If a set $A$ of sequences $x_n$ is bounded above by $G$ and if $x_n$ converges to $G$ then $G=sup A$

Question

We have that for every $n\in \Bbb{N}$ the sequence $x_n \in A$, and since $A$ is bounded above by $G$ $$x_n\leq G$$ $$x_n - G \leq 0$$

Lets assume there is another upper bound for $A$ called $G'$, such that $G'<G$ - it follows that $0<G-G'$. Lets look at $x_n - G=0$ first. $$x_n - G = 0$$$$x_n=G$$ But we know that $G'<G$, therefore $G'<x_n$. So $G'$ is not an upper bound for $A$. Next we have that $x_n-G<0$, and we can write that $$\vert x_n - G \vert=-(x_n -G)$$ We know $x_n\rightarrow G$ as $n\rightarrow \infty$. According to the definition of a converging sequence, there is a natural number $N<n$, such that $$\vert x_n - G \vert < G - G'$$ $$-(x_n - G) < G - G'$$ $$-x_n + G < G - G'$$ lets reduce $G$ from both sides and we can conclude that $$-x_n<-G'$$ $$x_n>G'$$

Thus $G'$ is not an upper bound of $A$. Now since any number $G'<G$ can not be an upper bound for $A$ it must be that $G$ is the least upper bound. Correct?

Answer 1

Note that $x_n-G$ could be $0$ some of the time and strictly negative the rest of the time. That case is not really covered as you write it - it should at least be commented on.

However, this solution is pretty messy and a much easier way of showing this is:

$$ G-x_n=G-\sup_{n\in\mathbb{N}} x_n+\sup_{n\in \mathbb{N}}x_n-x_n\geq G-\sup_{n\in\mathbb{N}} x_n\geq 0, $$ since the supremum is the least upper bound. Thus, if $G-\sup_{n\in \mathbb{N}}x_n\neq 0$, define $\varepsilon=\frac{G-\sup_{n\in \mathbb{N}}x_n}{2}$. Then, $$ |G-x_n|=G-x_n>\varepsilon $$ for all $n$ and we conclude that $G$ cannot be the limit of $(x_n)_{n\in\mathbb{N}}$.

Answer 2

Remember the two defining properties of supremum of a set:

• no member of the set exceeds its supremum.
• every number less than the supremum is exceeded by at least one member of the set.

This is the way one must understand the definitions of terms in one's language with minimal use of mathematical symbols.

Now check whether these two properties are satisfied by $G$ for set $A$. First we are given that no member of $A$ exceeds $G$ (this is the meaning of $A$ is bounded above by $G$). Thus it remains to verify only the second property for $G$. And for that we choose a number $G'<G$ and try to find a member, say $x_m$, of $A$ such that $x_m>G'$. Can we find such a member?

Yes! Just note that $x_n\to G$ and therefore if we choose $\epsilon=G-G'>0$ we have a positive integer $N$ such that $|x_n-G|<\epsilon $ for all $n>N$. In other words we have $$ G-\epsilon<x_n<G+\epsilon$$ for all $n>N$. Since $G-\epsilon=G'$ it follows that $x_n>G'$ for all $n>N$. Thus we have found many members of $A$ which exceed $G'$ and we are done with the proof.