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On Page 65, Set Theory, Jech(2006), Collection Principle is formulated as follows:

$\forall{X}\exists{Y}(\forall{u}\in{X})[\exists{v} \psi(u, v, p) \to(\exists{v\in{Y}}) \psi(u, v, p)]$
($p$ is a parameter)

Even with help of some explanations below it, I still have difficulty understanding it.

Here's how far I understand it.

$X$ is an arbitrary set of index. $u$ is a generic element of $X$. $Y$ is a set whose element is determined by $X$. If there exists $v$ that satisfy $ \psi(u, v, p)$, then $v$ belongs to $Y$.

My interpretation deviates from the textbook, which is as follows:

Let $C_u=\{v:\psi(u, v, p)\}$, if $C_u \neq \emptyset$, then $C_u \cap Y \neq \emptyset$.

Another question of mine is if so, is it correct to replace $v$ in $(\exists{v\in{Y}}) \psi(u, v, p)]$ with other arbitrary variable, say, $w$? Then the Collection Principle should be modifies as: $\forall{X}\exists{Y}(\forall{u}\in{X})[\exists{v} \psi(u, v, p) \to(\exists{w\in{Y}}) \psi(u, w, p)]$

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A word of advice, if you ever intend to reach forcing (or algebraic topology, of something like that) avoid talking about generic elements. :-) –  Asaf Karagila Dec 5 '12 at 7:01
    
@AsafKaragila: Thank you for your advise. I learned that word from those economists who see the economy as an aggregation of many "generic" firms and consumers. –  Metta World Peace Dec 5 '12 at 7:18
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1 Answer 1

up vote 1 down vote accepted

$X$ is an arbitrary set of index. $u$ is a generic element of $X$. $Y$ is a set whose element is determined by $X$. If there exists $v$ that satisfy $ \psi(u, v, p)$, then $v$ belongs to $Y$.

No, this isn’t quite right: it actually says that if there is some $v$ that satisfies $\psi(u,v,p)$, then there is is some $v\in Y$ that satisfies $\psi(u,v,p)$. It need not be true that every $v$ satisfying $\psi(u,v,p)$ belongs to $Y$. The explanation in the book is right: if $C_u$ is the set of $v$’s that satisfy $\psi(u,v,p)$, and if $C_u\ne\varnothing$ (i.e., if there is at least one $v$ that satisfies $\psi(u,v,p)$, then $C_u\cap Y\ne\varnothing$ (i.e., there is at least one $v$ in Y that satisfies $\psi(u,v,p)$.

The answer to your last question, however, is yes:

$$\forall{X}\exists{Y}(\forall{u}\in{X})[\exists{v} \psi(u, v, p) \to(\exists{w\in{Y}}) \psi(u, w, p)]$$

says exactly the same thing as the original version.

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Thank you very much. I'm wondering if there is some principle to replace variable with others. Could you please explain it a little bit, or tell me where I can find an answer? –  Metta World Peace Dec 5 '12 at 6:24
    
@MettaWorldPeace: If you quantify over a variable, you can change the name of that variable within the quantified expression. That’s what you did here: you changed $v$ to $w$ throughout the expression $(\exists v\in Y)\psi(u,v,p)$. Keywords to look for in this connection are free variable and bound variable. –  Brian M. Scott Dec 5 '12 at 6:29
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