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Defined $X*Y ≡ \{ z\in\mathbb R:\text{ there exist }x\in X , y\in Y\text{ such that }z = xy \}$.

For $|X|,|Y| = \aleph_0$ I have to prove that $|X*Y|=\aleph_0$.

Ideas?

Thanks in advance.

Brian M. Scott
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adamco
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2 Answers2

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I assume $X,Y\subseteq\mathbb{R}$? You obviously have that $X\ast Y$ is infinite, but you also have that $X\times Y\to X\ast Y:(x,y)\mapsto xy$ is a surjection, and so $\#(X\ast Y)\leqslant \#(X\times Y)=\aleph_0$.

Alex Youcis
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You have an obvious surjective map $f:X\times Y\to X*Y$. Therefore $$|X*Y| \le |X|\times |Y|= \aleph_0.\aleph_0=\aleph_0.$$

On the other hand, for any $y\in Y$, $y\ne 0$ you have a bijection $X\to X*\{y\}$, $x\mapsto xy$. (You know that $Y$ contains a non-zero number, since $|Y|=\aleph_0$, hence you can choose some element $y\in Y\setminus\{0\}$.) Since $X*\{y\}\subseteq X*Y$ you get $$|X*Y| \ge |X*\{y\}| = |X| = \aleph_0.$$

For the proof of $\aleph_0.\aleph_0=\aleph_0$ see e.g. Bijecting a countably infinite set $S$ and its cartesian product $S \times S$, How does one get the formula for this bijection from $\mathbb{N}\times\mathbb{N}$ onto $\mathbb{N}$?, The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable and Proving the Cantor Pairing Function Bijective

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Martin Sleziak
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