(Let R represent the real numbers) Let D be a nonempty set and suppose f: D --> R and g: D-->R. Define f+g: D-->R by (f+g)(x) = f(x)+g(x). If f(D) and g(D) are bounded above then PROVE sup[(f+g)(D)] is less than or equal to sup f(D) + sup g(D).

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- Oct 13th 2008, 04:17 AMGoldendoodleMomSupremum
(Let R represent the real numbers) Let D be a nonempty set and suppose f: D --> R and g: D-->R. Define f+g: D-->R by (f+g)(x) = f(x)+g(x). If f(D) and g(D) are bounded above then PROVE sup[(f+g)(D)] is less than or equal to sup f(D) + sup g(D).

- Oct 13th 2008, 07:21 AMPlato
From the given we know that these exist: $\displaystyle \alpha = \sup \left( {f(D)} \right)\,\& \,\beta = \sup \left( {g(D)} \right)$.

$\displaystyle \left( {\forall z \in D} \right)\left[ {f(z) + g(z) \leqslant \alpha + \beta } \right]

$.

This means that $\displaystyle {\alpha + \beta }$ is an upper bound of $\displaystyle \left[ {f + g} \right](D)$.

Can you finish?