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Let f, g, be real-valued functions defined on a nonempty set X satisfying Range f and Range g are bounded subsets of $\mathbb{R}$. Prove the following...

A) If $f(x) \leq g(x)$ for all $ x \in X $ then $sup\left \{ f(x): x \in X \right \}\leq sup\left \{ g(x): x \in X \right \}$

This is what I have for my proof...

Let $f(x) \leq g(x)$ for all $ x \in X $

Since g is a bounded set then by definition of bounded g has an upper bound so that $f(x) \leq g(x) \leq sup g(x)$

Thus f is bounded above by sup g so $supf(x) \leq supg(x)$ for all $x \in X$

Therefore, $sup\left \{ f(x): x \in X \right \}\leq sup\left \{ g(x): x \in X \right \}$

This is what I have thought to do to prove this, but is this even correct? If there are mistakes within my proof could someone explain/show what should be written.

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  • $\begingroup$ Your proof is pretty nice. I think your second line should be a bit more specific when talking about the supremum instead of just an upper bound. And for your second to last line you may want to justify that step a little bit more from the definition of supremum. $\endgroup$ – lordoftheshadows Feb 24 '17 at 4:51
  • $\begingroup$ That proof looks good to me! I think the latter half of your second to last line and your last line are really the same thing. I prefer the statement of the last line as the "for all $x \in X$ in the second to last line seems superfluous because $\sup$ is clearly about a set, not a single point $f(x)$ for a single $x$. $\endgroup$ – FalafelPita Feb 24 '17 at 4:57
  • $\begingroup$ Does this look better for the proof/any other changes?.... Let $ f(x)≤g(x)f(x)≤g(x)$ for all $x∈X$. Since g is a bounded set then by definition of bounded it is bounded above. Thus g has an upper bound and since g is a nonempty set then there exists a supremum so that $f(x)≤g(x)≤supg(x)$. Therefore, f is bounded above by sup g so that $sup{f(x):x∈X}≤sup{g(x):x∈X}$ since g is an upper bound. $\endgroup$ – user6259845 Feb 24 '17 at 5:12

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