Why is the axiom of union harmless?

The axiom of union says:

Let $M$ be a set of sets. Then $\bigcup M$ exists.

The axiom of replacement says:

Let $M$ be a set and replace every element $x\in M$ with another object $x'$. Then $\{x'\mid x\in M\}$ is a set.

Most mathematicians believe that ZFC is consistent. But I wonder if it is possible to construct a set that is as big as the set of all sets or russel set by using axiom of replacement and axiom of union again and again. Why do mathematicians blindly believe that these constructions aren't contradictory?

Is there an heuristic argument that the axiom of union and the axiom of replacement can't construct "sets that are too big to be sets" (that is "contradictory sets" or "proper classes") together?

Note: If one uses an axiom system then one eventually thinks this axiom system isn't contradictory. So my question goes to everyone who uses ZFC: Why are you sure that ZFC isn't contradictory?

• I don't know much about ZFC, to be hones. Not really my field. However, I love the "blindly believe". It implies the "If I don't know why this is true, then surely nobody really knows it's true" level of thinking that is on the rise all over the world. – 5xum Mar 2 '16 at 15:04
• On my side, I'm not sure that ZFC isn't contradictory. By the way for such a topic, I don't know what means to be sure. I just think it is likely to be the case... and I live with it to do math! – mathcounterexamples.net Mar 2 '16 at 15:48
• What does it mean "again and again" here? A finite number of times? – Marco Disce Mar 2 '16 at 16:12
• Yes, it means a finite number of times – asdfadsfasdfasdf Mar 2 '16 at 16:22
• I'm not entirely convinced that $\operatorname{ZFC}$ is consistent. I'm not even sure that $\operatorname{PA}$ is. However, studying a theory and pushing its boundaries might very well be the best way to either prove its inconsistency or otherwise develop a heuristic reason to believe in it. So either way, it doesn't matter. If $\operatorname{ZFC}$ is consistent, it is - in my opinion - a pleasant theory to study. If it's inconsistent, we should eventually be able to find out, pinpoint the reason for its inconsistency, alter it accordingly and try again. – Stefan Mesken Mar 2 '16 at 17:01

If I understand you correctly, you ask for a heuristic argument why replacement and unions do not end up with proper classes. Perhaps the following helps.

The union of finitely many sets is again finite. Likewise, if we take a finite set and replace every element by another, we get again a finite set. So we see that replacement and finite unions cannot create infinite sets from finite ones. The key here is that we only allow finite unions, because if an infinite union of finite sets can quite easily be infinite.

Likewise, taking replacements and unions over sets of sets should always give back a set, and never a proper class. Of course if one allows the union to range over a proper class (for example taking the union of all sets at once) we can get a proper class, but as long as the index over which we do the union is a set, the result should be a set again.

Edit in responce to the comment below: At the very least the above shows that the collection of finite sets satisfies the axioms of replacement and union. So if one is willing to accept the existence of finite sets, then one also accepts that these axioms are non-contradictory. But the analogy also illuminates why the proposed method of taking unions and replacements over and over again does not necessarily lead to sets that are to big. Certainly, taking more and more unions leads to bigger and bigger sets, but there is no reaso to expect these sets to ever become "too big".

I hope this helps.

• Where is your argument? "Likewise" is not a convincing argument. And why are you talking about finite sets? Can you please read my question carefully. I know that the union of finitely many finite sets yields a finite set. But again, this is not my question. – asdfadsfasdfasdf Mar 2 '16 at 16:33
• @asdfadsfasdfasdf I have added a paragraph. Of course, we can't really expect a proof of the consistency of ZF in a math.se post, so there is an upper limit at how 'convincing' an argument can be. But if you give some more details about why unions and replacements would lead to "sets that are to big to be sets", I can try to adjust my answer. – Marc Paul Mar 2 '16 at 17:02

One way to protect against problems that occur with "set of all sets" is to protect against situations where you get:

$$S_1=S_{n+1}\in S_{n}\in\cdots \in S_2\in S_1$$

We'll say that "M is okay" if $M$ doesn't have this problem and every one of its elements is "okay." This is an intuitive definition, not a definition in ZFC.

If $M$ is okay, we assert that (again, intuitively) $\bigcup M$ is okay.

All the elements of $\bigcup M$ are "okay" because they are elements of elements of $M$, and "okay" was defined so that these are okay.

So, can we have $\bigcup M\in S_{n}\in\cdots\in S_2\in \bigcup M$? Then $S_2$ is seen to be "not okay," too, because $S_2\in \bigcup M \in \cdots\in S_2$. But $S_2\in S\in M$ for some $S\in M$, by the "definition" of $\bigcup M$. So, since $S_2$ is not "okay," then $S$ is "not okay," and hence M is "not okay."

So the axiom of union doesn't let us create stuff that is "not okay" from stuff that is "okay."

So the intuition is that we are trying to avoid loops in the set membership relationship, because that is what has always caused us paradoxes in the past. Union does not allow us to create new loops, so it is not a problem for those kinds of inconsistencies.

The axiom of replacement seems trickier, to my mind.

• Your argument isn't satisfying me. Famously, there are non-wellfounded models of ZFC because of the compactness theorem. The problem is not a situation where $S_1=S_{n+1}\in S_{n}\in\cdots \in S_2\in S_1$ but a situation where there is a set of all sets that are not members of themselves. – asdfadsfasdfasdf Mar 2 '16 at 16:10
• I'm not saying that all models of ZFC are well-founded. And yes, the problem is the loop situation. You get the problematic sets by assuming first that there is a "set of all sets." While the paradox arises by taking a subset, the problem is the "set of all sets" problem. Union doesn't add to the problems, but that doesn't mean that every model will actually have all sets "nice." – Thomas Andrews Mar 2 '16 at 16:47
• @asdfadsfasdfasdf If you can't have loops with the relation $\in$ then the set that are members of themself don't exist and the "set of all sets that are not members of themselves" is just the set of all set. – Marco Disce Mar 2 '16 at 16:49
• @asdfadsfasdfasdf: You're confusing internal and external ill-foundedness. In any case, there is never a set of all sets which are not members of themselves. It's true that the class of sets which are members of themselves might be non-empty, if one discards the axiom of regularity, but that's not even remotely an argument why there is a set of all sets which are not members of themselves. – Asaf Karagila Mar 2 '16 at 17:05