Solution of the 9th Homework
نویسنده
چکیده
Proof. Let us first assume that c > 0. To prove that sup(cA) = c(sup A), we claim that sup(cA) ≤ c(sup A) and sup(cA) ≥ c(sup A). For the first inequality, let a′ ∈ cA be arbitrary. Then a′ = ca for some a ∈ A. But since a ≤ sup A, we have a′ = ca ≤ c(sup A). This shows that c(sup A) is an upper bound of cA, hence we have sup(cA) ≤ c(sup A). The reverse inequality also follows in a similar manner. (Or notice that c sup A = c sup(c−1cA) ≤ cc−1 sup(cA) = sup(cA).) Then inf(cA) = c(inf A) also follows in the same way. When c < 0, the proof goes in almost the same way, but what changes now is that multiplying c to an inequality reverses the order. I leave the detail of the proof to you.
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