Question about "or" in set theory

[xwolfhunter]

So I'm reading up on some set theory, and I came to the axiom of pairing. The book uses that axiom to prove/define a set which contains the elements of two sets and only the elements of those two sets. $~~B$ is the set which contains the elements and only the elements of sets $a$ and $b$.$$B=\{x: x=a~\mathsf{or}~x=b\}=\{a,b\}$$
I am hung up on how "or" is used here. I understand why it's formed the way it is, in order to ensure that unambiguously the set $B$ contains exactly the elements of both sets $a$ and $b$, but the "or" operator confuses me. $q~\mathsf{or}~z$ means "is true if either $q$,$~z$, or both $q$ and $z$ is true." This means that according to the above, I can't be sure which statement is true ($x=a$ or $x=b$), or whether they're both true, but of course it is implied in the statement that they are both true. I see why they can't use "and" there, because then by necessity $a=b$, but what is the exact definition of "or" here? It demonstrates different iterations of $x$, but what are the rules of this and why are both iterations true when it's the "or" operator which is used here?

I just need to know precisely what everything means, sorry :)

 

[mfb]

a and b are not sets if used in the context of your definition of B. They are elements of B.

"is true if either aa, b~b, or both aa and bb is true."

a and b are not logical values. The things that can be true or false are the equations x=a and x=b.

but of course it is implied in the statement that they are both true.

No it is not.
Here is an example with numbers:
$$B=\{x: x=2~\mathsf{or}~x=3\}=\{2,3\}$$
"B contains all elements x where (x is 2) OR (x is 3)".

 

[xwolfhunter]

a and b are not sets if used in the context of your definition of B. They are elements of B.
a and b are not logical values. The things that can be true or false are the equations x=a and x=b.
No it is not.
Here is an example with numbers:
$$B=\{x: x=2~\mathsf{or}~x=3\}=\{2,3\}$$
"B contains all elements x where (x is 2) OR (x is 3)".

If what you say is true then Mr. Paul Halmos should probably reconsider his career in mathematics. In his definition, $a$ IS a set, and SO IS $b$. I did misspeak when I said that $B$ contains the elements of $a$ and $b$, for $B$ is the set which contains as elements both sets $a$ and $b$ and nothing else. I'll quote his book in a second.

I know that $a$ and $b$ are not logical values as defined earlier. I should have used different things than $a$ and $b$, they were supposed to represent arbitrary statements. I'll edit that in a sec.

So here's what he says in the book, verbatim:

One consequence (in fact an equivalent formulation) [of the axiom of pairing] is that for any two sets there is a set that contains both of them and nothing else. Indeed, if $a$ and $b$ are sets, and if $A$ is a set such that $a \in A$ and $b \in A$, then we can apply the axiom of specification to $A$ with the sentence "$x=a~\mathsf{or}~x=b$." The result is the set$$\{x \in A:~x=a~\mathsf{or}~x=b\},$$and that set, clearly, contains just $a$ and $b$.

Thus both statements, according to him, are of necessity true; the set defined contains just $a$ and $b$, according to him, meaning it contains explicitly both, not "$a$ or $b$," which would be excessively useless. So where does the unspecification of the "or" get resolved? Because clearly according to him it does, and it does of necessity.

I guess the english equivalent would be "the set of all $x$es found in $A$ where $x$ is either the set $a$ or the set $b$." So is the resolution in the statement "all $x$es"? That begins to make sense to me. The "all $x$es" means that whenever the conditions applied to $x$ are true, then said $x$ is included. So I think actually that it is resolved, but what do you think?

The more I think about it the more it makes sense now.$$A = \{a,b,c\}$$$$B=\{x:x=a~\mathsf{or}~x=b\}$$So when determining which $x$es are to be included in $B$, we include all $x$es for which the conditions are true, and when $x=a$ or $x=b$, the conditions are met, but not when $x=c$. So the set $B$ contains $a$ and $b$ and nothing else.

Never mind, then, I guess all it took was a little thought to resolve. :)

 

[Mark44]

One consequence (in fact an equivalent formulation) [of the axiom of pairing] is that for any two sets there is a set that contains both of them and nothing else. Indeed, if $a$ and $b$ are sets, and if $A$ is a set such that $a \in A$ and $b \in A$, then we can apply the axiom of specification to $A$ with the sentence "$x=a~\mathsf{or}~x=b$." The result is the set$$\{x \in A:~x=a~\mathsf{or}~x=b\},$$and that set, clearly, contains just $a$ and $b$.

Where he says "if $a$ and $b$ are sets" looks like a typo to me. The usual style for the names of sets uses capital letters for the sets and lowercase letters for the elements of sets. Everywhere else in the quoted paragraph follows this style, except the phrase I picked out.