# If the von Neumann rank $x$ is less than that of $y$, is $x\in y$?

If $x$ and $y$ are sets such that $\newcommand{\rank}{\operatorname{rank}}\rank x < \rank y$ , then $x\in y$?

I understand that generally if $\rank x < \rank y$, then it should be true, but I do not think it is true for all possible sets $x$ and $y$.

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The rank of $y=\{\omega\}$ is infinite, but there is no $x$ whose rank is finite which is in $y$. What true is the converse, if $x\in y$ then the rank of $x$ is strictly less than the rank of $y$.
Again, recall the definition of rank. The rank of $y$ is the least ordinal larger than all the ranks of the elements of $y$. Your question, if true, would imply that any set which has an infinite rank would be infinite. In particular a singleton whose element is a set of infinite rank. This is obviously false, then.