Let $\{x_n\}_{n\in\mathbb{N}}$ and $\{y_n\}_{n\in\mathbb{N}}$ be two sequences in $[0,1]$. Consider the set,
$$ A = \left\{\frac{1}{2^n}|x_n-y_n| : n \geq 1\right\} $$
It's clear that $A$ is bounded above and non-empty and so $\alpha := \sup A$ exists. However, I'm having difficulty showing why $\alpha \in A$.
I tried proving that $A$ is closed, but I don't think this is necessarily true. We can see that $\inf A = 0$, but $0 \in A$ only if $x_n = y_n$ for some $n \geq 1$. But this doesn't have to be true.
So I don't think I can prove that $\alpha \in A$ using topological arguments; only first principles using the properties of the supremum. Any hints on how to start?
The supremum of $A$ is in $A$, because for $n$ large enough, the product $\frac{1}{2^n}|x_n - y_n|$ will be negligible. Thus, the $\sup$ will be the maximum of the earlier elements.
To make this precise, first note that if $x_n = y_n$ for all $n$, then $\sup A = 0$ and $0$ is contained in $A$. Otherwise, let $k$ be the minimal positive integer with $|x_k - y_k| \neq 0$. By the triangle inequality, $|x_n - y_n| \leq 2$ for all $n$, so we can pick $N$ so that for all $n > N$, we have $\frac{1}{2^n} |x_n - y_n| < \frac{1}{2^k}|x_k - y_k|$. It follows that $\sup A = \max_{n \leq N} \frac{1}{2^n} |x_n-y_n|$, so $\sup A \in A$.