# Question about Axiom of Specification

1. Sep 12, 2013

### Seydlitz

In page 6 of Naive Set Theory by Halmos, he introduces the definition of the axiom of specification, then sets up one example based on the axiom, in which he changes $S(x)$ to $x \not \in x$ to illustrate something. I understand that this mean $x$ doesn't belong in $x$.

Afterwards comes this excerpt,
How does one infer that if $B \in A$ then either $B \in B$ or $B \not \in B$. Is it like when I am shopping to the market $A$ and there's a basket $B$ there. If a product is in the market it is possible to have a product either in the basket or not in the basket i.e in the shelves.

Or is it by inferring from the definition like this:

$(*)$ $B \in B$ if and only if $(B \in A \text{ and } B \not \in B)$

But then clearly there's something wrong since it cannot be both true that $(B \in B$ and $B \not \in B)$? I mean yes if $B \in A$ then according to the definition $B \not \in B$ but then if we accept $B \in A$ such that $B \not \in B$, clearly according to the definition $B \in B$.

Could you guys explain this part further? I also think that I haven't fully appreciated this example. What does the writer want to show to us? That a set cannot have everything and there must be at least something that is not in the set, hence the Axiom of Specification?

Thank You

2. Sep 13, 2013

### verty

3. Sep 13, 2013

### Seydlitz

Can you specify further which part exactly? I think I already got the contradiction while writing the post, I just want to know his point behind showing this example.

4. Sep 14, 2013

### verty

Do you know the story of the three blind sages who came upon an elephant for the first time, one thought it was a tree, one thought it was a snake, etc? This is a little like that. Focus on boolean algebra and truth tables, that should direct you to what you need to know.

5. Sep 14, 2013

### lugita15

The fact that either $B \in B$ or $B \not \in B$ follows from the more general fact for any statement P, either P is true or P is false. That is the law of excluded middle.

6. Sep 14, 2013

### Seydlitz

I don't seem to understand how the law applies here, why it is has to be $B \in B$ or $B \not \in B$, when we want to know whether $B \in A$?

I've heard it once or two and I've searched the story in wikipedia. It only described the story from multiple viewpoints without any logical treatment or how does it relate to Boolean logic.

7. Sep 14, 2013

### lugita15

Well, the statement $B \in B$ is either true or it's false, and if it's false, that's equivalent to $B \not \in B$.

8. Sep 14, 2013

### Seydlitz

But what is the connection to $B \in A$?

9. Sep 14, 2013

### lugita15

I think you misunderstood what Halmos is saying. He's not using to $B \in A$ to prove that either $B \in B$ or $B \not \in B$. He's just assuming $B \in A$, and trying to show that that leads to a contradiction. And along the way, he's using the fact that no matter what, either $B \in B$ or $B \not \in B$.

10. Sep 14, 2013

### Seydlitz

Ok I got you now, but what is the point of this example actually, when he then proceeded to say that because of it, "nothing contains and everything" and "there is no universe."

11. Sep 17, 2013

### MrAnchovy

I haven't read the book so I am not sure what point the author is trying to make, but the reason we have the Axiom of Specification, and the reason we must always have B∉B is to avoid Russell's Paradox - without it, set theory is broken.

12. Sep 17, 2013

### SteveL27

And even with it ... it's still possible that set theory is broken. Just wanted to clarify that point.

13. Nov 4, 2013

### wyan

I have been struggling with this page myself for the last couple of days. What I believe is meant by that affirmation is: he has just proved that $B\not\in A$. Since the set $A$ is arbitrary, one could take $A$ to be the "set of all sets", should such a thing exist (the "Universe" he mentions). But by what he just proved, there is something that does not belong to $A$ (namely, the $B$ so constructed), thus $A$ could not be the "set of all sets" in the first place.