I have a question regarding the placement of a given product of infimums in relation to an arbitrary lower bound in the following question and would appreciate any clarifications. Thank you.
Let $S \subset \Bbb{R^+}$ and $T \subset \Bbb{R^+}$ be nonempty and bounded below.Prove that $P:=\{xy : x \in S \text{ and } y \in T\}$ is bounded from below and that $inf(P)=inf(S)*inf(T)$.
Attempt :
(i) $S$ and $T$ are nonempty subsets of $\Bbb{R^+}$ that are bounded below $\Rightarrow$ Define $\alpha=inf(S) \geq 0$ and $\beta=inf(T) \geq 0$ and note that $\alpha \leq x$ for all $x\in S$ and $\beta \leq y$ for all $y \in T$ $\Rightarrow$ $(xy) \geq \alpha y \geq \alpha \beta$ $\Rightarrow$ P is bounded below.
(ii) Having determined in (i) that P has a lower bound, proving that $inf(P)=inf(S)*inf(T)= \alpha \beta$ amounts to:
- assuming that c is a lower bound for P
- proving that $c \leq \alpha \beta$
Hence:
Suppose c is a lower bound for P $\Rightarrow$ $c \leq (xy)$ for all $x \in S$ and $y \in T$ $\Rightarrow$ Define $\gamma=inf(P)$ and distinguish between two cases [a] $c \leq 0$ and [b] $c \gt 0$.
[a] With $c \leq 0$, then $c \leq 0 \leq \alpha \beta \leq \gamma \leq (xy)$ and hence $c \leq \alpha \beta$ as required.Noting that $\alpha \beta \geq 0$
[b] With $c \gt 0$ and $c$ being a lower bound for P $\Rightarrow$ $c \leq (xy) \text{ for all } x \in S \text{ and } y \in T$ $\Rightarrow$ $c \over y$ $\leq x \text{ for all } x \in S$ $\Rightarrow$ $c\over y$ is a lower bound for $S$.$\Rightarrow$ $c \over y$ $\leq \alpha \leq x$ $\Rightarrow$ $c \leq \alpha y \leq xy$.
Now,can I simply conclude that $c \leq \alpha \beta \leq \alpha y \leq xy$ from the fact that c is some arbitrarily chosen lower bound? Is it not possible for $\alpha \beta \leq c$ instead?
Thank you for any explanations.
a = inf S; b = inf T; c = inf ST.
Pick x in S.
For all y in T, c <= xy, c/x <= y.
Thus c/x <= b, c/b <= x.
As for all x in S, c/b <= x, c/b <= a.
Thus c <= ab.
As ab is a lower bound for ST, ab <= c.