If a set has an upper bound, it has infinitely many upper bounds.

Question

Let $A$ be a subset of the real numbers, with $A \neq \emptyset$. Prove that if $x$ is an upper bound of $A$, then $A$ has infinitely many upper bounds.

This seems like something that is pretty obvious to me. But I'm not very good at coming up with proofs yet. How would you go about proving this? Maybe a proof by contradiction?

Thanks!

Answer 1

$S$ is subset of Real numbers

Let $x$ be the upper bound of $S$.

Let $s$ be an element of $S$. Then $\forall s$, $s<x$. Therefore $s<x+r$ where $r\geq0$ for all $r$ in Real numbers.

Hence, $S$ has infinitely many upper bounds.