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For subsets $X$ and $Y$ of $\mathbb{R}$, define: $$X*Y ≡ \{ z\in ℝ \mid\text{there exist }x\in X, y\in Y\text{ such that }z = xy \}.$$

Then define for $0\lt a\lt b$, $[a,b] = \{x\in\mathbb{R}\mid a\leq x\leq b\}$ and let $X=[a,b]$.

and they ask me to find $Y\subseteq \mathbb{R}$ such that $X*Y=X$.

any ideas?

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How about $Y=\{1\}$? – David Mitra May 3 '12 at 16:15
Because of formatting issues, it is not clear what is asked for. But if $Y=\{1\}$ then $XY=X$. Note that by the definition of set multiplication, $Y$ must be a subset of $\mathbb{R}$, not an element of $\mathbb{R}$. – André Nicolas May 3 '12 at 16:16
By the way, it should be $Y\subseteq \mathbb{R}$ rather than $Y\in \mathbb{R}$, since $Y$ appears to be a set, not a real number. – Tara B May 3 '12 at 16:18
$Y = \{1\}$ is the only solution, for if $Y = \emptyset$, then $XY = \emptyset \ne X$ and if $y\in Y$ with $y > 1$ then $XY \ni yb \not\in [a,b]$ as $yb > b$ and if $y \in Y$ with $y < 1$ then $ya < a$ and hence $ya \in XY\setminus X$. – martini May 3 '12 at 16:20
A simpler way to write the definition is $X*Y=\{xy \mid x\in X\text{ and }y\in Y\}$. – Michael Hardy May 3 '12 at 17:29
up vote 0 down vote accepted

Hint If $y$ is any element of $Y$, then $ay \in X*Y$ thus

$$a \leq ay \leq b$$

and $by \in X*y$ thus

$$a \leq by \leq b \,.$$

Since $a,b$ are positive, the inequalities $a \leq ay$ and $by \leq b$ tell you that there is only one possible value $y$ can take....

P.S. The problem is more interesting, and more complicated if $a,b$ have opposite signs. There are then multiple solutions for $Y$, namely all the subsets of some fixed interval...

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