# Finding the $\liminf$ of a sequence of sets

Is the given proof of the following correct?

Let $\{A_n\}$ be a sequence of sets, then: $$\liminf_{n\to\infty} A_n=\bigcup_{i=1}^\infty\bigcap_{j=i}^\infty A_j=\{\text{elements that belong to all but finitely many } A_i's\}$$

Pf.

($\rightarrow$) Let $$B_i=\bigcap_{j=i}^\infty A_j$$

Then, $$x\in B_i \implies x \in B_j \text{ } \forall j \gt i$$

So $$x \in \bigcup_{j\ge i} B_j$$

This implies that $x \in$ all but finitely many $A_i$'s.

($\leftarrow$) Suppose that $x \in$ all but finitely many $A_i$'s. Call the finite $A_i$'s: $\bar A_1, \bar A_2, \dots, \bar A_n$. Then we have:

$$x \in \bar A_{n+1}$$

Let $\bar A_{n+1} = A_m$ for some integer $m$. Then $x\in A_m$ and $x \in A_{m+1}$ and so on for all $j \gt m$. So $$x \in \bigcap_{j\ge m} A_j=B_j$$

Therefore, $$x \in B_1 \cup \dots \cup B_j \cup \dots = \liminf$$ and we have the result.

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