Zermelo-Frankel axiomatization of set theory was introduced as "solution" of resolving the many antinomies (among which Russell's paradox is probably the most famous) that plagued Cantor's naive set theory.
While ZF takes care of the old paradoxes, is it known whether ZF itself is paradox free ? Could, in principle, someone come up with a "new" paradox that would wreck havoc in ZF, the same way "old" paradoxes wrecked havoc in the naive set theory?
Yes, it is conceivable that there is a paradox in ZF. Indeed, by Godel's second incompleteness theorem, if ZF proves that ZF is consistent, then ZF is inconsistent! (Perhaps surprisingly, it is possible that ZF proves "ZF is inconsistent," and yet ZF is consistent!)
Now here's a very brief argument for the proposition "We'll never be certain that ZF is consistent": Since ZF cannot prove its own consistency (unless it is consistent), any proof of the consistency of ZF would have to take place in a stronger theory. This implies that we can basically never be certain that ZF is consistent: in order to believe a proposed proof that ZF is consistent, we would already have to believe that more-than-ZF is consistent in the first place.
Note that we don't need truth here, merely consistency: if ZF is inconsistent, then any theory proving the consistency of ZF must also be inconsistent, since that theory would be able to prove both "ZF is consistent" and "ZF is inconsistent" (by copying the assumed inconsistency in ZF).
But that argument's not exactly accurate! It is true that (assuming ZF is consistent) any proof of the consistency of ZF has to take place in some theory $T$ which is not a subtheory of ZF; however, $T$ itself might not contain ZF, either! $T$ and ZF could be incomparable theories. So this raises an interesting question:
Can ZF be proved consistent from a theory $T$, which is "uncontroversially consistent"?
OK, ignore for a second the phrase "uncontroversially consistent", which is pretty meaningless and really just pushes the problem back one step. There is an interesting idea here:
Take some very weak theory which we all agree is consistent (say, $I\Sigma_1$ - this is essentially a very weak subtheory of arithmetic).
Now, find some new axiom $\psi$ which, when added to $I\Sigma_1$, proves that ZF is consistent.
Finally, make an argument for the "obvious" truth of $\psi$.
This was historically the motivation for Gentzen's idea of proof-theoretic ordinals. The idea is as follows. Suppose we want to argue that a theory $S$ is consistent. Very vaguely, we can define a linear order $\alpha$, and show that the statement "$\alpha$ is a well-ordering", when added to $I\Sigma_1$, lets us prove (usually via some form of cut elimination) that $S$ is consistent. This statement, $\psi$, is then the uncontroversial axiom we're looking for . . .
. . . assuming we can look at $\varphi$ and say, "Yep, the linear order $L$ which $\varphi$ defines is obviously a well-ordering." So it all comes down to how confident we are that certain descriptions of linear orders actually correspond to well-orderings.
(Note: this is merely one kind of ordinal analysis which can be performed, the "$\Pi^1_1$ analysis" if I have my terminology straight. There are other analyses which are more useful in various different contexts. For simplicity, I'm sticking with this one.)
For example, Gentzen's original result was that the proof-theoretic ordinal of $PA$ is $$\epsilon_0=\omega^{\omega^{\omega^{...}}};$$ if you believe that's an ordinal, then you should believe that $PA$ is consistent. Now, $\epsilon_0$ is big, but with a bit of effort it's not too hard to get a good picture of what it looks like and an understanding of why it's well-founded. So this looks promising, right?
Well, I don't know about that. Personally, I already had more faith in the claim "$PA$ is consistent" than I did in the claim "$\epsilon_0$ is well-founded," and this discrepancy only gets worse as the theories in question get stronger (and in particular we're galaxies away from computing the proof-theoretic ordinal of ZF). Still, one's mileage may vary.
