Question: Sorry if it might be simple but I basically can't get my head around it.

The question asks me to:

> 
> If exists, construct a bijection between the sets $\mathbb R$ and $\mathbb R - \{0\}$
> 
> 
> 

But I don't know how to handle it, I thought about the existence of a relationship but the absence of $\{0\}$ in the second set makes me feel lost and can't prove the existence of the bijection.

Answer: 
Let's assume the set we're dealing with is finite. Let $S = \{s\_1, s\_2, \cdots, s\_n \}$, and fix the size we're considering as $k$.

There are $\frac{n!}{k! (n-k)!}$ total subsets of size $k$. Now each element of $S$ will be a member of some (but not all) of these subsets. In particular, if we force $s\_j$ to be in subset, there are $k-1$ remaining members to pick from the remaining $n-1$ members of $S$. Thus, there are exactly $\frac{(n-1)!}{(k-1)!(n-k)!}$ subsets of size $k$ containing $s\_j$.

Since this holds for any $j$, and muliplication in commutative, we find the value you're l0oking for, $V$ is given by:

$$V = \prod\_{j=1}^{n}s\_j^r$$
where $r$ is given by:
$$r = \frac{(n-1)!}{(k-1)!(n-k)!}$$

---

Applying this to your example above, we have $n =3$ and $k=2$, which yields $r = 2$. Then we have:
$$\begin{align}
V &= 1^2 \*2^2 \*3^2 \\ &= 1\*4\*9 \\ &=36
\end{align}$$
Consider the $k-$sets of $\{a\_1,\cdots,a\_n \}$ each element will occur $\binom{n-1}{k-1}$ times in a given set, so the product of the products of the sets will be $\color{red}{\left(\prod\_{i=1}^{n} a\_i \right)^{\binom{n-1}{k-1}}}$.