Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $i\in I$.

share|improve this question
@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
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.

share|improve this answer

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.

share|improve this answer
How the existence of function F garanteed? –  Katlus Jun 6 '12 at 13:38
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
@Katlus Apparently Martin Sleziak answer before I can. –  Giorgio Mossa Jun 7 '12 at 13:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.