2. Let D be a nonempty set and suppose that f: D R and g: D R.

Define the function f + g: D R by (f + g)(x) = f(x) + g(x).

Recall that sets f(D), g(D), and (f + g)(D) are the ranges of f, g, and f + g, respectively.

#2 (a) Fill in the blanks below to prove the following:

If f(D) and g(D) are bounded above, then (f + g)(D) is bounded above

and sup[(f + g)(D)] sup f(D) + sup g(D).

Proof:

Let D be a nonempty set, and suppose that f: D R and g: D R, and f(D) and g(D) are bounded above.

Since D is nonempty, f(D), g(D), and (f + g)(D) are nonempty.

Since f(D) and g(D) are nonempty and bounded above, by the ________________________(axiom/property),

sup f(D) and sup g(D) must exist and be finite.

Let u = sup f(D) and v = sup g(D).

Then ______ u for all x in D and g(x) _____ for all x in D.

Thus (f + g)(x) = f(x) + g(x) _________ for all x in ___, so (f + g)(D) is bounded above by __________.

Since (f + g)(D) is nonempty and bounded above, by the ___________________________(axiom/property), sup[(f + g)(D)] must exist and be finite.

So, the least upper bound for (f + g)(D) must be less than or equal to __________ (same as earlier blank) .

Thus, sup[(f + g)(D)] sup f(D) + sup g(D).

#2(b). Consider this statement:

If f(D) and g(D) are bounded above, then sup[(f + g)(D)] = sup f(D) + sup g(D).

Find a counterexample to show that this statement is false.

[Thus you will be showing that back in part (a), " " could not be replaced by "=". ]

Briefly explain your counterexample.