5
$\begingroup$

I am just beginning to learn the ZF axioms of set theory, and I am having trouble understanding the Axiom of Foundation. What exactly does it mean that "every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets." In particular, how can $y$ be an element of a set $x$, and also be disjoint from it (I have seen this be called the epsilon-minimal element). My intuitive understanding of sets (which is obviously wrong) tells me that if $y$ is an element of $x$, then the intersection of $\{y\}$ and $x$ should be $\{y\}$. How can it be $\varnothing$?

$\endgroup$
5
$\begingroup$

The axiom isn't saying that $x\cap \{y\}=\emptyset$. It's saying $x\cap y=\emptyset$. Keep in mind that in ZFC everything is a set, including the elements of other sets.

$\endgroup$
4
$\begingroup$

You need to understand that $y$ and $\{y\}$ are two different things.

Suppose $x = \{ \{ 1,2,3,4\}, \{2,3,4,5\},\{5,6,9,10\}\}.$

Then $x$ has three members. One of those members is $\{1,2,3,4\}$. That set has four members. One of those members is $2$. That member, $2$, is not a member of $x$. Similarly each of the other four members of $\{1,2,3,4\}$ fails to be one of the three members of $x$. So the intersection of $x$ with any of its members is empty.

Now consider the set $$ x=\Big\{ \quad\varnothing,\quad \big\{\varnothing\big\},\quad \big\{\varnothing, \{\varnothing\}\big\},\quad \big\{\varnothing, \{\varnothing\}, \{\varnothing, \{\varnothing\}\}\big\}\quad \Big\}. $$ This set has four members. One of those members is $\big\{\varnothing, \{\varnothing\}, \{\varnothing, \{\varnothing\}\}\big\}$. Another is $\{\varnothing\}$. Those two sets do have a member in common. In this case $x$ has only one member that does not intersect $x$.

$\endgroup$
  • $\begingroup$ Thank you for the clear explanation and example. I get it now! $\endgroup$ – user147686 May 4 '14 at 13:22
  • 1
    $\begingroup$ If I may ask, which member of $x$ doesn't intersect it? Complete math beginner here (came here after reading something about Haskell; I just want a pointer to which element you mean). $\endgroup$ – Noein Apr 5 '15 at 20:21
  • 2
    $\begingroup$ @Noein : $\varnothing$ is a member of $x$, and $\varnothing$ does not intersect $x$, i.e. there is no member of $x$ that is also a member of $\varnothing$. That is because $\varnothing$ has no members at all. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 5 '15 at 23:16
  • $\begingroup$ It's 3 am over here and this makes sense... scary. I sort of get it now, though. There must be (a) 'most elementary composition part(s)' for a set; something that can't be broken down further into a kind that can belong to the big set, or something like that. $\endgroup$ – Noein Apr 6 '15 at 0:27

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.