Let $S subseteq mathbb{R}$ have a supremum $x$. Show that there is a sequence $(x_n)$ in $S$ that converges to $x$. Code Answer

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I understand how to prove this using the approximation property of the supremum and invoking the squeeze theorem of limits. However, upon reading answers for similar questions, I think there’s a more direct way of proving this without using the squeeze theorem. I’m having trouble understanding the logic though. I was wondering if I could receive some guidance. Here is what I have so far:

Since $x$ is the least upper bound of $S$, for all $n in mathbb{N}$ we can pick an $x_n in S$ such that $x_n > x – frac{1}{n}$. Since $x$ is an upper bound of $S$, $x geq x_n$ for all $n in mathbb{N}$. Therefore we have $x geq x_n > x-frac{1}{n}$ for all $n in mathbb{N}$.

How do I go from here to showing that $(x_n)$ converges to $x$ using the definition of convergence (without using squeeze theorem)?

Answer

For every fixed $epsilon >0$, there is an $Ninmathbb{N}$ such that $frac{1}{n}<epsilon$ for all $ngeq N$ (this follows by the Archimedean property). Thus, for all $ngeq N$, $|x_{n}-x|=x-x_{n}<x-(x-frac{1}{n})=frac{1}{n}<epsilon$.

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