Given $x$ a set, we define $\cup x$ to be the sets of those sets $z$ for which $\exists y \in x$ such that $z \in y$ (for example, $x \cup y = \cup \{x,y\}$).

Using this, find all sets $x$ such that $\cup x = \emptyset$.

I was thinking of having $x = \emptyset$ and as well $x = \{\emptyset\}$. But I'm sure there other sets $x$ missing. How can we find all sets $x$ such that the union is $\emptyset$? I'm not sure how to go around using the above definition.

Thanks for the help.


You are right.

To show that no other $x$ satisfies $\bigcup x=\varnothing$, you need to show that if $x\neq\varnothing$ and $x\neq\{\varnothing\}$, then there is some $y$ such that $z\in\bigcup x$.

But what does it mean that $z\in\bigcup x$? It means that for some $y\in x$ such that $z\in y$. Under what conditions that is true?


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