## 3.1. Sequences

### Proposition: Sup and Inf of bounded Sequences

If a sequence is
bounded above, then
If a sequence is
bounded below, then

*c = sup(a*is finite. Moreover, given any_{j})*> 0*there exists at least one integer*k*such that*a*(see picture)._{k}> c -

*c = inf(a*is finite. Moreover, given any_{j})*> 0*, there exists at least one integer*k*such that*a*(see picture)._{k}< c +

### Proof:

We will only prove the first statement, leaving the second one involving*inf*as an exercise.

Recall that the *sup* is the smallest of all upper bounds. Since
the sequence is assumed to be bounded above, the *sup* must be
less than or equal to that upper bound, by definition. Therefore the
*sup* can not be positive infinity. But the *sup*
must also be at least as big as the first number of the sequence, hence
the *sup* can not be negative infinity. Therefore, the
*sup* must be bounded.

The second statement can be proved by contradiction, which means that we
first need to figure out the contra-positive of the statement to be proved
(which is the only hard part of the proof).
We need to turn every *"take any"* to *"there exists"*, any
*"at least one"* to *"for all"*, and switch the inequality.

But that opposite statement implies immediately that

Original statement:- Suppose that
c = sup(a. Then_{j})given any> 0there existsat least one integerksuch thata_{k}> c -

Opposite of that statement:- Suppose that
c = sup(a. Then_{j})there exists one> 0such thatfor allintegerskwe havea_{k}< c -

*c -*would be a new upper bound for the sequence

*{a*, and clearly

_{j}}*c - < c*. But then

*c*could not have been the original

*sup*of the sequence, because the

^{ must be the smallest one of all upper bounds. Therefore we have our contradiction and the original statement is proved. Next | Previous | Glossary | Map }