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Let $x = 1$. Is it valid to define sets like $Y = \{ x | 1 = 1 \} = \{ 1 \}$ and $Z = \{ x | 1 \neq 1 \} = \emptyset$?

What I want to know: Are we allowed to define sets like $\{ y | A(z) \}$ where $A(z)$ is a condition that is independent from $y$ and $y$ is defined somewhere outside the set (intensional definition)? Normally we define sets like $\{ y | A(y) \}$ where $A(y)$ is a condition depending on $y$.

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  • $\begingroup$ I see no issue here - the constant function is still a function, and similarly the condition $A(x): 1\neq 1$ is still a condition. $\endgroup$
    – user1729
    Aug 20, 2014 at 10:04
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    $\begingroup$ Your $Y$ would contain all sets, so doesn't form a set in ZFC. But in general there's no need for $x$ to occur free in the condition $A$. Your $Z$ is a good example, and follows from the axiom of separation. That is, $Z = \{x : 1\not= 1\} = \{x\in y: 1 \not = 1\}$. $\endgroup$
    – user104955
    Aug 20, 2014 at 10:11
  • $\begingroup$ Thanks, but I defined „Let $x=1$.“ outside the set ($x$ is not free). It’s not wrong to define such thing outside a set, right? I could also write $\{ 0 | 1 = 1 \} = \{ 0 \}$ or $\{ 0 | 1 \neq 1 \} = \emptyset$. $\endgroup$
    – Ronny
    Aug 20, 2014 at 10:17
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    $\begingroup$ That's fine. The "$x$" inside the brackets is bound. So, when you say "$Y = \{x: 1=1\}$", what you mean is "there is a set $y$ such that for all $x$, $x$ is in $y$ iff $1 = 1$; and $Y = y$". Any use of $x$ outside this sentence is thus independent of its use inside, and thus independent of its use in "$\{x:\phi\}$". Once the definition of "$\{x:\phi\}$" is made clear, the question is really just about the underlying logic and the interaction between free and bound variables in various contexts. $\endgroup$
    – user104955
    Aug 20, 2014 at 13:05
  • $\begingroup$ Tanks! I think your Bound-Argument is right. If you create an answer I will mark it as answer. $\endgroup$
    – Ronny
    Aug 21, 2014 at 14:38

2 Answers 2

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In the framework of axiomatic set theory, you cannot do that as you'll need some axiom to justify the existence of a set and in cases like these you'd use the axiom of separation for which you'd need another set. In other words, if $Y$ is a set, then $\{x \in Y \mid A(x)\}$ is a set, no matter what $A(x)$ is.

But there's no general axiom that makes something like $\{x \mid A(x)\}$ a set.

$\{x \mid 1 = 1\}$ would be the class (note: class, not set) of all sets. And even something like $\{x \mid x \neq 42\}$ would result in a proper class. See Russell's paradox.

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  • $\begingroup$ Thanks, but I defined „Let $x=1$.“ outside the set. It’s not wrong to define such thing outside a set, right? I could also write $\{ 0 | 1 = 1 \} = \{ 0 \}$ or $\{ 0 | 1 \neq 1 \} = \emptyset$. $\endgroup$
    – Ronny
    Aug 20, 2014 at 10:15
  • $\begingroup$ @Ronny: I missed the $x=1$ part, but the intensional notation $\{x \mid A(x)\}$ only makes sense if $x$ is a free variable. It's like: What is $\sum_{k=0}^{42}k^2$ if I defined $k$ to be $0$? $\endgroup$
    – Frunobulax
    Aug 20, 2014 at 10:20
  • $\begingroup$ Well, in my dissertation I have a case where I need this: I want to replace the structure $ X = \begin{cases} \{S\} & \text{for } P_1 \neq \emptyset \wedge P_2 \neq \emptyset \\ \emptyset & \text{else} \end{cases}$ by $ X = \{ S | P_1 \neq \emptyset \wedge P_2 \neq \emptyset \}$ where $P_1, P_2$ are sets and $S$ is something defined outside. $\endgroup$
    – Ronny
    Aug 20, 2014 at 10:29
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    $\begingroup$ @Ronny: Why do you want to replace it? The first definition (with the cases) seems crystal-clear to me. The other one I wouldn't understand if I read it in a paper. I think you shouldn't insist on using private non-standard notation in your dissertaion. $\endgroup$
    – Frunobulax
    Aug 20, 2014 at 10:32
  • $\begingroup$ Okay :-) Thanks, I think I will use the more standard notation. $\endgroup$
    – Ronny
    Aug 20, 2014 at 12:18
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It's fine to use "$x$" outside of "$\{x:\phi\}$" in the way you do. The reason is that the "$x$" in "$\{x:\phi\}$" is bound. So, for instance, when you say "$Y = \{x: 1= 1\}$", what you mean is "there is a set $y$ such that for all $x$, $x$ is in $y$ iff $1=1$; and $Y=y$". Any use of "$x$" outside this sentence doesn't fall within the scope of "for all $x$" and is thus independent of the "$x$" in "$Y = \{x:1=1\}$". In general, terms of the form "$\{x:\phi\}$" involve "$x$" bound in this way, and uses of "$x$" outside of them will be independent of uses inside them.

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