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$$x\in \bigcap \emptyset \iff (\forall b \in \emptyset)\,x\in b$$ Can it be drawn that $\forall x,\,x\in \bigcap \emptyset$ using the right hand side of above? How to show $(\forall b \in \emptyset)\,k\in b$ for an arbitrary element $k$?

op ol
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  • Can you any $b$ with $b\in\varnothing$ and $k\notin b$? – drhab Feb 04 '22 at 08:02
  • @drhab I read some posts in this MSE. Most of them use your approach. Assume $k \not\in b$ then it leads contradiction. That makes sense but I want to start at the definition and prove it directly. Is it possible? – op ol Feb 04 '22 at 08:08
  • Op ol, have you seen the duplicate? Your question is answered there – FShrike Feb 04 '22 at 08:15
  • I think it is a matter of accepting the Boolean concept: if a statement is false then its negation is true (tertium non datur). – drhab Feb 04 '22 at 09:03

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