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If |f(x)−g(x)| < ε for all x∈A, then Sup(of A)F−sup(of A)G ≤ ε.

I am trying to prove this statement. And I tried to implement a similar idea as Suprema Proof. (Suprema proof: prove $\sup(f+g) \le \sup f + \sup g$)

However, the idea that Supremum need not necessarily be in contained in the domain gives a little more room in the second part. (strict inequality turns into less than or equal to). I need a little hint on how to tweak with this idea of supremum.

Joon
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$f(x) <g(x)+\epsilon$ for all $x \in A$. This implies that $f(x) <\sup_{y \in A} g(y)+\epsilon$. This in turn means that $\sup_{y \in A} g(y)+\epsilon$ is an upper bound for $\{f(x): x\in A\}$. Hence $\sup_{y \in A} f(y) \leq \sup_{y \in A} g(y)+\epsilon$.

Kavi Rama Murthy
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  • Thank you for the answer. I definitely understood the idea. Can I assume f(x) > g(x) without loss of generality here. – Joon Dec 26 '19 at 06:05
  • @Joon No need to assume that $f(x)>g(x)$. If $|a| <\epsilon$ then $a <\epsilon $ whether $a$ is positive or not. [$a=f(x)-g(x)$ in our case]. – Kavi Rama Murthy Dec 26 '19 at 06:14