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Why $\liminf A_n=\{\omega\mid\omega \text{ that are in all but finitely many $A_n$}\}\ ?$

To me, $$\liminf A_n=\bigcup_{n\in\mathbb N}\bigcap_{m\geq n}A_m,$$

and thus, $$\omega \in\liminf A_n\iff \exists n\in\mathbb N: \forall m\geq n, \omega \in A_m$$ what would be $$\liminf A_n=\{\omega\mid\omega \text{ that are in all $A_n$ except a finite number of $A_n$}\}.$$ What's wrong in my interpretation ?

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    The only thing (linguistically) wrong is the words "that are" added in the set builder notation; this would read "... is the set of all $\omega$ for which $\omega$ that are in ..." -- To add a third semi-natural language variant: $\liminf A_n=\{\,\omega\mid\omega\text{ in almost all }A_n\,\}$ – 2017-01-16

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Your interpretation is correct. The set of $\omega$ which are contained in all $A_n$ except a finite number of $A_n$ is equal to the set of $\omega$ which are contained in all but finitely many $A_n$.

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    Why these two sets are equals ? – 2017-01-16
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    "contained in all but finitely many" means "contained in all but a finite number of them" means "contained in all except a finite number of them" – 2017-01-16