I was reading a proof of a theorem that goes like this:
Let $A$ be an infinite set of cardinality $k$ and $A^{<\omega}$ the set of finite sequences of elements of A. Then $\operatorname{card}(A^{<\omega})=k$.

In the proof, we note that $A^{<\omega}=\bigcup\limits_{n \in \mathbb{N}} \underbrace{A\times...\times A}_{n}$. So at some point, we have to show $\operatorname{card}(A \times A)=k$, and for this the author uses the fact that $\operatorname{card}(A \times A)=\operatorname{card}(\operatorname{card}(A) \times \operatorname{card}(A))$.
I am not super sure why this is the case. My thoughts: We have by definition a bijection $A \leftrightarrow \operatorname{card}(A)$. So we have a bijection $A \times A \leftrightarrow \operatorname{card}(A) \times \operatorname{card}(A)$. So in particular since these sets are bijective they have the same cardinal $\operatorname{card}(A \times A)=\operatorname{card}(\operatorname{card}(A) \times \operatorname{card}(A))$. Is that it?
So from this we can prove $\operatorname{card}(\underbrace{A\times...\times A}_{n})=k$ $\forall n$.

Then the author says $k\le \operatorname{card}(\bigcup\limits_{n \in \mathbb{N}} \underbrace{A\times...\times A}_{n}) \le \operatorname{card}(A)\cdot\omega$ where the $\cdot$ is the cardinal product. Why is that?


So far, so good. Next you need $k^2=k$ to prove by induction that each of the products has cardinality $k$. Then their union has at most size $k\aleph_0\le k^2=k$. But the proof there are at least $k$ sequences is trivial.

  • $\begingroup$ In general, what bound do we have : $\operatorname{card}(\bigcup\limits_{i \in I} B_i) \le ?$ $\endgroup$ – roi_saumon Jan 23 at 23:56
  • 1
    $\begingroup$ @rol_saumon We have lower bound $\max_i\operatorname{card}(B_i)$, upper bound $\sum_i\operatorname{card}(B_i,)$. $\endgroup$ – J.G. Jan 24 at 6:02

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