Let $D_{n}$ be a set with $2^{n}$ elements for $n=1,2,...$. Let $A = \bigcup_{n=1}^{\infty}D_{n}$, and let $B = \prod_{n=1}^{\infty}\{0,1\}$. Let $A_{k} = \bigcup_{n=1}^{k} D_{n}$, and let $B_{k} = \prod_{n=1}^{k} \{0,1\}$.

I'm familiar with the proofs that $A$ is countable and $B$ is uncountable, but it seems odd that the "partial union" $A_{k}$ has at least as many elements as the "partial product" $B_{k}$ for each natural number $k$.

Intuitively, why does the limit of the union yield a set with a lower cardinality than that of the limit of the product?

  • 3
    $\begingroup$ Union is adding, product is multiplying. $\endgroup$ – copper.hat Mar 9 '14 at 7:39

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