Let $S \subset\mathbb{R}$ and $u$ an upper bound of $S$. If $u \in S$, then $\sup S= u$.

Is my proof correct?

Suppose $u \in S$.

  1. $u$ is an upper bound

  2. Suppose there exists another upper bound $v \in S$. Since $v$ is an upper bound $v>u$, a contradiction. So $v \leq u$.

I am confused since from the definition of supremum, for all upper bound $d$, $d$ must be greater than or equal to $u$, but I have shown otherwise, which is $v \leq u$.

  • $\begingroup$ If the main goal of your question is to ask specifically about your proof {as opposed to asking for any proof of this claim), then you should probably make this clearer by using the (proof-verification) tag, $\endgroup$ Commented Aug 24, 2017 at 16:16

1 Answer 1

  1. Suppose there exists another upper bound v $\in$ S. Since v is an upper bound v>u, a contradiction. So v $\leq$ u.

That's not ok. First of all, you didn't prove that $v\leq u$, because if $v$ is an upper bound, then all you really have is $v\geq u$. Second of all, you have no reason to assume that $v\in S$.

What you want to prove is that

if there exists some $v$ such that $v$ is an upper bound for $S$, then $v\geq u$.

There is no demand that $v\in S$. $v$ can be a real number.

So, restart your proof, and start with

"Suppose $v\in\mathbb R$ is an upper bound for $S$"

And you have to finish with $v\geq u$

  • $\begingroup$ @MashedPotato Looks good to me. Just a short piece of advice: Write entire mathematical expressions in dollar signs. So, don't write s$\leq$v (which produces "s$\leq$v"), but rather $s\leq v$ (producing a much nicer $s\leq v$). $\endgroup$
    – 5xum
    Commented Aug 24, 2017 at 15:18
  • $\begingroup$ Advice taken! Thank you so much $\endgroup$ Commented Aug 24, 2017 at 15:19

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