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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? –  Katlus Jun 6 '12 at 12:48
    
X denotes product of disjoints sets and $\prod$ denotes product of sets with no constrain –  Katlus Jun 6 '12 at 12:50
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I’ve never encountered that convention. –  Brian M. Scott Jun 6 '12 at 13:11
    
@Brian: Hajnal and Hamburger use somewhat similar convention in their book, see p.28 and p.30 –  Martin Sleziak Jun 6 '12 at 13:17

2 Answers 2

up vote 2 down vote accepted

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? –  Katlus Jun 6 '12 at 13:38
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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.) –  Martin Sleziak Jun 6 '12 at 13:39
    
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. –  Martin Sleziak Jun 6 '12 at 13:42
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@Katlus Apparently Martin Sleziak answer before I can. –  Giorgio Mossa Jun 7 '12 at 13:54

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