Characteristic property of supremum on a set of the form $\{x_n: n\in \mathbb{N}\}$

Question

Let $(x_n)$ be real numbers sequence. If $\sup \{x_n:n\in\mathbb{N}\}=1$, from the characteristic property of supremum it holds that for every $\varepsilon>0,$there exists $n\in\mathbb{N}$ such that $1-\varepsilon<x_n$. If we let $\varepsilon$ be $1/n$ we get $x_{k_n}$ such that $1-1/n<x_{kn} , \forall n.$ Could we suppose that the elements $x_{k_n}$ we collected are a subsequence of $(x_n)$? Or is there any tric to get a subsequence of $(x_n)$ that $x_{k_n}\to 1$? Thanks.

Answer 1

I’m not exactly sure if you can call it a subsequence—- Consider the sequence $1,0,-1…$ where $x_{n+1}=x_n-1$, clearly the supremum of this sequence is $1$, whence you cannot find a subsequence that converges to $1$.
Given a sequence $x_n$, let $n_1,n_2…$ be positive integers where $n_1<n_2…$, then the sequence $x_{n_k}$ is called a subsequence of the original sequence. Picking the same element over and over again does not satisfy the definition of a subsequence.

Answer 2

Let's say that $0\notin \Bbb N$ for convenience. Given a sequence $(x_n)_{n\in\Bbb N}$, a subsequence is, by definition, its composition with a strictly increasing function $\Bbb N\to\Bbb N$. If, for instance, your original $x_n$ is $x_n=\frac1n$, then there is no subsequence $\left(x_{n_k}\right)_{k\in\Bbb N}$ such that $x_{n_k}>\sup\{x_n\,:\, n\in\Bbb N\}-\frac1k$ for all $k$.