Is there such a classification as Co-Paradox?

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So, my line of thinking is that a set that contains all sets that do not contain themselves is a paradox. And the opposite of that is a set that contains all sets that contain themselves, and, while it not decidable if it contains itself, neither state leads to a contradiction.

So, I would ignorantly say that the Co-Paradox of the set of all set that do not contain themselves is itself not a Paradox.

Is this close to something in mathematics? Are there Co-Paradoxes that are also Paradoxes?

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the concept of paradox is not a mathematical concept. The meaning of the word paradox is something that exhibits a strong counter intuitive behaviour. Something that looks ok but upon closer inspection reveals itself to be highly problematic, to the point that it is totally hopeless.

Russell's paradox is a seemingly legit definition of a set, i.e., the set of all sets that do not contain themselves as elements, which upon closer inspection leads to an unrepairable situation; no answer to the question "does that set contain itself as an element?" is consistent, and thus the question is unresolved.

The paradox you describe above is similarly a seemingly legit definition of a set, i.e., the set of all sets that do contain themselves as elements, which upon closer inspection reveals to be problematic; the question "does that set contain itself as an element?" does not seem to be answerable from the definition of the set, and thus the question is unresolved.

Notice however that the paradox you describe is weaker than Russell's. Russell's paradox leads to a contradiction. The situation you describe is not a contradiction. It just shows that no answer to the question immediately leads to a contradiction, and thus none can be immediately discarded. However, it leaves open the possibility that a more clever argument can resolve the question. Of course, you will have to carefully axiomatize your paradox in order to really answer such a question.

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Proposition (There exists no universe)

Say $A$ is an arbitrary set (or set of discourse or universe).

Then there exists a set $B$ that is not in $A$.

Proof: Say $A$ is a set. Define $$B:=\{x\in A | x\not\in x\}$$

Then $B$ can not be in $A$:

Assume $B\in A$.

Then we have either

$B\in B$ and therefore by definition of $B$ it follows that $B\not\in B$

or we would have

$B\not\in B$ and therefore by definition of $B$ it follows that $B\in B$

so in any case a contradiction. $\diamond$


The point here is that the set $A$ was chosen arbitrarily and what we showed is, that we can always find something not in that set $A$. So this means nothing contains all or u could also say more spectacular there is no universe (universe meaning universe of discourse).