# Math for Non-Geeks: How to prove existence of a supremum or infimum

Fehler: Aktuelle Seite wurde in der Sitemap nicht gefunden. Deswegen kann keine Navigation angezeigt werden

## General procedure

How do we find the supremum/infimum of a set? there are some strategies:
* Visualize the set: What does the set look like? Try to draw it. You may use a paper, a computer or your imagination!
* Make a hypotheses: 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?
* 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.
* 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:
* Prove that $s$ is an upper bound of $M$ : This is done showing $y\leq s$ for all $y\in M$.
* 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:
* Prove that ${\tilde {s}}$ is a lower bound of $M$ : This is done showing $y\geq {\tilde {s}}$ for all $y\in M$.
* 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:
* Prove that $m$ is an upper bound of $M$: This is done showing $y\leq m$ for all $y\in M$.
* 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:
* Prove that ${\tilde {m}}$ is a lower bound of $M$: This is done showing $y\geq {\tilde {m}}$ for all $y\in M$.
* Prove that ${\tilde {m}}\in M$.

## Examples for determining supremum/infimum
### Finite sets

With finite sets of real numbers, determining the infimum and supremum is simple. These sets must always have a maximum (greatest number) and a minimum (smallest number). The maximum of the set is automatically supremum and the minimum is automatically infimum of the set.
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:
* $M_{1}=\{4,\,72,\,-5,\,99,\,42\}$
* $M_{2}=\{(-1)^{k}\cdot 2^{-k}:1\leq k\leq 5\land k\in \mathbb {N} \}$
* $M_{3}=\{\pi ,\,{\sqrt {2}},\,1\}\cap \{{\sqrt {2}},\,-1,\,4\}$
Solution:
* The supremum of $M_{1}$ is its biggest element $99$ . The infimum is the smallest element $-5$.
* There is $\{(-1)^{k}\cdot 2^{-k}:1\leq k\leq 5\land k\in \mathbb {N} \}=\left\{-{\tfrac {1}{2}},\,{\tfrac {1}{4}},-{\tfrac {1}{8}},{\tfrac {1}{16}},-{\tfrac {1}{32}}\right\}$. So the supremum is ${\tfrac {1}{4}}$ and the infimum is $-{\tfrac {1}{2}}$.
* 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

The determination of the infimum and supremum for intervals is quite simple, because the lower boundary point is always the infimum and the upper boundary point is always the supremum:
Theorem (supremum and infimum of intervals)
Let $I$ be an interval.So there are $a,b\in \mathbb {R}$ with $a<b$, such that $I$ takes one of the following 4 forms:
* $I=[a,b]$
* $I=(a,b]$
* $I=[a,b)$
* $I=(a,b)$
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: $a\leq x\leq b$. So we know that $a$ is a lower bound and $b$ is an upper bound. That means, $a\leq \inf I$, $b\geq \sup I$ . 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 calue 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 interval $a\leq x$ , 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 $I$. The there is no $x\in I$, such that $x<{\tilde {a}}$. We distinguish two cases:
Case 1: ${\tilde {a}}>b$ (small interval)
Let $x:={\tfrac {a+b}{2}}$. We can estimate ${\begin{aligned}&a={\frac {a+a}{2}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ a<b\right.}\\[0.3em]<&{\frac {a+b}{2}}=x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ a<b\right.}\\[0.3em]<&{\frac {b+b}{2}}=b\end{aligned}}.$ 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: ${\tilde {a}}\leq b$ (large interval)
Now, take $x:={\tfrac {a+{\tilde {a}}}{2}}$. By our assumptions, ${\begin{aligned}&a={\frac {a+a}{2}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ a<{\tilde {a}}\right.}\\[0.3em]<&{\frac {a+{\tilde {a}}}{2}}=x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ a<{\tilde {a}}\leq b\right.}\\[0.3em]<&{\frac {b+b}{2}}=b\end{aligned}}.$ So $a<x<b$ and we have $x\in I$.
But since $a<{\tilde {a}}$ we have $x={\tfrac {a+{\tilde {a}}}{2}}<{\tfrac {{\tilde {a}}+{\tilde {a}}}{2}}={\tilde {a}}$. 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 $x:={\tfrac {a+b}{2}}$ for ${\tilde {b}}<a$ and $x:={\tfrac {b+{\tilde {b}}}{2}}$ for ${\tilde {b}}\geq a$, we will get a violation of this bound.
Alternatively, one may use the properties of suprema and Infiuma 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:
* $M_{1}=(-1,3]$
* $M_{2}=\{x\in \mathbb {R} :2\leq 2x\leq 8\}$
* $M_{3}=\{x\in \mathbb {R} :|x-1|<4\}$
* $M_{4}=\{\sin(x):x\in \mathbb {R} \}$ (here you may use, what you have learned in school about the sine function )
* The supremum is $3$ and the infimum is $-1$.
* Since $2\leq 2x\leq 8\iff 1\leq x\leq 4$ , we have that $M_{2}=[1,4]$. This is an interval with infimum 1 and supremum 4.
* $M_{3}$ 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 $5$.
* $M_{4}$ 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:
* $[a,b]\cap \mathbb {Z}$
* $(a,b]\cap \mathbb {Z}$
* $[a,b)\cap \mathbb {Z}$
* $(a,b)\cap \mathbb {Z}$
It helps a lot to know that the sets are all finite:
* $[a,b]\cap \mathbb {Z} =\{x\in \mathbb {Z} :a\leq x\leq b\}=\{a,a+1,a+2,\ldots ,b-1,b\}$
* $(a,b]\cap \mathbb {Z} =\{x\in \mathbb {Z} :a<x\leq b\}=\{a+1,a+2,\ldots ,b-1,b\}$
* $[a,b)\cap \mathbb {Z} =\{x\in \mathbb {Z} :a\leq x<b\}=\{a,a+1,a+2,\ldots ,b-2,b-1\}$
* $(a,b)\cap \mathbb {Z} =\{x\in \mathbb {Z} :a<x<b\}=\{a+1,a+2,\ldots ,b-1\}$
(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.
* $\sup([a,b]\cap \mathbb {Z} )=b$ and $\inf([a,b]\cap \mathbb {Z} )=a$
* $\sup((a,b]\cap \mathbb {Z} )=b$ and $\inf((a,b]\cap \mathbb {Z} )=a+1$
* $\sup([a,b)\cap \mathbb {Z} )=b-1$ and $\inf([a,b)\cap \mathbb {Z} )=a$
So there is always a supremum or an infimum. 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 $\sup \emptyset =-\infty$ and the improper infimum $\inf \emptyset =\infty$. However, those are no real numberes, i.e. no proper suprema/ infima. For $a+1\neq b$, the set is non-empty and we have supremum $b-1$ and infimum $a+1$.

### Sequences

Suprema and infima of sequence elements are often required in mathematics. As those elements form a set, the supremum and infimum is well-defined. The set will have often infinitely, but always countably many elements. Let's start with an example:
Exercise (Sets of sequence elements)
What are the supremum and the infimum of the set $M=\left\{5+{\tfrac {2}{n}}:n\in \mathbb {N} \right\}$. 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:
* $5+{\tfrac {2}{1}}=7\implies 7\in M$
* $5+{\tfrac {2}{2}}=6\implies 6\in M$
* $5+{\tfrac {2}{3}}=5{\tfrac {2}{3}}\implies 5{\tfrac {2}{3}}\in M$
* $5+{\tfrac {2}{4}}=5{\tfrac {1}{2}}\implies 5{\tfrac {1}{2}}\in M$
So the set $M$ is of the form $M=\left\{7,6,5{\tfrac {2}{3}},5{\tfrac {1}{2}},5{\tfrac {2}{5}},\dots \right\}$, 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 isbounded 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 amount. So we have to show two things:
* $7\in M$
* $7\geq x$ for all $x\in M$
The number $7$ is definitely an elements of $M$ , namely for $n=1$ there is $5+{\tfrac {2}{n}}=7$. In order to prove that $7$is an upper bound of $M$ , we need to show $7\geq 5+{\tfrac {2}{n}}$. This can be achieved by equivalent transformations:
${\begin{aligned}7&\geq 5+{\tfrac {2}{n}}\\2&\geq {\tfrac {2}{n}}\\2n&\geq 2\\n&\geq 1\end{aligned}}$
Now $n\geq 1$ is an obvious statement for natural numbers. But in the proof we must go the opposite way: Since we want to show the inequality $7\geq 5+{\tfrac {2}{n}}$, we have to start at $n\geq 1$ and gradually transform this inequality into $7\geq 5+{\tfrac {2}{n}}$. 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
* $5\leq x$ for all $x\in M$
* For all $y>5$ there is an $x\in M$ with $x<y$
Um auch zu zeigen, dass $5$ kein Minimum ist, haben wir außerdem zu beweisen, dass $5\notin M$. Zunächst muss ein Beweis for $5\leq 5+{\tfrac {2}{n}}$ for all $n\in \mathbb {N}$ gefunden werden:
In order to show that in addition, $5$ is not a minimum, we also have to prove that $5\notin M$. First of all, we have to find a proof to $5\leq 5+{\tfrac {2}{n}}$ for all $n\in \mathbb {N}$:
${\begin{aligned}5&\leq 5+{\tfrac {2}{n}}\\0&\leq {\tfrac {2}{n}}\end{aligned}}$
Now $0\leq {\tfrac {2}{n}}$ is an obviously true statement, since ${\tfrac {2}{n}}$ is positive. So in the later proof we can get to the inequality $5\leq 5+{\tfrac {2}{n}}$ starting from $0\leq {\tfrac {2}{n}}$ by performing the above equivalent transformation backwards (i.e. adding $5$ on both sides).
Let 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 $x=5+{\tfrac {2}{N}}$ and $N\in \mathbb {N}$ such that $y>x=5+{\tfrac {2}{N}}$. 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 $N\in \mathbb {N}$:
${\begin{aligned}y&>5+{\tfrac {2}{N}}\\y-5&>{\tfrac {2}{N}}\\{\tfrac {y-5}{2}}&>{\tfrac {1}{N}}\end{aligned}}$
Because of $y>5$, we have $y-5>0$ and ${\tfrac {y-5}{2}}>0$. The Archimedian axiom now guarantees us that we can find a suitable $N$, because according to the Archimedian axiom the fraction ${\tfrac {1}{N}}$ becomes smaller than any positive real number.
Finally, we need to show that $5is\notin M$. Or put in different words, $5\neq 5+{\tfrac {2}{n}}$ is valid for all $n\in \mathbb {N}$. But because ${\tfrac {2}{n}}>0$ there is $5+{\tfrac {2}{n}}>5$ and therefore $5\neq 5+{\tfrac {2}{n}}$. 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 $5+{\tfrac {2}{n}}=7$. So $7\in M$.
Proof step: $7$ is an upper bound of $M$
For all $n\in \mathbb {N}$ there is
${\begin{aligned}&&n&\geq 1\\&\implies &2n&\geq 2\\&\implies &2&\geq {\tfrac {2}{n}}\\&\implies &7&\geq 5+{\tfrac {2}{n}}\end{aligned}}$ 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
${\begin{aligned}&&0&\leq {\tfrac {2}{n}}\\&\implies &5&\leq 5+{\tfrac {2}{n}}\end{aligned}}$
So $5$ is smaller or equal than 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, ${\tfrac {y-5}{2}}>0$ and by the Archimedean axiom, there is an $N\in \mathbb {N}$ with ${\tfrac {y-5}{2}}>{\tfrac {1}{N}}$. There is
${\begin{aligned}&&{\tfrac {y-5}{2}}&>{\tfrac {1}{N}}\\&\implies &y-5&>{\tfrac {2}{N}}\\&\implies &y&>5+{\tfrac {2}{N}}\end{aligned}}$
Since $5+{\tfrac {2}{N}}\in M$ 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 ${\tfrac {2}{n}}>0$ and hence ${\tfrac {2}{n}}+5>5$. 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
$M=\left\{{\frac {1}{1+x^{2}}}:x\in \mathbb {R} \right\}$
How to get to the proof?
Let in the following be $f:\mathbb {R} \rightarrow \mathbb {R} :x\mapsto {\tfrac {1}{1+x^{2}}}$. We proceed as follows:
Step 1: Visualize the set $M$.
The function $f(x)={\tfrac {1}{1+x^{2}}}$ drawn in a diagram looks as follows:
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 $f(0)={\tfrac {1}{1+0^{2}}}=1$, $1$ 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 or equal 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 ${\tfrac {1}{1+0^{2}}}=1$ we can directly see that $1\ mustbeinM$. 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:
${\frac {1}{1+x^{2}}}\leq 1$
Let us transform this inequality by means of equivalent transformations:
${\begin{array}{lrl}&{\frac {1}{1+x^{2}}}&\leq 1\\[0.3em]\iff \ &1&\leq 1+x^{2}\\[0.3em]\iff \ &0&\leq x^{2}\end{array}}$
We already know that the last inequality is fulfilled for all $x\in \mathbb {R}$ . Since we have only used equivalent transformations, we can later recover the inequality ${\frac {1}{1+x^{2}}}\leq 1$.
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, ${\tfrac {1}{1+x^{2}}}$ 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 $x\in \mathbb {R}$ with
${\frac {1}{1+x^{2}}}<y$
Let us transform this inequality by means of equivalent transformations:
${\begin{array}{lrl}&{\frac {1}{1+x^{2}}}&<y\\\iff \ &1&<y\cdot (1+x^{2})\\\iff \ &1&<y+y\cdot x^{2}\\\iff \ &1-y&<y\cdot x^{2}\\[0.5em]&&{\color {OliveGreen}\left\downarrow \ y{\text{ is positive}}\right.}\\[0.5em]\iff \ &{\frac {1-y}{y}}&<x^{2}\\\end{array}}$
In order to take the square root, ${\tfrac {1-y}{y}}\geq 0$ is required, i.e. $y\leq 1$. But for the further proof it is no problem to assume $0<y\leq 1$, because for $y>1$ the last inequality is always fulfilled. The square number $x^{2}$ is then always greater than the negative number ${\tfrac {1-y}{y}}$.
So consider $0<y\leq 1$. Then:
${\begin{array}{lrl}&{\frac {1-y}{y}}&<x^{2}\\\iff &{\sqrt {\frac {1-y}{y}}}&<|x|\\\end{array}}$
For $0<y\leq 1$ we need to choose an $x\in \mathbb {R}$ with $|x|>{\sqrt {\frac {1-y}{y}}}$. This $x$ automatically fulfils
${\frac {1}{1+x^{2}}}<y$
hence $y$ cannot by 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 ${\tfrac {1}{1+x^{2}}}=1$. So $1\in M$.
Proof step: $1$ is an upper bound of $M$
There is
${\begin{array}{lrl}&0&\leq x^{2}\\\implies \ &1&\leq 1+x^{2}\\\implies \ &{\frac {1}{1+x^{2}}}&\leq 1\\\end{array}}$
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 $x\in \mathbb {R}$:
${\begin{array}{lrl}&x^{2}&\geq 0\\\implies \ &1+x^{2}&>0\\\implies \ &{\frac {1}{1+x^{2}}}&>0\\\end{array}}$
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 $0<y\leq 1$ choose $x$ such that $|x|>{\sqrt {\tfrac {1-y}{y}}}$ . Then the following inequality is satisfied.
For any $y>0$ there is at least one real number $x$ with $x^{2}>{\tfrac {1-y}{y}}$. For real numbers of this kind we have
${\begin{array}{lrl}&{\frac {1-y}{y}}&<x^{2}\\\implies \ &1-y&<y\cdot x^{2}\\\implies \ &1&<y+y\cdot x^{2}\\\implies \ &1&<y\cdot (1+x^{2})\\&{\frac {1}{1+x^{2}}}&<y\\\end{array}}$
So $y$ cannot be a lower bound of $M$, which proves that $0$ is the largest lower bound of $M$.
Fehler: Aktuelle Seite wurde in der Sitemap nicht gefunden. Deswegen kann keine Navigation angezeigt werden