Raising the bar From minimum to infimum: Math is just a logical game

Apr 18

From minimum to infimum: Math is just a logical game

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 $1<a_n<a$ .

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.