Convergence and supremum

Question

Define X as the set of all sequences $\{x_n\}_{(n \in \mathbb{N})}$ of real numbers such that $\lim_{n \to \infty} x_n = 0$.

Question: How to show that if $\{x_n\} \in X$, then there is a $K \in \mathbb{N}$ such that $$ |x_K| = \sup \{|x_n|: n \in \mathbb{N}\}$$ (i.e. $x_K$ is an element of maximal absolute value), using the definition of convergence?

Comments: I am stuck at this problem and don't know how to find the connection between the supremum and convergence.

We define convergence as: The sequence will converge to x if for every ε > 0 we can find an N ∈ N that for all n≥N, then d(x_n,x) < ε

Answer 1

By definition of convergence to zero, $\forall \epsilon>0, \exists N>0 |n>N \implies |x_n|<\epsilon$. So there are at most $N$ terms where $|x_n|>1$ for a given sequence. A bounded, non-empty subset of the reals has a supremum by the Completeness Axiom. Suprema Are Unique because they are both upper bounds. If one is greater than the other, the lesser is a smaller upper bound, so the larger can't be the least upper bound. So two candidate suprema must be equal. Given a non-empty bounded subset of the reals, $A$ and supremum L, $\forall \epsilon>0, \exists a\in A : \ |a-L|<\epsilon$. Otherwise $a\in A, a\le L-\epsilon$. So $L-\epsilon$ is a smaller upper bound, a contradiction. This is the Approximation Property. Consider the set B' composed of all elements of B less $sup(B)$. If $0\notin B'$, then it has an infinite number of elements by applying the Approximation property with diminishing values of $\epsilon$. It's finite, so $0\in B'$. So $sup(B)\in B$.

So a bounded, non-empty, finite set contains its supremum as its maximum element. To each sequence in the problem, we can associate this maximum element. Similar reasoning applies to infima, so we are guaranteed a maximum absolute value for any of these sequences.

Let integers $m,n>0 : \{x_n\}_m=\frac{m}{n} $. In other words, this indexes a set of sequences with $m$ having maximum element $m$. All these sequences converge to $0$. While each set has a supremum for it's absolute value, those suprema have no finite supremum.

Answer 2

If every element in the sequence equals $0$, then any element will work as your $x_K$.

If there are any non-zero elements, pick one, and let its absolute value be $\epsilon$. There is some tail of the sequence in which every element has absolute value less than $\epsilon$. The initial segment, preceding that tail, has finitely many elements, so (at least) one of them has maximum absolute value.

Does that give you enough of a push in the right direction?