I was solving some exercises of set theory until I found one where I'm stuck . I don't know what to do when I have a power set with the relation "not element of".

This is the problem:

Let $y \in x \to y \notin B$ for every y. Prove that $x \notin P(B)$ or $x \in \{ \varnothing \}$

I was doing something like:

Let $p \in x$. Then $p \notin B$. Therefore $p \in B^c$. From that, $x \subset B^c$. Using the definition of power set, it means that $x \in P(B^c)$ should hold...

  • $\begingroup$ How can $y \in x$? $x$ is it a set? Should you write it like $X$? $\endgroup$ – hlapointe Oct 24 '15 at 18:28
  • $\begingroup$ "Let $y\in x \to y\notin B$ for every $y$" is kind of a strange phrasing. Presumably they mean "Assume that $y\in x \to y\notin B$ for every $y$" which just says that $x\cap B=\varnothing$ ... $\endgroup$ – hmakholm left over Monica Oct 24 '15 at 18:29
  • $\begingroup$ I edited the problem using your suggestions. $\endgroup$ – Arturo Oct 24 '15 at 18:39
  • 1
    $\begingroup$ @Arturo: Hmm, I thought that was a verbatim quote from the exercise ... $\endgroup$ – hmakholm left over Monica Oct 24 '15 at 18:42
  • $\begingroup$ @Arturo: If the original version was a direct quotation from the exercise, you should restore it. $\endgroup$ – Brian M. Scott Oct 24 '15 at 18:50

When the desired conclusion is a disjunction (‘then $A$ or $B$’), it’s often easiest to assume that one of the possibilities does not hold and try to show that in that case the other one must hold. Here I would assume that $x\in P(B)$ and try to show that this forces $x\in\{\varnothing\}$. It’s worth noting first that $x\in\{\varnothing\}$ is just a fancy way of saying that $x=\varnothing$.

Suppose, then, that $x\in P(B)$. Then $x\subseteq B$, so if $y\in x$, then $y\in B$. But we know that if $y\in x$, then $y\notin B$, so there cannot be a $y\in x$: its existence would lead immediately to a contradiction. But then $x=\ldots\;$?


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