Let $f: P(\Bbb{R})\to P(\Bbb{N})$, $f(X)=X\cap \Bbb{N}$.

  1. Find $|f^{-1}(\emptyset)|$.
  2. Prove that $|f^{-1}(\emptyset)|=|f^{-1}(\Bbb{N})|$.

I am having a difficulty solving 2., but this is what I did in 1. :

Logically, $f^{-1}(\emptyset)=\{A|\exists X\in P(\Bbb{R}) \space s.t \space A=(X\setminus X\cap \Bbb{N})\}$. It is clear that $f^{-1}(\emptyset)\subseteq P(\Bbb{R})$ and therefore $|f^{-1}(\emptyset)|\le \aleph $. Let us look at the singleton $\{a\}\subseteq P(\Bbb{R})$. $\forall 0<a<1,\{a\}\in f^{-1}(\emptyset)$ and therefore $|f^{-1}(\emptyset)| \ge \aleph$. By Cantor-Schroeder-Bernstein theorem $|f^{-1}(\emptyset)|=\aleph$.

I would really appreciate criticism on the my attempts and help with "2.".

  • 1
    $\begingroup$ Hint for 2: Look for a bijection $f^{-1}(\emptyset)\to f^{-1}(\Bbb{N})$. I think there's a fairly obvious one. $\endgroup$ – Simon S Feb 5 '15 at 18:29
  • $\begingroup$ Is it admissible to write a function that adds elements? $\endgroup$ – Donna Feb 5 '15 at 18:30
  • 1
    $\begingroup$ Provided it is a well-defined function and a member of the appropriate set, yes. $\endgroup$ – Simon S Feb 5 '15 at 18:35

So you are looking for a bijection from $f^{-1}(\emptyset)$ to $f^{-1}(\mathbb{N})$.

But what is $f^{-1}(\emptyset)$ really? It is the set containing all subsets of $\mathbb{R}$ that contain no natural numbers. That is, $A\in f^{-1}(\emptyset)$, if $A\cap\mathbb{N}=\emptyset$.

You need to map this set $A$ to a set in $f^{-1}(\mathbb{N})$.

But what is $f^{-1}(\mathbb{N})$? It is the set containing all subsets of $\mathbb{R}$ that contain all natural numbers, that is, $B\in f^{-1}(\mathbb{N})$, if $B\cap\mathbb{N}=\mathbb{N}$.

Can you think of a natural way of mapping the set $A$ that contains no natural numbers at all to a set $B$ that contains all the natural numbers?


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