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Apr 18

## From minimum to infimum: Math is just a logical game

The true Math is a continuous exercise in logic. A good teacher makes that logic visible and tangible. A genius teacher does the logic mentally and only displays the result. I am trying to be as good a teacher as humanly possible.

## Steps to develop a new definition

Step 1. Start with a simple example.

What is the minimum $m$ of the set $A=[1,\infty)$? In my class everybody says $m=1$.

Step 2. Give the most visual definition.

Here is a good candidate: the minimum $m$ of a set $A$ is its leftmost point.

Step 3. Formalize the definition you gave.

We said "its point". This means $m$ should be an element of $A$. "Leftmost" is formalized as $m\le a$ for any $a\in A$. Thus the formal definition should be:  the number $m$ is called a minimum of a set $A$ if

1) $m\le a$ for any $a\in A$ and

2) $m\in A$.

Step 4. Look at bad cases when the definition fails. Try to come up with a generalized definition that would cover the bad cases.

## Here is a bad case: meet the infinity

Let $A=(1,\infty)$. In my class, some students suggested $m=0$ and $m=2$. Can you see why neither is good? See the explanation below if you can't. I would try $m=1$. It still satisfies part 1) of the above definition but does not belong to $A$. This is the place to stretch one's imagination.

Statement 1. No element of $A$ satisfies the formal definition above.

Proof. Take any element $a\in A$. Then it is larger than 1 and the number $a_1=\frac{a+1}{2}$ is halfway between $a$ and 1. We have shown that to the left of any $a\in A$ there is another element of $A$. So no element of $A$ is leftmost.

In the game of chess, the one who thinks several moves ahead wins. A stronger version of Statement 1 is the following.

Statement 2. To the left of any $a\in A$ there are infinitely many elements of $A$.

Proof. Indeed, set $a_n=1+1/n,\ n=1,2,...$ The numbers $a_n$ approach 1 from the right. As $n$ increases, they become closer and closer to 1. For any given $a\in A$, infinitely many of these numbers will satisfy $1.

Step 5. When thinking about a new definition, try to exclude undesirable outcomes.

Since for $m=1$ condition 2) is not satisfied, one might want to omit it altogether. However, leaving only part 1) would give a bad result. For example, $m=0$ would satisfy such a "definition", and it would be bad for two reasons. Firstly, there is no uniqueness. We could take any number $m<1$. Secondly, if you take any $m<1$, in the interval $(m,1)$ there are no elements of $A$, so there is no reason to call such a number a minimum of $A$.

Definition. A number $m$ is called an infimum of $A$ (denoted $m=\inf A$) if

I) $m\le a$ for any $a\in A$ and

II) in $A$, there is a sequence $\{a_n\}$ approaching $m$.

The above discussion shows that $\inf (1,\infty)=1$.

Exercise 1. Repeat all of the above for $A=(-\infty,-1)$. First define the maximum of a set. Then modify the definition to obtain what is called a supremum.

Exercise 2. Let $A=(e,\pi)$, where $e=2.71828...$ is the basis of the natural log and $\pi=3.1415...$ is the ratio Length of circumference/Diameter of that circumference. Find the infimum and supremum of $A$. Simply naming them is not enough; you have to prove that both parts of the definitions are satisfied.

Exercise 3. The infimum is also defined as the largest lower bound. Can you prove equivalence of the two definitions?

Exercise 4. The supremum is also defined as the least upper bound. Can you prove equivalence of the two definitions?

Exercise 5. Consider any two sets and their union. What is the relationship between the infimums of the three sets?