# $\sum_{b\prec a}\sigma_b\prec\prod_{b\prec a}\delta_b$ [duplicate]

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Suppose $a$ is a cardinal number and $\lbrace\sigma_b\rbrace_{b\prec a}$ and $\lbrace\delta_b\rbrace_{b\prec a}$ are two families of cardinal numbers such that for all $b\prec a$, $\sigma_b\prec\delta_b$. please help me to prove that $$\sum_{b\prec a}\sigma_b\prec\prod_{b\prec a}\delta_b$$ or any hint. Thanks.

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## marked as duplicate by Asaf Karagila, Daniel Rust, Rebecca J. Stones, amWhy, M TurgeonOct 6 '13 at 15:02

Hint: Pick pairwise disjoint sets $\{ A_b \}_{b < a}$ with $| A_b | = \sigma_b$, and sets $\{ B_b \}_{b < a}$ with $| B_b | = \delta_b$, and let $f : \bigcup_{b < a} A_b \to \prod_{b < a} B_b$ be arbitrary. For each $b < a$ consider the function $f_b : A_b \to B_b$ defined by $$f_b ( x ) = \text{the }b^{\text{th}}\text{ coordinate of }f(x).$$ What do we know about each of these functions $f_b$? Can you now build an element of $\prod_{b < a} B_b$ which is not in the range of $f$?
Oh, yes there not exist an onto function from $\bigcup_{b<a} A_b$ to $\prod_{b<a}B_b$. Thanks very much. –  user97619 Oct 6 '13 at 11:48