13. Let f : P ({1,2,3,4}) → {0,1,2,3,4} be the function that maps each subset S ⊆ {1,2,3,4} to its cardinality, f (S)= |S|. Is f onto or 1-1? After working through this, I was sure that function f was one-to-one and not onto, but according to my answer key its onto but not one-to-one. I don't exactly understand how though. Here's my thought process:
Function f maps elements of P{(1, 2, 3, 4)} to the set {0, 1, 2, 3, 4}
Possible subsets of P({1,2,3,4)} are: {∅,{1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4} = 16 total subsets, each denoted as set S.
Every subset set has elements that can ALL be uniquely mapped to elements in the set {0,1,2,3,4}, but 0 in this set can never be mapped to any elements of S.
So, I would think the function is one-to-one, but not onto.
Can someone explain where I'm going wrong? Thanks! :D
 A: It's not the elements that are being mapped.  It's the sets themself.
The function is is so that $f({1,3,4})=3$ because $\{1,3,4\}$ has three elements.  And the function is so that $f(\{1,2,4\})=3$ because $\{1,2,4\}$ has three elements.  And $f(\{1,3\}) =2$ and $f(\{2,4\}) =2$.
And $f(\emptyset) = 0$ because $\emptyset$ has $0$ elements.
If $f$ were one-to-one that would mean every subset will have a different cardinality.  Is that true? And if $f$ were not onto that would mean there is some number that represents a cardinality that no subset has....
So is it one-to-one?  Is it onto?

  One-to one:
$f(A) =f(B) \iff |A| = |B|$.  As there are MANY cases of $|A| =|B|$ without $A = B$ (for example $|\{1\}| = |\{3\}| = 1$ and $|\{1,3\}| = |\{1,2\}| = 2$ etc. $f$ is not one to one:


 Onto:
$f(A) = k\iff |A|=k \iff A$ has $k$ elements.
$f(A) = 0$ if $|A|$ has $0$ element.  $\emptyset$ has $0$ elements so $f(\emptyset) = 0$.
$f(A) = 1$ if $|A|$ has $1$ element.  $\{1\}, \{2\}, ... $ etc. all have one element so $f(\{1\} = f(\{2\}=.... = 1$ and $f(A) =1$ does happen.
 Similarly $f(A) =2$ or $f(A) = 3$ or $f(A) =4$ if $A$ has two, three or four elements.  I'll leave it as "obvious" that, indeed, there are sets with $2,3$ or $4$ elements.  (Ex. $\{1,2\}$ or $\{1,2,3\}$ or $\{1,2,3,4\}$.
 So... for every value $0,1,2,3,4$ there are sets with those number of elements and so $f$ of those sets will be equal to $0,1,2,3,4$ and so every value is mapped to, so it is onto.

A: $f(\emptyset)=0$, the function is onto. And definitely is not 1-1, because $|P(\{1,2,3,4\})|>5$.
