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I want to show that if $\{X_\beta : \beta<\lambda^+\}$ is an enumeration of $[\lambda\times\lambda^+]^{\leq\lambda}$ then, for all $i<\lambda$, the set $\{(X_\beta)_i:\beta<\lambda^+\}$ is an enumeration of $[\lambda^+]^{\leq\lambda}$ where $(X_\beta)_i=\{\alpha<\lambda^+ : (i,\alpha)\in X_\beta\}$.

Let $i<\lambda$ fixed.

Let $\beta<\lambda^+$. Evidently, the cardinal of $(X_\beta)_i$ is less than $\lambda$ so $\{(X_\beta)_i:\beta<\lambda^+\}\subseteq [\lambda^+]^{\leq\lambda}$.

Let $X\subseteq\lambda^+$ such that $|X|\leq\lambda$. Let $\{x_j : j<\gamma\}$ with $\gamma\leq\lambda$ an enumeration of $X$. Then the set $\{(i;x_j) : j<\gamma\}\subseteq \lambda\times\lambda^+$ is of cardinal less than $\lambda$ so there exists a $\beta<\lambda^+$ such that it is equal to $X_\beta$. But then $X=(X_\beta)_i$ and $\{(X_\beta)_i:\beta<\lambda^+\}=[\lambda^+]^{\leq\lambda}$.

Does my proof is correct and does an enumeration is always construct with disjoints sets ?

regards.

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In the third paragraph you mean that the cardinality of $(X_\beta)_i$ is less than or equal to $\lambda$, but otherwise it looks fine. –  Brian M. Scott May 5 '12 at 9:10
    
Sorry: I didn't notice the last question before. No: here we have enumerations of all sets of a given type. –  Brian M. Scott May 5 '12 at 9:39
    
Yes, I meant that the cardinality of $(X_\beta)_i$ is less or equal to $\lambda$. Thanks for your answer. –  Marc Moretti May 5 '12 at 11:53
    
I'm sorry but the last question is wrong. I wanted to ask if an enumeration is constructed with distincts sets and not disjoints of course. –  Marc Moretti May 5 '12 at 12:39
    
An enumeration does not have to be one-to-one; in particular, I don't think that $\{(X_\beta)_i:\beta<\lambda^+\}$ is necessarily a 1-1 enumeration of $[\lambda^+]^{\le\lambda}$. –  Brian M. Scott May 5 '12 at 17:20
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