2
$\begingroup$

Apparently it is false, that $M \cap \mathcal{P}(M) = \emptyset$ for all sets $M$, if I am to believe my source material. However, if I am not missing some convention, this cannot be true:

$A \cap B := \{ x | x \in A \land x \in B\}$

And

$\mathcal{P} := \{ m | m \subseteq a \}$

And hence

$\forall x \in A. \forall m \in \mathcal{P}(A). x \neq m$, as an element of $A$ is of a different type than a set of elements of $A$.

I don't think I am missing something here, but I am not inclined to assume my source is wrong. Is it though?

$\endgroup$
  • $\begingroup$ If $M=\{ \emptyset \}$ then $P(M)=\{ \emptyset , \{ \emptyset \} \}$ and $M \cap P(M)=\{ \emptyset \}$ $\endgroup$ – Sami Fersi Nov 1 '17 at 15:42
  • 4
    $\begingroup$ It all boils down to what you mean by 'type' and what foundations you're using. In the usual (set theoretic) foundations of mathematics, there is only one type: sets. As such, saying two things are of 'different types' is false. There are settings where there is more validity to your argument, but since this is a set theory question, I won't go into those... $\endgroup$ – Clive Newstead Nov 1 '17 at 15:46
3
$\begingroup$

Each $\mathcal{P}(M)$ contains an empty set. $\{\}$. If $\{\}\in M$ then $\emptyset$ is a common element of $M$ and $\mathcal{P}(M)$, hence they're not disjoint.

Also if $M$ is non-empty, then it has some item $m\in M$ and the power set has its singleton as an element: $$\{m\} \in \mathcal{P}(M)$$ If $M$ has that singleton as element together with $m$ itself: $m\in M$ and $\{m\}\in M$, then the singleton is a common element of $M$ and $\mathcal{P}(M)$.

$\endgroup$
2
$\begingroup$

The reason is that set theory is a first-order theory, and first-order logic is untyped. Namely, there is a universe and everything in the universe is "the same".

It is true, you can introduce predicates in the form of relations, or define such relations given your language (and theory). But the general idea is that the elements of the universe are just that.

In set theory, there are no types and we call all the objects of the universe "sets".

This is one of the reasons some people feel like type theory could provide a better foundations for mathematics, as it reflects more closely some of the naive ways we think about mathematics.

(While it is indeed the case, set theory can internalize type theory in fairly straightforward ways for the most part, so there is no "right" or "wrong" approach, only a matter of convenience. And though set theory is untyped, this is not necessarily a bad thing. Your computer is untyped as well, remember that all the things that you might write in any programming language end up being just a bunch of "pass current" and "block current".)

$\endgroup$
  • $\begingroup$ +1 (and beat me to it, dang): this is the crucial observation, that types play no role in (usual, ZFC-style) set theory. $\endgroup$ – Noah Schweber Nov 1 '17 at 16:28
1
$\begingroup$

Let $M=\{\emptyset,\{\emptyset\}\}$. Then $\mathcal P(M)=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\emptyset,\{\emptyset\}\}\}$ and $M \cap \mathcal P(M)=M$

$\endgroup$
  • $\begingroup$ Same for $M=\{\emptyset\}$. $\endgroup$ – CiaPan Nov 1 '17 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.