Definition of supremum and infimum using epsilons ?

[AxiomOfChoice]

Is this what it is:

"For every $\epsilon > 0$ there exists $x\in A$ such that $x \leq \inf A + \epsilon$."

...and similarly for the supremum?

 

[LCKurtz]

No, that's not quite it. Any number y which is greater than all numbers in A would satisfy your definition.

 

[snipez90]

Well in the first definition the OP said "there exists", so really the concern of upper bounds isn't really important. This is a useful proposition, and sometimes the inequality is strict, though that doesn't matter all that much.

 

[LCKurtz]

What "first definition"? Are you talking about something other than:

"For every $$\epsilon > 0$$ there exists $$x\in A$$ such that $$x \leq \inf A + \epsilon .$$"?

While that is true about the infinimum, it won't do for the definition for the reason I gave.

 

[John Creighto]

What "first definition"? Are you talking about something other than:

"For every $$\epsilon > 0$$ there exists $$x\in A$$ such that $$x \leq \inf A + \epsilon .$$"?

While that is true about the infinimum, it won't do for the definition for the reason I gave.

That's correct. We must also state that inf A is a lower bound.

 

[snipez90]

Yeah I realized I was thinking of the theorem that states that if L is a lower bound for a set A in R, then L = inf A iff for every epsilon > 0, there is an x in A with x - L < epsilon. My apologies, of course it's not a definition.