Infimum and Supremum

The number of images of the function shown is limited, so the function is also limited.

In mathematics , the terms supremum and infimum as well as the smallest upper bound and the largest lower bound appear when investigating semi-ordered sets . The supremum is clearly an upper bound that is smaller than all other upper bounds. Correspondingly, the infimum is a lower bound that is greater than all other lower bounds. If a supremum or infimum exists, it is clearly determined. The concept is used in various modifications in almost all mathematical sub-areas .

Definitions

Suprema (and Infima) of sets

Intuition

The supremum is the smallest upper bound of a set.

The supremum (in German "supreme") of a set is related to the maximum of a set and is - clearly speaking - an element that lies "above" all or "beyond" (above) all other elements. The expression “ above the other” is intended to indicate that the supremum does not have to be the greatest element “ among the others”, but can also be outside ( “beyond” ) the crowd. And because there can be several elements that correspond to this view, for reasons of clarity the smallest element is chosen which has this property; the element, so to speak, which is “closest” or “immediately” above all others - the supremum thus denotes something “immediately above”. Elements that are above all elements of a set, but not necessarily in a direct way, are called upper bounds . This then results in the definition of the supremum as the smallest upper bound of a set.

The infimum (German "lower limit") of a set is defined analogously, as "immediately below " or the largest lower limit .

In real life

This view can easily be transferred to sets of real numbers (as subsets of the real numbers): Let

${\displaystyle X:=\{x\in \mathbb {R} :x<2\}\subseteq \mathbb {R} }$

the set of real numbers less than 2. Then 2 is the supremum of (in ). Because 2 is an upper bound of , since it is greater than or equal (actually actually greater) than every element of - that is, it is "above". But in contrast to the number 4, which is also an upper bound, there is no number smaller than 2, which is also an upper bound of . Hence, 2 is the smallest upper bound of , hence Supremum. ${\displaystyle X}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle X}$ ${\displaystyle X}$ ${\displaystyle X}$ ${\displaystyle X}$

By making a small change, the relationship between Supremum and Maximum becomes clear. The maximum is namely the largest element " among all elements" of a set:

Obviously has no maximum, because for every real number there is a real number that is greater than , e.g. B. with the choice . The number 2 as a supremum is larger than all elements of , but is not in , because it is not really smaller than itself. Now let's look at the crowd ${\displaystyle X}$ ${\displaystyle a<2}$ ${\displaystyle b<2}$ ${\displaystyle a}$ ${\displaystyle b={\tfrac {a+2}{2}}}$ ${\displaystyle X}$ ${\displaystyle X}$

${\displaystyle X':=\{x\in \mathbb {R} :x\leq 2\}\subseteq \mathbb {R} }$,

so 2 is the maximum of , since it is less than or equal to itself and there is no greater number than 2 that is less than or equal to 2. At the same time, however, 2 is also a supremum of, as it was of , since the same conditions are fulfilled as there. ${\displaystyle X^{\prime }}$ ${\displaystyle X^{\prime }}$ ${\displaystyle X}$

In fact, every maximum is always also a supremum. It is therefore customary not to define the term maximum in an elementary way, but to name it as a special case of the supremum, if this is itself an element of the set whose supremum it represents. - The same applies to the minimum.

In general

However, upper and lower bounds as well as suprema and infima can not only be considered on the real numbers, but also generally on semi-ordered sets . The formal definitions are as follows:

If a semi-ordered set with partial order and a subset of then applies: ${\displaystyle M}$ ${\displaystyle \leq }$ ${\displaystyle T}$ ${\displaystyle M}$

Upper bound

An element is called the upper bound of if holds for all . ${\displaystyle b\in M}$ ${\displaystyle T}$ ${\displaystyle x\leq b}$ ${\displaystyle x\in T}$

Lower bound

Analogously, the lower bound of is called if holds for all . ${\displaystyle b}$ ${\displaystyle T}$ ${\displaystyle b\leq x}$ ${\displaystyle x\in T}$

quantity limited upwards or downwards

If there is an upper (lower) bound of , then it is called bounded above (below) . ${\displaystyle T}$ ${\displaystyle T}$

upwards or downwards unlimited amount

If it is not restricted upwards (downwards), then upwards (downwards) is called unrestricted . ${\displaystyle T}$ ${\displaystyle T}$

limited amount

${\displaystyle T}$is called restricted if it is restricted upwards and downwards, otherwise unrestricted or not restricted . That is, is unlimited (or non-restricted ) when either above or below or above and below is not limited. If it is to be expressed that a set is unrestricted both upwards and downwards , the set must be expressly described as unrestricted upwards and downwards . ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$

Supremum

An element is called the supremum of if there is a smallest upper bound of . ${\displaystyle b\in M}$ ${\displaystyle T}$ ${\displaystyle b}$ ${\displaystyle T}$

Infimum

It is called the infimum of if it is a greatest lower bound of . ${\displaystyle T}$ ${\displaystyle T}$

If the set of real numbers is then: ${\displaystyle M}$

• Is bounded above and not empty, then has a smallest upper bound (proof idea below) and it is called the upper limit or supremum of  - in signs . ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle \sup(T)}$
• If it is bounded downwards and not empty, then it has a largest lower bound (analogous proof) and it is called the lower limit or infimum of  - in signs . ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle \inf(T)}$
• If there is an upper limit and the supremum of in is included, the supremum is also called the maximum of , in signs . ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle \max(T)}$
• If restricted downwards and the infimum of in is included, the infimum is also called the minimum of , in signs . ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle \min(T)}$
• If there is no upper limit, one writes: (see infinity ). The symbol + ∞ is not a real number and also not the supremum of in the sense defined here: the supremum value is precisely the formal notation that there is no supremum, see also expanded real numbers . Occasionally in this context it is also referred to as an “improper supremum”. ${\displaystyle T}$ ${\displaystyle \sup T=+\infty }$
  ${\displaystyle T}$ ${\displaystyle \infty }$ ${\displaystyle +\infty }$
• Is down indefinitely, to write analog: . ${\displaystyle T}$ ${\displaystyle \inf T=-\infty }$

Suprema (and Infima) of illustrations

General illustrations

The concept of the supremum on sets is also applied to mappings (functions). Because the image of a picture is always a lot . Namely for a picture

${\displaystyle f\colon X\rightarrow Y}$

the amount

${\displaystyle f(X):=\{f(x):x\in X\}=\{y\in Y:y=f(x){\text{ für ein }}x\in X\}}$

the so-called element images , d. H. the pictures of the individual elements from below the figure . ${\displaystyle X}$ ${\displaystyle f}$

${\displaystyle f(X)}$is also called the function image . ${\displaystyle f}$

If it is a semi-ordered set, then the supremum of auf  - if it exists in - is defined by ${\displaystyle Y}$ ${\displaystyle f}$ ${\displaystyle X}$ ${\displaystyle Y}$

${\displaystyle \sup f:=\sup _{x\in X}f(x):=\sup f(X)=\sup\{f(x):x\in X\}}$.

The supremum of a function is thus defined as the supremum of the image set of . The infimum of on is defined analogously . ${\displaystyle f}$ ${\displaystyle f}$ ${\displaystyle f}$ ${\displaystyle X}$

The defining property of the supremum can be formulated as a monotonous Galois connection between and : for all and applies ${\displaystyle \sup \dashv \Delta }$ ${\displaystyle \sup \colon Y^{X}\to Y}$ ${\displaystyle \Delta \colon Y\to Y^{X}}$ ${\displaystyle y\in Y}$ ${\displaystyle f\in Y^{X}}$

${\displaystyle \sup f\leq _{Y}y\Longleftrightarrow f\leq _{Y^{X}}\Delta (y)}$.

Here is equipped with the pointwise order and . ${\displaystyle Y^{X}}$ ${\displaystyle \Delta (y)(x):=y}$

Applies analogously . ${\displaystyle \Delta \dashv \inf }$

Follow as pictures

If you take a sequence of elements as a figure ${\displaystyle a_{1},a_{2},a_{3},\ \ldots }$ ${\displaystyle Y}$

${\displaystyle f\colon \mathbb {N} \rightarrow Y}$

on - so according to

${\displaystyle a_{1}:=f(1),\ a_{2}:=f(2),\ a_{3}:=f(3),\ \ldots }$

- the definition of the supremum (infimum) of mappings immediately results in the definition of the supremum (infimum) of a sequence - if it exists in. ${\displaystyle (a_{n})}$ ${\displaystyle Y}$

properties

Uniqueness and existence

Is an upper bound of , and so also is an upper bound of . Conversely , if there is no upper bound of and , then there is also no upper bound of . The same applies to lower bounds. ${\displaystyle b}$ ${\displaystyle T}$ ${\displaystyle c>b}$ ${\displaystyle c}$ ${\displaystyle T}$ ${\displaystyle c}$ ${\displaystyle T}$ ${\displaystyle b<c}$ ${\displaystyle b}$ ${\displaystyle T}$

The supremum of is (if it exists) uniquely determined. The same is true for the infimum of . ${\displaystyle T}$ ${\displaystyle T}$

It is possible that a subset of a semi-ordered set has several minimal upper bounds, i. H. upper bound, so that every smaller element is not an upper bound. However, as soon as it has more than one minimum upper bound, there is no smallest upper bound; H. no supremum, from . One example is the set with the partial order . Here has the two minimum upper bounds and . ${\displaystyle T}$ ${\displaystyle M}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle M=\{a,\ b,\ c,\ d\}}$ ${\displaystyle \{a<c,\ b<c,\ a<d,\ b<d\}}$ ${\displaystyle T=\{a,\ b\}}$ ${\displaystyle c}$ ${\displaystyle d}$

Properties related to an epsilon environment

Be a non-empty subset of the real numbers, then also holds for that ${\displaystyle X}$

• Supremum by : ${\displaystyle X}$

1. If so exists for all one , so that is. ${\displaystyle \sup X<+\infty }$ ${\displaystyle \epsilon >0}$ ${\displaystyle x\in X}$ ${\displaystyle (\sup X)-\epsilon <x}$
2. If so there exists for all one such that . ${\displaystyle \sup X=+\infty }$ ${\displaystyle k>0}$ ${\displaystyle x\in X}$ ${\displaystyle k<x}$

• Infimum of : ${\displaystyle X}$

1. If so exists for all one , so that is. ${\displaystyle \inf X>-\infty }$ ${\displaystyle \epsilon >0}$ ${\displaystyle x\in X}$ ${\displaystyle x<(\inf X)+\epsilon }$
2. If so there exists for all one such that . ${\displaystyle \inf X=-\infty }$ ${\displaystyle k>0}$ ${\displaystyle x\in X}$ ${\displaystyle x<-k}$

Creation of convergent sequences

• Let be a non-empty subset of the real numbers with a supremum . Then a sequence can be created from suitably selected elements of which converges to . ${\displaystyle X}$ ${\displaystyle \sup X<+\infty }$ ${\displaystyle X}$ ${\displaystyle (x_{n})}$ ${\displaystyle \sup X}$

Proof: be a null sequence , is a constant sequence. With the calculation rules for limit values , the sequence “from below” converges to . Because of the above-mentioned section "property of the supremum with respect to an epsilon-around" the members exist a sequence with between and enclosed is. So how the enclosing consequences converges against . ${\displaystyle (\epsilon _{n}>0)}$ ${\displaystyle (b_{n}=\sup X)}$ ${\displaystyle (a_{n}=(\sup X)-\epsilon _{n})}$ ${\displaystyle \sup X}$ ${\displaystyle x_{n}}$ ${\displaystyle (x_{n}),}$ ${\displaystyle a_{n}=(\sup X)-\epsilon _{n}<x_{n}\leq \sup X=b_{n}}$ ${\displaystyle (a_{n})}$ ${\displaystyle (b_{n})}$ ${\displaystyle (x_{n})}$ ${\displaystyle \sup X}$

• Let be a non-empty subset of the real numbers with an infimum . Then a sequence can be created from suitably selected elements of which converges to . ${\displaystyle X}$ ${\displaystyle \inf X>-\infty }$ ${\displaystyle X}$ ${\displaystyle (x_{n})}$ ${\displaystyle \inf X}$

Proof: is a constant sequence, be a zero sequence . With the calculation rules for limit values , the sequence “from above” converges to . Because of the above-mentioned section "property of the infimum with respect to an epsilon-around" the members exist a sequence , with between and enclosed is. So how the enclosing consequences converges against . ${\displaystyle (a_{n}=\inf X)}$ ${\displaystyle (\epsilon _{n}>0)}$ ${\displaystyle (b_{n}=(\inf X)+\epsilon _{n})}$ ${\displaystyle \inf X}$ ${\displaystyle x_{n}}$ ${\displaystyle (x_{n})}$ ${\displaystyle a_{n}=\inf X\leq x_{n}<(\inf X)+\epsilon _{n}=b_{n}}$ ${\displaystyle (a_{n})}$ ${\displaystyle (b_{n})}$ ${\displaystyle (x_{n})}$ ${\displaystyle \inf X}$

Remarks:

• Neither nor do they have to be monotonous . ${\displaystyle (\epsilon _{n})}$ ${\displaystyle (x_{n})}$

• If it is of finite power , then the supremum is a maximum (or the infimum is a minimum), and almost all are equal to the supremum (or infimum). ${\displaystyle X}$ ${\displaystyle x_{n}}$

Existence of the supremum for bounded subsets of the real numbers

The existence of the supremum for a bounded subset of the real numbers can be shown in several ways: ${\displaystyle M}$

A. On the one hand, the existence of supremum and infimum for bounded subsets of real numbers can simply be defined as an axiom . This requirement is often called the axiom of the supremum or the axiom of completeness .

B. If one starts from the axiom that every nesting of intervals defines exactly one real number, an interval nesting can serve to prove the existence of the supremum of , for which there is no upper bound of , but each is one. ${\displaystyle \sup M}$ ${\displaystyle M}$ ${\displaystyle ([a_{k},\,b_{k}]),\;k\in \mathbb {N} }$ ${\displaystyle a_{k}}$ ${\displaystyle M}$ ${\displaystyle b_{k}}$ ${\displaystyle \mathbf {(i)} }$

Such an interval nesting defines a number , and the sequences and converge to . Any one is bigger than almost all of them because of . Since every upper bound of is is . So there is an upper bound of . It remains to be considered whether there can also be an upper bound of . Because of almost all of them are larger than . Since there is no upper bound of , there is also none. So this is the claimed supremum of . - It remains to be shown that there is an interval nesting that satisfies condition (i). ${\displaystyle \sigma }$ ${\displaystyle (a_{k})}$ ${\displaystyle (b_{k})}$ ${\displaystyle \sigma }$ ${\displaystyle b>\sigma }$ ${\displaystyle \lim _{k\to \infty }b_{k}=\sigma }$ ${\displaystyle b_{k}}$ ${\displaystyle b_{k}}$ ${\displaystyle M}$ ${\displaystyle b\notin M}$ ${\displaystyle \sigma }$ ${\displaystyle M}$ ${\displaystyle \sigma '<\sigma }$ ${\displaystyle M}$ ${\displaystyle \lim _{k\to \infty }a_{k}=\sigma }$ ${\displaystyle a_{k}}$ ${\displaystyle \sigma '}$ ${\displaystyle a_{k}}$ ${\displaystyle M}$ ${\displaystyle \sigma '}$ ${\displaystyle \sigma }$ ${\displaystyle M}$ ${\displaystyle ([a_{k},\,b_{k}])}$

For this purpose, an interval sequence is defined recursively . For the first interval is an arbitrary number that is smaller than an arbitrary element of is an arbitrary upper bound . is the midpoint of the -th interval of the sequence. The limits of the following interval are, ${\displaystyle ([a_{k},\,b_{k}])}$ ${\displaystyle a_{1}}$ ${\displaystyle M}$ ${\displaystyle b_{1}}$ ${\displaystyle M}$ ${\displaystyle c_{k}={\frac {a_{k}+b_{k}}{2}}}$ ${\displaystyle k}$ ${\displaystyle [a_{k+1},\,b_{k+1}]}$

• if no upper limit of is ; ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle a_{k+1}=c_{k},b_{k+1}=b_{k}}$
• if an upper bound of is . ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle a_{k+1}=a_{k},b_{k+1}=c_{k}}$

The following applies to such a sequence of intervals: is an upper bound of , not. At the transition from to an interval limit replaces the upper bound of is if and only if it is the upper bound of ; but if there is no upper bound of , then an interval limit replaces that is also not such. So every but not an upper bound of , and the interval sequence satisfies condition (i). - It remains to be shown that there is interval nesting. ${\displaystyle b_{1}}$ ${\displaystyle M}$ ${\displaystyle a_{1}}$ ${\displaystyle [a_{k},b_{k}]}$ ${\displaystyle [a_{k+1},b_{k+1}]}$ ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle c_{k}}$ ${\displaystyle b_{k}}$ ${\displaystyle a_{k}}$ ${\displaystyle M}$ ${\displaystyle ([a_{k},\,b_{k}])}$ ${\displaystyle ([a_{k},\,b_{k}])}$

Assertion : is monotonically increasing . ${\displaystyle (a_{k})}$ ${\displaystyle \Leftrightarrow \forall k:a_{k+1}\geq a_{k}.\mathbf {(1)} }$

Proof : There is nothing to prove for. For follows : . ${\displaystyle a_{k+1}=a_{k}}$ ${\displaystyle a_{k+1}=c_{k}}$ ${\displaystyle b_{k}>a_{k}}$ ${\displaystyle a_{k+1}={\frac {a_{k}+b_{k}}{2}}>{\frac {a_{k}+a_{k}}{2}}=a_{k}}$

Claim : is monotonically decreasing . ${\displaystyle (b_{k})}$ ${\displaystyle \Leftrightarrow \forall k:b_{k+1}\leq b_{k}.\mathbf {(2)} }$

Proof : There is nothing to prove for. For follows : . ${\displaystyle b_{k+1}=b_{k}}$ ${\displaystyle b_{k+1}=c_{k}}$ ${\displaystyle a_{k}<b_{k}}$ ${\displaystyle b_{k+1}={\frac {a_{k}+b_{k}}{2}}<{\frac {b_{k}+b_{k}}{2}}=b_{k}}$

Assertion : , is a Nullfollge. . - proof : ${\displaystyle (d_{k})}$ ${\displaystyle d_{k}=b_{k}-a_{k}}$ ${\displaystyle \mathbf {(3)} }$

• If is not an upper bound of , is ; ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle d_{k+1}=b_{k+1}-a_{k+1}=b_{k}-c_{k}={\frac {2b_{k}}{2}}-{\frac {a_{k}+b_{k}}{2}}={\frac {b_{k}-a_{k}}{2}}={\frac {d_{k}}{2}}}$
• if is an upper bound of is . ${\displaystyle c_{k}}$ ${\displaystyle M}$ ${\displaystyle d_{k+1}=b_{k+1}-a_{k+1}=c_{k}-a_{k}={\frac {a_{k}+b_{k}}{2}}-{\frac {2a_{k}}{2}}={\frac {b_{k}-a_{k}}{2}}={\frac {d_{k}}{2}}}$

So all can also be written, and is because of a (geometric) zero sequence. ${\displaystyle d_{k}}$ ${\displaystyle d_{k}=d_{1}\cdot \,\left({\tfrac {1}{2}}\right)^{k-1}}$ ${\displaystyle (d_{k})}$ ${\displaystyle \left|{\tfrac {1}{2}}\right|<1}$

With (1), (2) and (3) an interval nesting, q. e. d. ${\displaystyle ([a_{k},\,b_{k}])}$

C. An equivalent formulation for the existence of the supremum is the axiom of intersection , according to which every Dedekind intersection is generated by a real number.

Examples

Real numbers

The following examples relate to subsets of the real numbers.

• ${\displaystyle \sup\{1,2,3\}=3}$
• ${\displaystyle \sup\{x\in \mathbb {R} :0<x<1\}=\sup\{x\in \mathbb {R} :0\leq x\leq 1\}=1}$
• ${\displaystyle \sup\{x\in \mathbb {Q} :x^{2}<2\}={\sqrt {2}}\notin \mathbb {Q} }$
• ${\displaystyle \sup\{(-1)^{n}-{\tfrac {1}{n}}:n\in \mathbb {N} \}=1}$
• ${\displaystyle \sup \mathbb {Z} =+\infty }$
• ${\displaystyle \sup\{a\}=\inf\{a\}=\max\{a\}=\min\{a\}=a\quad \forall \,a\in \mathbb {R} }$
• ${\displaystyle \sup\{a+b:a\in A\land b\in B\}=\sup A+\sup B}$
• ${\displaystyle \sup -A=-\inf A}$or , where ${\displaystyle -\sup A=\inf -A}$ ${\displaystyle -A:=\{-x\in \mathbb {R} :x\in A\}}$

Other semi-ordered sets

On , each non-empty upwards or downwards limited subset of a supremum and infimum. If you consider other sets on which order relations are defined, this is not mandatory: ${\displaystyle \mathbb {R} }$

• The set of rational numbers is totally ordered with respect to the natural order . The amount is limited upwards, for example, by the number , but has no supremum in . ${\displaystyle \mathbb {Q} }$ ${\displaystyle \{x\in \mathbb {Q} :x^{2}<2\}\subset \mathbb {Q} }$ ${\displaystyle 1{,}42\in \mathbb {Q} }$ ${\displaystyle \mathbb {Q} }$
• In any semi-ordered set , each element is both the lower and upper bound of the empty set . Hence the largest element of and the smallest. However, the largest and smallest elements do not have to exist: in the set of natural numbers with the usual order there is no infimum, and it is . ${\displaystyle (A,\leq )}$ ${\displaystyle \emptyset }$ ${\displaystyle \inf \emptyset }$ ${\displaystyle A}$ ${\displaystyle \sup \emptyset }$ ${\displaystyle \mathbb {N} =\{1,2,3,\dots \}}$ ${\displaystyle \emptyset }$ ${\displaystyle \sup \emptyset =1}$
• In the set partially ordered with regard to inclusion , the set is limited both by the element and by an upper limit. However, a supremum, i.e. a smallest upper bound of , does not exist in . ${\displaystyle {\mathcal {X}}:=\{\{1\},\{2\},\{1,2,3\},\{1,2,4\}\}}$ ${\displaystyle M:=\{\{1\},\{2\}\}\subset {\mathcal {X}}}$ ${\displaystyle \{1,2,3\}\in {\mathcal {X}}}$ ${\displaystyle \{1,2,4\}\in {\mathcal {X}}}$ ${\displaystyle M}$ ${\displaystyle {\mathcal {X}}}$

See also

• In measure theory, the concept of the essential supremum is derived from the term supremum , which, for example, plays an important role in the theory of spaces . ${\displaystyle L^{p}}$
• The investigation of partially ordered sets, in which there is a supremum and an infimum for every two-element subset, is the subject of lattice theory .

literature

• Stefan Hildebrandt: Analysis 1. Springer 2005, ISBN 3-540-25368-8 .

Web links

Wikibooks: Math for non-freaks: Supremum and Infimum  - learning and teaching materials

Wikibooks: Math for non-freaks: determining and proving supremum and infimum  - learning and teaching materials

Commons : Infimum and supremum  - collection of images, videos and audio files

Individual evidence

1. ↑ Interval nesting # Convergence of the limit sequences of an interval nesting
2. ↑ The train of thought is a complete induction .
3. ↑ More on the convergence of certain geometric sequences here .