Proving elements near supremum exist in set S

[Eclair_de_XII]

So, $x-\epsilon$ is not an upper bound for $S$ because there exists an $a∈S$ such that $a>x-\epsilon$.

...Oh, so because $x-\epsilon$ is not an upper bound for $S$, then it is implied that there must exist an element $a∈S$ greater than $x-\epsilon$?

 

[PeroK]

So, $x-\epsilon$ is not an upper bound for $S$ because there exists an $a∈S$ such that $a>x-\epsilon$.

...Oh, so because $x-\epsilon$ is not an upper bound for $S$, then it is implied that there must exist an element $a∈S$ greater than $x-\epsilon$?

Yes!

 

[Eclair_de_XII]

Thanks for bearing with me and helping me finish my homework. I'm kind of apprehensive about how I'll handle the rest of the class, to be honest...

 

[LCKurtz]

So, $x-\epsilon$ is not an upper bound for $S$ because there exists an $a∈S$ such that $a>x-\epsilon$.

...Oh, so because $x-\epsilon$ is not an upper bound for $S$, then it is implied that there must exist an element $a∈S$ greater than $x-\epsilon$?

I like the fact that you italicized the word "must" above. That's the first inkling you have given that you actually get it. Good for you. That's what this forum is here for.