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Properties of supremum and infimum – Serlo

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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

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First, let us introduce the following abbreviations:

Definition

For all sets and all we define:

Now, for the supremum and infimum the following rules apply (where and and ). In the following it is always assumed that the supremum or the infimum exists.

Rules for the supremum

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  • , if holds.
  • for
  • , if and contain only non-negative elements.
  • There is a sequence in with .

Question: Why does not hold? Find a counterexample!

A counterexample is .

Question: Why does not hold? Find a counterexample!

Let and . Then and hence , but and , so .

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

Question: Why does not hold? Find a counterexample!

We set . As a function, we choose and . Then , so

Rules for the infimum

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  • for
  • , if and contain only non-negative elements.
  • There is a sequence in with .

Proof of the rules

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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

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Theorem

Let be a non-empty, bounded set. Then, there is .

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 is non-empty, so it must contain an element, say . Since is an upper bound, there is . Analogously . That means and hence .

Estimating the supremum of subsets

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Theorem

In case , there is .

Proof

By definition of the supremum, must be in upper bound of . Since , it is also an upper bound of , i.e. for all there is . The supremum of is now the exactly the smallest upper bound of . So it must be lower or equal to .


Supremum and union

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Theorem

There is

Proof

If , then there is or . By definition of the supremum or . In particular, that means . Therefore is an upper bound of .

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

Supremum and intersection

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Theorem

There is

Proof

There is and , so by "Estimating the supremum of subsets" we have and , which means . 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

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Theorem

There is

Proof

For all there is . Multiplication of the inequality with gives exactly . But since all elements of are of this form, is an upper bound for . However, since the supremum of is the smallest of all upper bounds, follows.

Consider now any small . According to the definition of the infimum, is then no longer a lower bound of . This means that some exists with . Multiplying this inequality by gives . But now, is an element of , so cannot be an upper bound of . Since our was arbitrary, we have that is indeed the smallest upper bound.

Hint

From this rule, we get that under multiplication with , supremum and infimum interchange like and .

Supremum and multiplication with a (non-negative) number

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Theorem

For there is

Proof

If there is not much to show, because and .

We can therefore assume that . For all we can assume . Multiplication of the inequality with gives even . But since all elements of are of this form, is an upper bound for . However, since the supremum of is the smallest of all upper bounds, follows (i.e. we have an upper bound).

Do we also have the smallest upper bound? Let be given and define . This is allowed because we assume . According to the definition of the supremum, is then no upper bound of . This means that a exists so that . Multiplying this inequality by gives

But now is an element of , so cannot be an upper bound of . Since is arbitrary, we know that equality holds and we have indeed found the smallest upper bound.

Supremum and sums

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Theorem

There is

Proof

Each element has the form for some and some . According to the definition of the supremum, there is and . Adding the two inequalities gives . So is an upper bound for . But since the supremum of is the smallest of all upper bounds, .

Now, is also the smallest upper bound of ? Let be given and define . According to the definition of the supremum is then no upper bound of and no upper bound of . This means that an and a exist such that and hold. By adding the two inequalities one obtains

But now, is an element of , so cannot be an upper bound of . Since is arbitrary, we know that equality holds, so we have indeed found the smallest upper bound.

Supremum and products

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Theorem

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

Proof

Each element has the form for some and some . According to the definition of the supremum and . Multiplying the two inequalities gives . So is an upper bound for . But since the supremum of is the smallest of all upper bounds, follows.

Mind the special case or , which implies or , because all elements from and were assumed to be non-negative, i.e. greater than or equal to zero. Here, we immediately have .

In the following, we can hence assume and . Now, have we also found the smallest upper bound? Let be given. Then, following the previous theorems, we can define and .

According to the definition of the supremum, is no upper bound of and is no upper bound of . This means that an and a exist such that and . Multiplying both inequalities, we obtain

Carefully note the sign in the last step. Here was used. Now is an element in , so cannot be an upper bound of . Since our was arbitrary, the desired equality holds and we indeed have the smallest upper bound.

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

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Theorem

There is

Proof

There is

Existence of a sequence in with

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Theorem

If exists, then there is a sequence in with .

Proof

We recall the second part of the definition of the supremum, the epsilon definition: For all there is an with .

Consequently, for all there is an with . Since all are from , also applies. Thus for all we have:

By the squeeze theorem, we then have .