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

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\$.