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How do i prove $\sum a_i$ is equipotent with a subset of $\prod a_i$ ?? I seems obviously true but its actually hard to prove it... $\{a_i\mid i\in I\}$ is a set of cardinals and $a_i$ is a cardinal for each $i\in I$.

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    @Asaf I learnt that product symbol you wrote and X are different.. Am i wrong?2012-06-06
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    X denotes product of disjoints sets and $\prod$ denotes product of sets with no constrain2012-06-06
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    I’ve never encountered that convention.2012-06-06
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    @Brian: Hajnal and Hamburger use somewhat similar convention in their book, see [p.28](http://books.google.com/books/about/Set_Theory.html?id=pf0Rmrv-eDYC&pg=PA28) and [p.30](http://books.google.com/books/about/Set_Theory.html?id=pf0Rmrv-eDYC&pg=PA30)2012-06-06

2 Answers 2

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I assume that $a_i\ge 2$ for each $i$.

I would first try to show that $$a+b\le a\cdot b \tag{1}$$ whenever $a,b\ge 2$.

Then I would try to continue by transfinite induction - i.e. I would assume that $I$ can be well ordered, which means I can work with cardinals $a_\gamma$ for $\gamma<\alpha$.

Inductive step in the transfinite induction:

a) Non-limit ordinals: If we know that $\sum\limits_{\gamma<\alpha} a_\gamma<\prod\limits_{\gamma<\alpha} a_\gamma$ then $$\sum_{\gamma<\alpha+1} a_\gamma=\sum_{\gamma<\alpha} a_\gamma + a_\alpha \le \prod_{\gamma<\alpha} a_\gamma + a_\alpha \overset{(1)}\le \prod_{\gamma<\alpha+1} a_{\gamma}.$$

b) Limit ordinals: Suppose that $\alpha=\sup\{\beta; \beta<\alpha\}$. Then $$\sum_{\gamma<\alpha} a_\gamma = \sup_{\beta<\alpha} \sum_{\gamma<\beta} a_\gamma \le \sup_{\beta<\alpha} \prod_{\gamma<\beta} a_\gamma \le \prod_{\gamma<\alpha} a_\gamma.$$


You can find a different proof (without using transfinite induction) of a slightly more general result as Theorem 1.6.7a) in the book Michael Holz, Karsten Steffens, E. Weitz: Introduction to Cardinal Arithmetic, p.61. The second part of this theorem is König's theorem, which is a very useful result.

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Consider your family of cardinals $\langle a_i \rangle_{i \in I}$ then $$\sum a_i = \left \{ (x,i) \mid x \in a_i, i \in I\right\}$$ and $$\prod a_i = \left\{ f \colon I \to \bigcup a_i \mid \forall i \in I \ f(i) \in a_i\right\}$$

Consider the function

$$F \colon \sum a_i \to \prod a_i$$ where for each $(x,i) \in \sum a_i$ we have $F(x,i) \colon I \to \bigcup a_i$ such that $F(x,i)(j)=0$ if $j \ne i$ and $F(x,i)(i)=x+1$ (here by $x+1$ we mean the ordinal successor $x \cup \{x\}$). Now given a pair $(x,i),(y,j) \in \sum a_i$ if $F(x,i)=F(y,j)$ then $$x + 1 = F(x,i)(i) = F(y,j)(i)$$ and so $j=i$, otherwise $F(y,j)(j) = 0 = x+1$ which cannot be true.

Because $$x+1 = F(x,i)(i) = F(y,i)(i) = y+1$$ by properties of ordinals we must have that $x=y$.

So if $F(x,i)=F(y,j)$ then $(x,i)=(y,j)$ and so $F$ is an injective function.

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    How the existence of function F garanteed?2012-06-06
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    I guess ineff assumes that all $a_i$'s are infinite cardinals, so $x+1$ belongs to $a_i$ whenever $x$ belongs to $a_i$. (If we represent cardinals as ordinals, see e.g. [here](http://en.wikipedia.org/wiki/Von_Neumann_cardinal_assignment).)2012-06-06
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    BTW it is worth noticing that this is somewhat similar to the diagonal argument used in the proof of Cantor's theorem. If each $a_i$ would be equal to $2=\{0,1\}$ we could map 0 to 1 and 1 to 0 on the $i$-th coordinate.2012-06-06
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    @Katlus Apparently Martin Sleziak answer before I can.2012-06-07