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If $P$ is a property (with parameter $p$), then for any $X$ and $p$ there exists a set $Y = \{u \in X : P(u, p)\}$ that contains all those $u \in X$ that have property $P$.

–Jech, "Set theory"

I have studied Jech - introduction to set theory. (but I dropped off)

in the book for any set $X$ there exist $\{x\in X : P(x)\}$.

I can understand this. but multiple parameters are fuzzy for me.

please give some example on above axiom.

*I am not native English speaker. but I am still learning. and in Korea, there aren't many good axiomatic book. so I am reading English.

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2 Answers 2

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Example.

Let $X$ be some set.

Then $a=\{u\in X\mid \varnothing \notin u\}$ is the set of all elements of $X$ that do not contain the empty set (works as parameter here) as element.

According to the axiom scheme of separation $a$ is a set.

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  • $\begingroup$ thank you for answer. In this case, Is empty set a parameter p? $\endgroup$ May 11, 2018 at 8:47
  • $\begingroup$ Yes, the empty set works as parameter here. $\endgroup$
    – drhab
    May 11, 2018 at 8:59
  • $\begingroup$ Although the empty set itself can be expressed directly. So we don't really need is as a parameter. But if $Y$ is another set, then $X\cap Y$ is a good example of a parameterized version of this. $\endgroup$
    – Asaf Karagila
    May 11, 2018 at 11:38
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In the formal language of Set Theory there are no parameters. For a formula $P$ whose free variables (if any) are all among $v_0,...,v_n$ we have $$\forall X\; \forall v_1,...,v_n\;\exists Y \;\forall v_0 (v_0\in Y\iff (v_0\in X\land P)).$$A parameter is really just an abbreviation

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  • $\begingroup$ thank you for answer. but I think i should study elementary mathematical logic for set theory. $\endgroup$ May 11, 2018 at 17:17
  • $\begingroup$ An axiom must be a sentence, which means it cannot have any free variables.... that is, each occurrence of a variable must be "governed" by $\exists$ or $\forall. $ $\endgroup$ May 11, 2018 at 21:31

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