Prove that $\sup Z=\frac {\sup X}{\inf Y}$

Question

Let $X$ and $Y$ be two nonempty bounded sets of positive real numbers.

Define $Z=\{\frac{x}{y}:\space x\in X,y\in Y\}$. If $\inf Y>0$, prove that $\sup Z=\frac {\sup X}{\inf Y}$

This is one of the questions I encountered in the first mathematical analysis course. I noticed that $\frac{x}{y}$ is bounded above by $\frac {\sup X}{\inf Y}$. Then I assume to the contrary that $w=\sup Z<\frac {\sup X}{\inf Y}$, and want to come up with a contradiction. In particular, I thought about finding an $\frac{x_0}{y_0}$ where $\frac{x_0}{y_0}>\frac{w+\frac{\sup X}{\inf Y}}{2}$. However, I cannot find it out explicitly.

Can someone help me? Thanks in advance!

Answer 1

The 'missing' part of your argument can be completed as follows.

Let $\{x_n\}$ be an increasing sequence of elements of $X$ converging to $\sup X$.

Let $\{y_n\}$ be a decreasing sequence of elements of $Y$ converging to $\inf Y$.

Then $\{x_n/y_n\}$ is an increasing sequence of elements of $Z$ converging to $\sup X/\inf Y$.

Hence $\sup Z\ge \sup X /\inf Y$.

Answer 2

Note that

\begin{align} \sup Z &= \sup_{x\in X, y\in Y} \frac{x}{y} \\ &= \sup_{x\in X} x \, \sup_{y\in Y} \frac{1}{y}\\ & = \sup X \, \sup \frac{1}{Y}, \end{align}

Where the notation in the last line should be evident.

So we need to show that, when $\inf Y > 0$

$$ \sup \frac{1}{Y} = \frac{1}{\inf Y}. $$

You can think about it for a bit or look at this question for the solution.