Math Help - Another Axiom of Separation Qns

1. Another Axiom of Separation Qns

Let $\phi (v)$ be a formula of the languages of set theory. Let X be a set. Then the following is a set: $\{a \in X|\phi [a] \}$, ie all the elements of X with the property $\phi$ form a set.

Write down a formula $\phi (v)$ saying that "v is a set which has exactly three elements".

I would like to check whether it is possible to represent the above statement as this: $\exists x_1 \exists x_2 \exists x_3 ((x_1 \in v \wedge x_2 \in v \wedge x_3 \in v) \wedge \forall y (y \in v \longrightarrow (y = x_1 \vee y = x_2 \vee y = x_3 ))$

Thanks in advance.

2. Re: Another Axiom of Separation Qns

No, that only says that v has at least 1 element and at most 3 elements. You need to include formulation that x, y, and z are different from one another.

Also, you're missing a right parenthesis on the end.

3. Re: Another Axiom of Separation Qns

Originally Posted by MoeBlee
No, that only says that v has at least 1 element and at most 3 elements. You need to include formulation that x, y, and z are different from one another.

Also, you're missing a right parenthesis on the end.
Hey,

Can I write
$\exists x_1 \exists x_2 \exists x_3 ((x_1 \in v \wedge x_2 \in v \wedge x_3 \in v) \wedge (x_1 \not= x_2 \not= x_3) \wedge \forall y (y \in v \longrightarrow (y = x_1 \vee y = x_2 \vee y = x_3)))$?

Thanks in advance.

4. Re: Another Axiom of Separation Qns

Now you've only said that v has at least two elements and at most three elements. But you're getting close.

5. Re: Another Axiom of Separation Qns

Hey,

Am I right to represent it as such:
$\exists x_1 \exists x_2 \exists x_3 ((x_1 \in v \wedge x_2 \in v \wedge x_3 \in v) \wedge (x_1 \not= x_2 \wedge x_2 \not= x_3 \wedge x_1 \not= x_3) \wedge \forall y(y \in v \longrightarrow (y = x_1 \vee y = x_2 \vee y = x_3)))$?

Thanks in advance.

6. Re: Another Axiom of Separation Qns

Yep. That's it.