Definition of one-way functions: randomly-chosen point vs every point Notation: $\Sigma^k$ is the set of $k$-strings on the alphabet $\Sigma$; $\Sigma^\ast$ is the set of all finite dimensional strings on the alphabet $\Sigma$.



In the context of computer-science a strong one-way function is generally defined as follows:

A function $f:\Sigma^{\ast}\longrightarrow\Sigma^\ast$ where $\Sigma=\{0,1\}$is a strong one-way function if the following conditions holds:


$f$ can be calculated in polynomial time.
If $M$ is a probabilistic Turing machine that inverts $f$, then for every polynomial $q:\mathbb N\longrightarrow\mathbb N$ there is a $k_q\in\mathbb N$ such that  for every $k&gt;k_q$ we have that
$$\textrm{Pr}\bigg(M\left(f(x)1^k\right)\in f^{-1}(f(x))\bigg)\le \frac{1}{q(k)}$$
for $x$ randomly chosen in $\Sigma^k$.




Many book emphasize the fact that $x$ is randomly chosen amongst all strings of length $k$. But what is the difference between the condition 2) stated as above and a condition $2)'$ where the phrase "for $x$ randomly chosen in $\Sigma^k$" is replaced by "for every $x\in\Sigma^k$"? Generally if a property is true for a generic element of some set, then we conclude that the same property holds for all the elements of the set. Why this importante on the random choice of the string $x$? To be more precise, when I read "for $x$ randomly chosen in..." can I interpret this sentence as "for any $x$ in..."? It seems that the answer is NO, but I don't understand the reason.

(This is the Wiki definition ).
The reason you are looking at a random choice of $x$ is that, otherwise, there is no probability space any longer. Put differently, if we require that for every x there should not be an algorithm that given an image $f(x)$ can output a preimage, we have a problem, because there is such an algorithm. For every $x$ consider the algorithm $M_x$ that on any input simply outputs $x$. Thus, we have just defined for every $x$ an algorithm that (on input $f(x)$) inverts the function.