We define the limit inferior and limit superior of a sequence $(A_n:n\in \mathbb N)$ of subsets of a set $X$ by setting

(1) $$\begin{split}\lim \inf _{n\rightarrow\infty} A_n&=\bigcup_{n\in\mathbb N} \bigcap _{k\ge n} A_k\\ &= \{x\in X:x\in A_n \text{ for all but finitely many } n\in\mathbb N\}\end{split}$$

(2) $$\begin{split}\lim \sup_{n\rightarrow \infty} A_n &= \bigcap_{n\in\mathbb N}\bigcup _{k\ge n} A_k\\ &=\{x\in X:x\in A_n \text{ for infinitely many } n\in \mathbb N\}\end{split}$$

Now for my own expansion of their definition:

$$\begin{split}\lim\inf_{n\rightarrow\infty} A_n &= \left(\bigcap_{k=1,2,3,...} A_k \right)\cup\left(\bigcap_{k=2,3...} A_k \right)\cup...\end{split}$$

I do not see how the definition of liminf as "$x \in A_n$ for all but finitely many $n$" could work. If it is in finitely many $A_k$, then the number will not appear in the intersection of any $\cap A_k$ and thus not appear in the overall union. What gives?

However, I do see how limsup definition works, if

$$\begin{split}\lim\sup_{n\rightarrow\infty} A_n &= \left(\bigcup_{k=1,2,3,...} A_k \right)\cap\left(\bigcup_{k=2,3...} A_k \right)\cap...\end{split}$$

Thus there must be no end point to when the number stops appearing to appear in the limsup.


1 Answer 1


In English the phrase "all but" has two different meanings. However, in mathematics, "$x \in A_n$ for all but finitely many $n$" means "$x \in A_n$ is false for finitely many $n$." See also this answer.

In particular, if $N$ is larger than the finitely many $n$ for which "$x \in A_n$" is false, then $x \in \bigcap_{k = N, N+1, \ldots} A_k$.

  • 1
    $\begingroup$ I totally misread the phrase "all but finitely many." $\endgroup$
    – Jellyfish
    Jun 22 at 3:16

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