bound, since we cannot choose some upper bound smaller than $\sup B$ being above all elements in $B$ (or $A\cup B$). So in the first case, $\sup B=\max\{\sup A,\sup B\}$ is the smallest upper bound. In the second case, we just swap the names of $A$ and $B$, which exactly leads us to the first case and finishes the proof.

Supremum and intersection

Theorem

There is

$\sup(A\cap B)\leq \min\{\sup A,\sup B\}$

Proof

There is $A\cap B\subseteq A$ and $A\cap B\subseteq B$, so by "Estimating the supremum of subsets" we have $\sup(A\cap B)\leq \sup A$ and $\sup(A\cap B)\leq \sup B$, which means $\sup A\cap B\leq \min\{\sup A,\sup B\}$. This is already the proof. Be careful: there are cases, where the above inequality is strict, i.e. the supremum is NOT the minimum!

Supremum and multiplication with $-1$

Theorem

There is

$\sup(-A)=\sup(\{-x:x\in A\})=-\inf(A)$

Proof

For all $x\in A$ there is $\inf(A)\leq x$. Multiplication of the inequality with $-1$ gives exactly $-x\leq -\inf(A)$. But since all elements of $-A$ are of this form, $-\inf(A)$ is an upper bound for $-A$. However, since the supremum of $-A$ is the smallest of all upper bounds, $\sup(-A)\leq -\inf(A)$ follows.

Consider now any small $\varepsilon >0$. According to the definition of the infimum, $\inf(A)+\varepsilon$ is then no longer a lower bound of $A$. This means that some $x\in A$ exists with $x<\inf(A)+\varepsilon$. Multiplying this inequality by $-1$ gives $-x>-\inf(A)-\varepsilon$. But now, $-x$ is an element of $-A$, so $-\inf(A)-\varepsilon$ cannot be an upper bound of $-A$. Since our $\varepsilon$ was arbitrary, we have that $\sup(-A)=-\inf(A)$ is indeed the smallest upper bound.

Hint

From this rule, we get that under multiplication with $-1$, supremum and infimum interchange like $\sup(A)=\sup(-(-A))=-\inf(-A)$ and $\inf(A)=-(-\inf(A))=-\sup(-A)$.

Supremum and multiplication with a (non-negative) number

Theorem

For $\lambda \geq 0$ there is

$\sup(\lambda A)=\sup(\{\lambda x:x\in A\})=\lambda \sup(A)$

Proof

If $\lambda =0$ there is not much to show, because $\lambda \cdot A=0\cdot A=\{0\}$ and $\sup(\{0\})=0=0\sup(A)$.

We can therefore assume that $\lambda >0$. For all $x\in A$ we can assume $x\leq \sup(A)$. Multiplication of the inequality with $\lambda$ gives even $\lambda x\leq \lambda \sup(A)$. But since all elements of $\lambda \cdot A$ are of this form, $\lambda \sup(A)$ is an upper bound for $\lambda \cdot A$. However, since the supremum of $\lambda \cdot A$ is the smallest of all upper bounds, $\sup(\lambda \cdot A)\leq \lambda \sup(A)$ follows (i.e. we have an upper bound).

Do we also have the smallest upper bound? Let $\varepsilon >0$ be given and define $\varepsilon ':={\frac {\varepsilon }{\lambda }}$. This is allowed because we assume $\lambda >0$. According to the definition of the supremum, $\sup(A)-\varepsilon '$ is then no upper bound of $A$. This means that a $x\in A$ exists so that $x>\sup(A)-\varepsilon '$. Multiplying this inequality by $\lambda$ gives

$\lambda x>\lambda \sup(A)-\lambda \varepsilon '=\lambda \sup(A)-\lambda {\frac {\varepsilon }{\lambda }}=\lambda \sup(A)-\varepsilon$

But now $\lambda x$ is an element of $\lambda \cdot A$, so $\lambda \sup(A)-\varepsilon$ cannot be an upper bound of $\lambda \cdot A$. Since $\varepsilon$ is arbitrary, we know that equality $\sup(\lambda \cdot A)=\lambda \sup(A)$ holds and we have indeed found the smallest upper bound.

Supremum and sums

Theorem

There is

$\sup(A+B)=\sup(\{x+y:x\in A\land y\in B\})=\sup(A)+\sup(B)$

Proof

Each element $z\in A+B$ has the form $z=x+y$ for some $x\in A$ and some $y\in B$. According to the definition of the supremum, there is $x\leq \sup(A)$ and $y\leq \sup(B)$. Adding the two inequalities gives $z=x+y\leq \sup(A)+\sup(B)$. So $\sup(A)+\sup(B)$ is an upper bound for $A+B$. But since the supremum of $A+B$ is the smallest of all upper bounds, $\sup(A+B)\leq \sup(A)+\sup(B)$.

Now, is $\sup(A)+\sup(B)$ also the smallest upper bound of $A+B$? Let $\varepsilon >0$ be given and define $\varepsilon ':={\frac {\varepsilon }{2}}$. According to the definition of the supremum $\sup(A)-\varepsilon '$ is then no upper bound of $A$ and $\sup(B)-\varepsilon '$ no upper bound of $B$. This means that an $x\in A$ and a $y\in B$ exist such that $x>\sup(A)-\varepsilon '$ and $y>\sup(B)-\varepsilon '$ hold. By adding the two inequalities one obtains

$x+y>\sup(A)+\sup(B)-2\varepsilon '=\sup(A)+\sup(B)-2{\frac {\varepsilon }{2}}=\sup(A)+\sup(B)-\varepsilon$

But now, $x+y$ is an element of $A+B$, so $\sup(A)+\sup(B)-\varepsilon$cannot be an upper bound of $A+B$. Since $\varepsilon$ is arbitrary, we know that equality $\sup(A+B)=\sup(A)+\sup(B)$ holds, so we have indeed found the smallest upper bound.

Supremum and products

Theorem

If $A$ and $B$ only contain non-negative elements, we have that

$\sup(A\cdot B)=\sup(\{x\cdot y:x\in A\land y\in B\})=\sup(A)\cdot \sup(B)$

Proof

Each element $z\in A\cdot B$ has the form $z=x\cdot y$ for some $x\in A$ and some $y\in B$. According to the definition of the supremum $x\leq \sup(A)$ and $y\leq \sup(B)$. Multiplying the two inequalities gives $z=x\cdot y\leq \sup(A)\cdot \sup(B)$. So $\sup(A)\cdot \sup(B)$ is an upper bound for $A\cdot B$. But since the supremum of $A\cdot B$ is the smallest of all upper bounds, $\sup(A\cdot B)\leq \sup(A)\cdot \sup(B)$ follows.

Mind the special case $\sup(A)=0$ or $\sup(B)=0$, which implies $A=\{0\}$ or $B=\{0\}$, because all elements from $A$ and $B$ were assumed to be non-negative, i.e. greater than or equal to zero. Here, we immediately have $\sup(A\cdot B)=0$.

In the following, we can hence assume $\sup(A)>0$ and $\sup(B)>0$. Now, have we also found the smallest upper bound? Let $\varepsilon >0$ be given. Then, following the previous theorems, we can define $\varepsilon _{A}:={\frac {\varepsilon }{2\sup(B)}}$ and $\varepsilon _{B}:={\frac {\varepsilon }{2\sup(A)}}$.

According to the definition of the supremum, $\sup(A)-\varepsilon _{A}$ is no upper bound of $A$ and $\sup(B)-\varepsilon _{B}$ is no upper bound of $B$. This means that an $x\in A$ and a $y\in B$ exist such that $x>\sup(A)-\varepsilon _{A}$ and $y>\sup(B)-\varepsilon _{B}$. Multiplying both inequalities, we obtain

$x\cdot y>(\sup(A)-\varepsilon _{A})\cdot (\sup(B)-\varepsilon _{B})=\sup(A)\cdot \sup(B)-{\frac {\varepsilon }{2\sup(B)}}\cdot \sup(B)-{\frac {\varepsilon }{2\sup(A)}}\cdot \sup(A)+\varepsilon _{A}\cdot \varepsilon _{B}=\sup(A)\cdot \sup(B)-{\frac {\varepsilon }{2}}-{\frac {\varepsilon }{2}}+\varepsilon _{A}\cdot \varepsilon _{B}=\sup(A)\cdot \sup(B)-\varepsilon +\varepsilon _{A}\cdot \varepsilon _{B}>\sup(A)\cdot \sup(B)-\varepsilon$

Carefully note the $>$ sign in the last step. Here $\varepsilon _{A}\cdot \varepsilon _{B}>0$ was used. Now $x\cdot y$ is an element in $A\cdot B$, so $\sup(A)\cdot \sup(B)-\varepsilon$ cannot be an upper bound of $A\cdot B$. Since our $\varepsilon$ was arbitrary, the desired equality $\sup(A\cdot B)=\sup(A)\cdot \sup(B)$ holds and we indeed have the smallest upper bound.

Supremum of the sum is smaller/equal the sum of suprema

Theorem

There is $\sup(f+g)(D)=\sup(\{f(t)+g(t):t\in D\})\leq \sup f(D)+\sup g(D)$

Proof

There is

${\begin{aligned}&\sup(f+g)(D)\\[0.3em]=\ &\sup(\{f(t)+g(t):t\in D\})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \sup A\leq \sup B{\text{ for }}A\subseteq B\right.}\\[0.3em]\leq \ &\sup(\{f(t)+g(s):t,s\in D\})\\[0.3em]=\ &\sup(\{a+b:a\in f(D),b\in g(D)\})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \sup(A+B)=\sup A+\sup B\right.}\\[0.3em]=\ &\sup f(D)+\sup g(D)\end{aligned}}$

Existence of a sequence $(a_{n})_{n\in \mathbb {N} }$ in $A$ with $\lim _{n\to \infty }a_{n}=\sup A$

Theorem

If $\sup A$ exists, then there is a sequence $(a_{n})_{n\in \mathbb {N} }$ in $A$ with $\lim _{n\to \infty }a_{n}=\sup A$.

Proof

We recall the second part of the definition of the supremum, the epsilon definition: For all $\epsilon >0$ there is an $a\in A$ with $a+\epsilon >\sup A$.

Consequently, for all $n\in \mathbb {N}$ there is an $a_{n}\in A$ with $a_{n}+{\tfrac {1}{n}}>\sup A$. Since all $a_{n}$ are from $A$, $a_{n}\leq \sup A$ also applies. Thus for all $n\in \mathbb {N}$ we have:

${\frac {1}{n}}>\sup A-a_{n}\geq 0$

By the squeeze theorem, we then have $\lim _{n\to \infty }a_{n}=\sup A$.