Check if a function $f : \mathscr P(\Bbb R) \rightarrow [0, \infty[$ exists with the following properties:
$i) \ f(\emptyset) = 0,$
$ii) \ f(\bigcup_{k=1}^\infty A_k)$ = $\sum_{k=1}^\infty f(A_k)$ for disjoint sets $A_k$,
$iii) \ f(A + c) = f(A)$ for every $A \subset \Bbb R$ and $c \in \Bbb R,$
$iv) \ f([0, 1]) = 1.$
Do it the following way:
We want to call $x$ and $y$ equivalent iff $x - y \in \Bbb Q$ for $x, y \in [0, 1].$ We choose exactly one representative from every equivalence class. Let $V \subset [0, 1]$ be the set of every representative that we chose before.
Now let $\{q_1, q_2, \ ... \}$ be an enumeration of every rational number in $[-1, 1]$. Show that the sets
$V_k := q_k + V := \{q_k + v: v \in V \}, k \in \Bbb N$
are disjoint to each other.
That was the first task. I have no question about that one, but about the second task:
Show that
$1 \le f(\bigcup_{k=1}^\infty V_k) \le 3.$
What buggs me is the fact that we are actually supposed to show that this function $f$ does exist. There is even a last step where we are required to conclude it.
Now, I'm not sure whether I have to use the properties defined before or not. I wouldn't assume so, but since the function doesn't have any other properties, I'd guess that I have to do it.
If you want to show, that such a fuction $f$ does not exist, you suppose to the contrary that such a function exists. Then its your turn to derive a contradiction.