Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? $\{x:P(x)\}$ Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? The axiom that says that
if $\varphi$ is a property, then there exists a set 
$Y = \{X: \varphi(X)\}$ of all
elements having property $\varphi$.
If not then what are other paradoxes that result from that axiom?
 A: In addition to the paradoxes mentioned by Asaf, there is also the infinite family:
$$\begin{align}
\varphi(x) & = \lnot\exists y.x\in y \land y\in x\\
\varphi(x) &= \lnot\exists y\exists z.x\in y \land y\in z \land z\in x\\
&\vdots\end{align}$$
Somewhat different is Curry's paradox: $$\varphi_Y(x) = (x\in x)\to Y$$Then following the same pattern of reasoning that gives Russell's paradox (that is, let $X = \{x\mid \varphi_Y(x)\}$ and ask if $X\in X$), in this case we can conclude $Y$.  But since $Y$ was completely arbitrary, we can prove anything at all by this method.
A: No.
The Russell paradox looked at $\varphi(x)$ to be $x\notin x$. But we can use any of the "it cannot be a set" paradoxes.


*

*Burali-Forti paradox, define an ordinal to be a transitive set $A$ such that $(A,\in)$ is well-ordered. This can be expressed in a formula in the language of set theory, and $\{A\mid A\text{ is an ordinal}\}$ cannot be a set, since it is well-ordered by $\in$ and it is transitive, so it will have to be a member of itself and contradict the well-ordering.

*Cantor's paradox, consider just $\varphi(x)$ to be "$x$ is a set", which is really just $x=x$ in the context of set theory where everything is a set. Then Cantor's paradox says there is no bijection between a set and its power set; but every set has a power set. In particular the set $X=\{x\mid x\text{ is a set}\}$, but then $\mathcal P(X)\subseteq X$ so there is an injection which is a contradiction.

*The set of all singletons, consider the formula $\varphi(x)$ to be a statement saying that $x=\{y\}$ for some $y$. If $X=\{x\mid\varphi(x)\}$ is a set, then $\bigcup X$ is a set, and we reduce to the previous paradox.
There are other paradoxes as well. 
