# Real Analysis - Supremums

I'm currently learning about supremums but I'm having trouble understanding them.

I understand that for a number M to be the sup(S) it satisfy two conditions:

(1) M is an upper bound of S. (2) If M' is any upper bound of S then $M \leq M'$

The first condition is just saying that for all $s \in S$, $s \leq M$. Right?

I see why the second condition is true.. But I don't really understand how to use it.

I don't understand how to use it when evaluating problems dealing with supremums. How do we prove that for any upper bound $M'$ of set S, $M \leq M'$.



Also, I'm having trouble with the following example:

Let $a_n$ and $b_n$ be two bounded sequences of real numbers, show that: $$sup(a_n + b_n) \leq sup(a_n) + sup(b_n)$$

To understand how to use the supremum you need to see (and prove) exercises. In your case

Proposition 1. Let $A := (a_n)_{n=1}^\infty$ and $B := (b_n)_{n=1}^\infty$ be two bounded sequences of real numbers. If $C := (a_n + b_n)_{n=1}^\infty$, show that $\sup C \le \sup A + \sup B$.

Proof. Let $n \ge 1$ a positive integer. Since $a_n \leq \sup A$, $b_n \leq \sup B$, we have $a_n+b_n \leq \sup A + \sup B$. Thus $\sup A + \sup B$ is an upper bound for $C$.

Now suppose $c_n < \sup A + \sup B$ and define $\epsilon := \sup A + \sup B - c_n > 0$. Then $\epsilon/2 > 0$. Since $\sup A - \epsilon/2 < \sup A$, there exists $a_n$ such that $\sup A - \epsilon/2 < a_n$. Similarly, there exists $b_n$ such that $\sup B - \epsilon/2 < b_n$. Then $c_n = \sup A + \sup B - \epsilon = (\sup A - \epsilon/2) + (\sup B - \epsilon/2) < a_n + b_n$. So $c_n$ is not an upper bound for $C$. Therefore, $\sup A + \sup B$ is the least upper bound.

Also you can try to prove:

Proposition 2. Let $h > 0$ a positive real number and let $E$ a subset of real numbers. Show that if $E$ has a supremum, there exists $x \in E$ such that $x > \sup E - h$. (Hint: prove by contradiction.)

Proposition 3. There exists a positive real number $x$ such that $x^2 = 2$. (Hint: define the set $E := \{x \in \mathbf R : x \ge 0 \;\;\text{and}\;\; x^2 < 2\}$ and prove by contradiction.)

You can show that $\forall \epsilon >0$,$M-\epsilon$ is not an upper bound. That is $\exists s \in S$,such at $s>M-\epsilon$

For your second question, you just need to show $\sup(a_n) + \sup(b_n)$ is an upper bound of all $a_n + b_n$. Hence the inequality must hold by definition.