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Let $A$ and $B$ be subsets of $\mathbb{R}$ such that $\sup(A)$ and $\inf(B$) exist and $\sup (A) < 0$. Set $C$ is defined as:

$$C =\{ 1/a + b : a \in A, b \in B\}$$

Prove that $\inf(C)$ exists.

Here's what I know:

  1. Since $\sup (A) < 0$, I know that $a$ $\in$ $A$ is negative $\implies$ $1/a$ is negative.

  2. $A$ is bounded above and $B$ is bounded below (by completeness).

  3. The smallest value for $1/a$ and $\inf(B)$ give us $\inf(C)$

I'm having trouble with using this information to set up a formal proof with progression. I'm also not completely confident with my assumptions.

Any tips?

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If $A$ is bounded above by $0$ and $\sup(A)<0$, then $$ \inf\Bigl\{\frac{1}{a}:a\in A\Bigr\}=\frac{1}{\sup(A)} $$ because the function $x\mapsto 1/x$ is decreasing on $(-\infty,0)$ (a direct proof via the definitions is also possible).

Now prove that if $X$ and $Y$ are bounded below, then $$ \inf\{x+y:x\in X, y\in Y\}=\inf(X)+\inf(Y) $$

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Write $s:=\sup(A)$ and $r:=\inf(B)$.

Let $a\in A$, then $a\leq s\leq 0$ implies that $1/a\geq 1/s$.

Let $b\in B$, then $b\geq r$.

For $c\in C$ write $c=1/a+b$ for some $a\in A$ and $b\in B$. It follows that $c\geq 1/s+r$ and so $1/s+r$ is a lower bound for $C$, hence $\inf(C)$ exists.

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  • $\begingroup$ The last sentence isn't true. Suppose $A = \{-1/2, -1/3, -1/4, \dots\}$ and $B = \{17\}$. Then $C$ is unbounded below. $\endgroup$ Feb 7 '18 at 23:13
  • $\begingroup$ Also, if $a < 0 < s$, then we do not have $1/a \ge 1/s$. $\endgroup$ Feb 7 '18 at 23:16
  • $\begingroup$ Thanks for the corrections. Edited the answer. $\endgroup$
    – Yanko
    Feb 8 '18 at 10:37

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