I was wondering if someone would be so kind to provide a very simple explanation of $\limsup$ and $\liminf$ of a sequence of sets. For a sequence of subsets $A_n$ of a set $X$ we have $$\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$$ and $$\liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right).$$ But I am having a hard time imagining what that really means unions of intersections and intersections of unions I think maybe causing the trouble. I read the version on Wikipedia but that didn't resolve this either. Any help would be much appreciated.

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    $\begingroup$ A proof of what? $\endgroup$ Commented Feb 10, 2012 at 21:51
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    $\begingroup$ These are the definitions. Are you asking for help understanding them? $\endgroup$ Commented Feb 10, 2012 at 22:00
  • $\begingroup$ Thanks it should have been an explanation, although the book tries to provide a proof, which i could n't follow. Hence the question. $\endgroup$ Commented Feb 10, 2012 at 23:32
  • $\begingroup$ $\bigcap_{N=1}^\infty \bigcup_{n\ge N} (-\frac{1}{n}, \frac{1}{n})=\{0\}$. What are some more interesting examples? $\endgroup$
    – user398843
    Commented Mar 7, 2018 at 22:14

6 Answers 6


A member of $$ \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n $$ is a member of at least one of the sets $$ \bigcap_{n\ge N} A_n, $$ meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc. That means it's a member of all except finitely many of the $A$.

A member of $$ \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n $$ is a member of all of the sets $$ \bigcup_{n\ge N} A_n, $$ so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc. That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later. That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.

  • $\begingroup$ I think in the of sup explanation the should be U signs between the subsets, but the rest of the answer was well explained. $\endgroup$ Commented Feb 10, 2012 at 23:39
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    $\begingroup$ fixed ${{{{{}}}}}$ $\endgroup$ Commented Feb 11, 2012 at 0:51
  • $\begingroup$ thanks for this wonderful explanation! $\endgroup$
    – under root
    Commented Mar 27, 2014 at 19:09
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    $\begingroup$ what really clarified things for me was the "That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to." Thanks! $\endgroup$
    – Diego
    Commented Jan 8, 2016 at 0:10
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    $\begingroup$ How do we know that if $x \in \lim \inf A_n$ then it's a member of all except finitely many of the A's? Can it not be a member of all except an infinitely large number of A's? $\endgroup$ Commented May 17, 2020 at 13:11

I just came up with this mnemonic story:

There is a company with employes and one day a whole bunch $\mathcal A$ of them get fired. They become beggars and have to live on the street. One day, the local church decides to start to give out free food for them every week. In the $n$th week, $A_n$ are the people who show up (this is a sequence of $\mathcal A$-people: $\forall n\ (A_n\subseteq \mathcal A)$. Yes, beggars never die).

  1. Some of the people eventually get a new job and never show up at the church again.

  2. Others are too proud and try not to be seen around all the time, but they need to eat so they always come back eventually.

  3. Lastly there are the people who have low self-esteem; they feel inferior, and at one point, they don't care anymore and start to get their food from the church each week.

$\lim \sup A_n$ are all the people who don't get another job. (Categories 2 and 3). Thus, $\lim \inf A_n^C$ are all the people who eventually get a new job. (Category 1)

$\lim \inf A_n$ are the people who become weekly regulars. (Category 3)

Clearly $\lim \inf A_n \subseteq \lim \sup A_n$.

A series converges implies all the people who can't get another job eventually swallow their pride and become regulars too: $\lim \inf A_n$ = $\lim \sup A_n$. We call this the limit of $A_n$.

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    $\begingroup$ Publish this!!!!!! $\endgroup$
    – user198044
    Commented Jul 30, 2015 at 16:53
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    $\begingroup$ You mean to say lim sup consist of people of second and third category as you described and lim inf consist of third category of people. Right ? $\endgroup$
    – user268307
    Commented Dec 15, 2015 at 10:46
  • $\begingroup$ @Member Edited. $\endgroup$
    – BCLC
    Commented Apr 23, 2018 at 15:29
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    $\begingroup$ @Nikolaj-K nice answer. but I have a problem. we know that if $x \in \lim \inf A_n$ then $x$ belongs to all but finitely many $A_i$s. how this can be explained by your example. $\endgroup$
    – thomson
    Commented Jun 23, 2018 at 17:26
  • $\begingroup$ @thomson I believe the argument goes like this. Category 3 is people who eventually come every single week. That implies they belong to infinitely many $A_n$. (An infinitely long tail of $A_n$, specifically.) Since $A_n$ is countable, an (infinite) tail must have a finite head. So for the people in category 3, each person is missing from (at least some of) the the finite head of $A_n$, since that's how it was constructed. Hence people in category 3 are only missing from finitely many $A_n$. $\endgroup$
    – kdbanman
    Commented Jan 15, 2020 at 15:27

(This is a late answer, but hopefully it may add another perspective to the discussion.)

When one thinks about the problem of defining the limit of a sequence of sets, there are two easy cases: if the sequence is increasing, or if it's decreasing.

For example, when defining improper integrals $\iint_D$ in multivariable calculus, one looks at sequences $D_1 \subseteq D_2 \subseteq D_3 \subseteq \dotsb$ which exhaust $D$, meaning that

  • each $D_i$ is a subset of $D$, and
  • each element of $D$ is contained in $D_n$ for all sufficiently large $n$ (in other words, $D=\bigcup_{n=1}^\infty D_n$).

In this situation, students often ask if they can write $D_n \to D$, and the answer is usually “well, we haven't really defined limits of sets in this course, but if we had...”. So it's pretty intuitive that for an increasing sequence of sets, the limit should be defined as the union of all sets in the sequence.

Similarly, for a decreasing sequence $E_1 \supseteq E_2 \supseteq E_3 \supseteq \dotsb$, it's natural to define the limit as the intersection : $E_n \to E$ as $n\to\infty$, where $E=\bigcap_{n=1}^\infty E_n$.

Now, for an arbitrary sequence of sets $(A_1,A_2,A_3,\dots)$, we can squeeze it between an increasing sequence $(D_n)$ and a decreasing sequence $(E_n)$, like this: $$ \begin{array}{lcl} D_1 = A_1 \cap A_2 \cap A_3 \cap \dotsb &\quad\subseteq\qquad A_1 \qquad &\subseteq \qquad E_1 = A_1 \cup A_2 \cup A_3 \cup \dotsb \\ D_2 = \phantom{A_1 \cap{}} A_2 \cap A_3 \cap \dotsb &\quad\subseteq\qquad A_2 \qquad &\subseteq \qquad E_2 = \phantom{A_1 \cup{}} A_2 \cup A_3 \cup \dotsb \\ D_3 = \phantom{A_1 \cap A_2 \cap{}} A_3 \cap \dotsb &\quad\subseteq\qquad A_3 \qquad &\subseteq \qquad E_3 = \phantom{A_1 \cup A_2 \cup{}} A_3 \cup \dotsb \\ & \qquad\vdots & \end{array} $$ Moreover, $(D_n)$ is the largest increasing sequence such that $D_n \subseteq A_n$ for all $n$, and $(E_n)$ is the smallest decreasing sequence such that $A_n \subseteq E_n$ for all $n$, so it makes sense to define $$ \liminf_{n\to\infty} A_n = \lim_{n\to\infty} D_n ,\qquad \limsup_{n\to\infty} A_n = \lim_{n\to\infty} E_n . $$ And if these lower and upper limits are equal, then we say that that's $\lim\limits_{n\to\infty} A_n$.

(What Willie Wong's answer says is that the same construction makes sense in any partially ordered set where you can take the least upper bound and greatest lower bound of infinitely many elements, if you replace $(\subseteq,\cap,\cup)$ by $(\le,\wedge,\vee)$.)

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    $\begingroup$ Highly recommend this answer; it gives complete and intuitive comment on this question. $\endgroup$ Commented Dec 28, 2018 at 11:29
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    $\begingroup$ Sorry to bother you, but could you elaborate on what you mean by largest increasing sequence $D_n$? Is that fact particularly important to your construction here? Did you prove it somewhere that I did not see? Besides that, I thought your answer was the most helpful one, so thank you! $\endgroup$
    – D.R.
    Commented Oct 7, 2019 at 6:28
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    $\begingroup$ @D.R.: It means that if $(C_n)$ is an increasing sequence such that $D_n \subseteq C_n \subseteq A_n$ for all $n$, then $C_n=D_n$ for all $n$. I didn't prove it since I thought it was fairly obvious. For suppose that $(C_n)$ fulfills the assumptions, but $D_m \subset C_m$ (strict inclusion) for some $m$, i.e., $C_m$ contains some $x$ such that $x \notin D_m = A_m \cap A_{m+1} \cap \dots$, in other words for some $k \ge m$ we have $x \notin A_k$. Then, since the sequence $(C_n)$ is increasing, that element $x$ has to belong to $C_k$ as well, but then $C_k \subseteq A_k$ fails; contradiction. $\endgroup$ Commented Oct 7, 2019 at 7:43

In terms of sets, we have the following interpretations:

  • $\displaystyle x\in\bigcup_{i\in I} A_i$ means that $x$ is in at least one of the $A_i$ sets.
  • $\displaystyle x\in\bigcap_{i\in I} A_i$ means that $x$ is in all of the $A_i$ sets.

So this means that

  1. $\bigcap_{N\ge1}\bigcup_{n\ge N} A_n$ are all elements somewhere in $A_N,A_{N+1},A_{N+2},\dots$, no matter how large N is. Being a member of this set is logically equivalent to being "in infinitely many of the $A_i$ sets".
  2. $\bigcup_{N\ge1}\bigcap_{n\ge N} A_n$ are all elements in every single one of $A_{N},A_{N+1},A_{N+2},\dots$ for some $N$. Being a member of this set is logically equivalent to being "in all but finitely many the $A_i$ sets".

Are you familiar with the real analysis definition of $$\limsup_{n\to\infty} x_n = \inf_{m\geq 0} \sup_{n\geq m} x_n~?$$

The same definition can be applied to any sequence of elements in a complete lattice. Now apply it to the power set $2^X$ of some base set $X$ with set inclusion as the partial order.

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    $\begingroup$ Do you expect an undergraduate or beginning graduate student to understand "complete lattice" ? $\endgroup$
    – user198044
    Commented Jul 30, 2015 at 16:53
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    $\begingroup$ @JackBauer: While I don't understand why you are asking about what undergraduate or graduate students can or cannot understand, I do expect a beginning graduate student to understand what a complete lattice is and how it applies to the discussion at hand after learning the definition from, say, Wikipedia. $\endgroup$ Commented Aug 13, 2015 at 15:33
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    $\begingroup$ @JackBauer, I learned basic lattice theory while I was an undergraduate, and didn't find it especially difficult, and I'm not especially bright. And yes, I do find it clarifying, in much the same way as basic category theory is quite clarifying. More undergraduates should learn basic lattice theory, I think. $\endgroup$ Commented Feb 28, 2016 at 10:06
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    $\begingroup$ @DonShanil: you are certainly entitled to your opinion, but my answer does not presuppose an understanding/familiarity with the notion of a complete lattice, which is why I provided a link to the Wikipedia article. The reader is encouraged to first learn what a complete lattice is and then apply the definition in that context. I happen to believe there is pedagogical value in providing all the resources one would need to discover the answer without making all details explicit. $\endgroup$ Commented Aug 15, 2017 at 13:47
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    $\begingroup$ @DonShanil: I want to further point out that my answer was posted after the wonderful (accepted) answer of Michael Hardy. The brevity is entirely intentional and the point of the answer is that there is something much more general at work (lattice theory) which unifies the various concepts involved. $\endgroup$ Commented Aug 15, 2017 at 13:56

Another way to think about it is via indicator functions: let $f_n$ be the indicator function of the set $X_n$ (i.e. $f_n(x)=1$ if $x\in X_n$ and 0 otherwise). Then it is not difficult to check that $\lim \sup f_n$ is the indicator function of $\lim \sup X_n$ and similarly for limit inferior.


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