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Let $\{A_n\mid n\in \mathbb{N}\}$ be disjoint family of sets such that $\forall n\in\mathbb{N},\; \aleph(A_n)<c$. Is it true that $\aleph(\bigcup_{n\in\mathbb{N}}A_n)=\sum_{n\in\mathbb{N}}\aleph(A_n)<c$?

Here, $c=2^{\aleph_0}$.

I've been trying for a while a can't seem to be able to prove it. If it's true, a proof or a guide for a proof would be highly appriciated.

Sebastián P. Pincheira
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  • @bof . $\aleph(A)$ is meant to mean the cardinal number of $A$. It is used on the text Topology by James Dugundji. It's ussage on the text made me think it was standard use. Apparently not. (In said text $\text{card}; X\leq\text{card}; Y $ is meant to mean that there is an inyection $X\to Y$ yet $\aleph(X)\leq \aleph(Y)$ is meant to mean that there is an order isomorphism from $\aleph(X)$ to a subset of $\aleph(Y)$. This two notions are proven to be equivalent). – Sebastián P. Pincheira Apr 25 '22 at 04:03
  • The fact that I don't recall seeing it before doesn't mean it's not standard. And I'm not going to argue with Dugundji. Evidently the experts here knew what it meant. – bof Apr 25 '22 at 05:41

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