3

Here is the question:

If a set, $A$, is finite, then $|A| < 2^{|A|} = |P(A)|$, and so there is no surjection from set $A$ to its powerset. Show that this is still true if $A$ is infinite.

Here is the proof:

We prove there is no surjection by contradiction: suppose there was a surjection f : A → P(A) for some set A. Let W ::= {x in A | x not in f(x)}. So by definition, (x in W) ←→ (x not in f(x)) for all x in A. But W ⊆ A by definition and hence is a member of P(A). This means W = f(a) for some a in A, since f is a surjection to P(A). So (x in f(a)) ←→ (x not in f(x)) for all x in A. Substituting "a" for "x" yields a contradiction, proving that there cannot be such an f.

My question is what exactly does the definition of W mean? I don't understand the condition of "x not in f(x)". Second, I'm trying to learn proofs and I'm struggling with how one would be able to know to create W, which seems to be important to this proof. Third, how does one know how to create auxiliary information such as W when proving things in general?

Kushal Bhuyan
  • 7,110
Dak Song
  • 151

1 Answers1

3

Since $f$ is a map from $A$ to its power set $P(A)$, for any $x\in A$, the image $f(x)\in P(A)$ which means $f(x)$, the image of $x$, is a subset of $A$. Now there are two possibilities : either $x$ is an element of the subset $f(x)$ or not. $W$ is defined to be the set of elements $x$ of $A$ with the property that $x$ is not an element of $f(x)$.

Now we know that $f$ is a surjection. Thus as a subset of $A$, $W$ is also an element of the image of $f$. Therefore there must be an element $a\in A$ such that $f(a)=W$. Now $f(a)=W=\{x\in A\mid x\not\in f(x)\}$.

Check that both $a\in f(a)$ and $a\not \in f(a)$ leads to a contradiction which implies there is no such surjection.

Edit : I think this answers the first two questions. I don't know much about set theory but I think to define such subsets to get a contradiction is a common trick which is also used in the proof of the nonexistence of the universal set.

Levent
  • 4,804