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**NonCommAlg** for every $\displaystyle i \in I$ put $\displaystyle Y_i=X_i \times \{i \}.$ see that $\displaystyle \text{card} \ \bigcup_{i \in I}X_i \leq \text{card} \ \bigcup_{i \in I} Y_i.$ clearly $\displaystyle \text{card} \ X_i = \text{card} \ Y_i$ and $\displaystyle Y_i \cap Y_j = \emptyset,$ for all $\displaystyle i \neq j.$ now let $\displaystyle J=I \times \mathbb{N}.$ then $\displaystyle \text{card} \ I = \text{card} \ J.$ we also have an

one-to-one map $\displaystyle f_i : Y_i \longrightarrow \mathbb{N},$ for every $\displaystyle i \in I.$ finally define $\displaystyle g : \bigcup_{i \in I} Y_i \longrightarrow J$ by $\displaystyle g(y)=(i,f_i(y)),$ for all $\displaystyle i \in I, \ y \in Y_i.$ see that $\displaystyle g$ is one-to-one and thus $\displaystyle \text{card} \ \bigcup_{i \in I} Y_i \leq \text{card} \ J. \ \Box$