Proving calculating Minhash
Min-hash are not just (standard) hash functions, but a family of functions $\cal H$, such that if you randomly pick one function $h\leftarrow \mathcal{H}$ out of the family, it will satisfy the requirement you have in your question.

Check 
"Min-wise independent permutations"
by Broder, Charikar, Frieze and Mitzenmacher, 2000. They define this notion and analyze it in a very simple way to follow. 

From a practical point of view, you usually have a bound on the size of the sets you use (say $2^{256}$, probably more than the number of atoms in the universe).
Then, a 256bits output is enough to have hash-functions that are permutations (and clearly, the set of all the permutations is a good family of min-hash functions..)
So the problem is not the output length, but how many functions are there in the min-hash family, so that if you pick one at random, you get the probability that any element is the minimum exactly $1/2^{256}$.

To overcome the huge set of hash-functions (there are $256! \gg 2^{256}$ permutations of 256 bits..), 
they relax the requirements on the probability of the min-hash and get an approximated notion of $\epsilon$-min-hash, which guarantees that
$$\Pr_{h\leftarrow {\cal H}}[h(A)=h(B)] = \frac{|A\cap B|}{|A\cup B|}\pm \epsilon$$
which is more useful from efficiency perspective (the set $\mathcal{H}$ is smaller and easier to get).