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Let $I_n = [0, 1-\frac{1}{n}], n \in \mathbb{N}$ be a sequence of intervals. Is the statement (1) or statement (2) true? \begin{align} \lim_{n \rightarrow \infty} I_n = [0,1] && (1) \\ \lim_{n \rightarrow \infty} I_n = [0,1) && (2) \end{align} where the expression $\lim_{n \rightarrow \infty} I_n$ shall be understood as $\bigcup_{n=1}^{\infty} I_n$, since $I_n$ is an isotonic sequence of sets. (Thanks to the comments that made this clear).

I am not really sure how to approach this question. On the one hand I think that I can somehow just write $\lim_{n \rightarrow \infty} [0, 1 - \frac{1}{n}] = [0, 1- \lim_{n \rightarrow \infty} \frac{1}{n}] = [0,1]$. On the other hand this raises the strange question "Are Intervals continuous?" and I dont know how to answer that question (Of course this question does not make sense, since continuity is a property of mappings not of sets. Just to avoid misunderstandings. The question should only emphasize the confusion I am having right now). Also I see that $\forall n \in \mathbb{N}$ we have $1-\frac{1}{n} <1$ but does it hold in the limit? Its a really basic question but I dont really know how to answer it.

Red
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    The expression $\lim_{n \rightarrow \infty} I_n$ makes no sense. An expression of the form $\lim_{n \rightarrow \infty} a_n$ only makes sense if $(a_n)_n$ is a sequence of real numbers (ore more generally elements of a topological space). – Smiley1000 Nov 01 '24 at 13:28
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    The (2) nis correct. Indeed, $1\notin I_n$ for all $n\in\mathbb N$. – Surb Nov 01 '24 at 13:28
  • @Smiley1000: $\lim_{n\to \infty }I_n$ make sense and is defined as $\bigcup_{n\in\mathbb N}[0,1-1/n]$. – Surb Nov 01 '24 at 13:30
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    @Surb I don't think it's a good idea to define limits of intervals. For example, what is $\lim_{n \rightarrow \infty} \left[ 0, 2 + (-1)^n \right]$? – Smiley1000 Nov 01 '24 at 13:31
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    @Surb If a union is meant, one should simply write the union. – Smiley1000 Nov 01 '24 at 13:31
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    @Smiley1000: In this case is doesn't make sense. For a collection of sets ${A_n}$, $\lim_{n\to \infty }A_n$ is well defined whenever $\limsup_{n\to \infty }A_n=\liminf_{n\to \infty }A_n$. And in this case $$\lim_{n\to \infty }A_n=\limsup_{n\to \infty }A_n=\liminf_{n\to \infty }A_n.$$ – Surb Nov 01 '24 at 13:32
  • @Smiley1000 Arbitrary sets can have limits. See wikipedia set-theoretic limit. – Jan Nov 01 '24 at 13:32
  • @Jan The note at the top of that page says: "This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed." – Smiley1000 Nov 01 '24 at 13:35
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    The point is that such limits can be defined nicely, and many textbooks do so. Also, the wikipedia page for ‘Limit of a sequence’ also has this disclaimer at the top of the page. – Jan Nov 01 '24 at 13:43
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    This question uses two bad terminology practices that lead students astray: limit of sequence of sets, and describing sets as continuous. As there is resistance to fixing these, I have voted to close. – JonathanZ Nov 01 '24 at 13:43
  • @JonathanZ You could also tell me how to write it properly, like the other persons in the comment section did and not just vote to close immediately. Also I did not describe sets as continuous, I stated that I ask myself this strange (!) question. "Strange" implying that something about the notion is off. Also there is no resistance on my side (why do you think there is?). I just asked the question how I conceived it. Obviously there are some misconceptions I am having here, otherwise I wouldnt ask. – Red Nov 01 '24 at 13:50
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    Thank you all for your comments! So what I am getting out of it is 1) limits of Intervals should only be defined in terms of their unions (or intersections) and only make sense if the liminf and limsup coincide. Furthermore, the limit I am interested in is [0,1) since the argument of 1 - 1/n < 1 for all n holds? – Red Nov 01 '24 at 13:54
  • Excellent! Questions can always be edited, even if they're closed, and a question can be reopened even if it's closed. Lots of people drop a question and never bother to fix the problems in it. If people see you engaging with the comments and making fixes they'll know that you will fix them and it's worth keeping this question open. – JonathanZ Nov 01 '24 at 13:57
  • The rule is that the union of any open subsets is always open and that the intersection of a finite number of open subsets is open. It is the opposite for closed sets: the intersection of any closed subsets is closed and the union of a finite number of closed subsets is closed. There is no general result for the union of an infinite number of closed sets. For example, a singleton is always closed, and any set (including open ones) is the union of the singletons containing its element: $A=\bigcup_{a\in A}{{a}}$. – Serge Ballesta Nov 01 '24 at 14:03
  • @Smiley1000 The concept of limits of sequences of sets is well known and often used in measure theory. See, for example, Measure theory from Paul Halmos. However, I agree that the OP wrote that apparently to just to represent the union, which is a bad idea. – jjagmath Nov 01 '24 at 15:22
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    @JonathanZ First of all, limits of sequences of sets is not a bad terminology practice, it's a standard notion from set theory/measure theory - I suggest you open your favourite textbook and find it there. Second of all, since when are questions of confused students asking for help the authoritative source of knowledge for other students?? The suggestion that the asker should refrain from describing their line of thought because it could lead other students astray is totally ridiculous, and a close vote for that reason is a cherry on top. – Adayah Nov 02 '24 at 08:59
  • Well, I learned from Alliprantis and Burkinshaw, and I don't recall every seeing it used (though that was a while ago). I most often see the concept being a source of confusion for beginners, and this thread started of with incorrect definitions in both question and comments. If a closure threat is what it takes to get things updated, so be it. – JonathanZ Nov 02 '24 at 14:33

2 Answers2

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Since $I_n$ is a nondecreasing sequence of sets (i.e. $I_1 \subseteq I_2 \subseteq \ldots$), we have $$\liminf_{n\rightarrow\infty}I_n= \limsup_{n\rightarrow\infty}I_n=\lim_{n\rightarrow\infty}I_n = \bigcup_{n=1}^{\infty}I_n.$$ But $1 \notin I_n$ for each $n$, so $\bigcup_{n=1}^{\infty}I_n = [0,1)$.

As requested, you can find the definition of a set theoretic limit below, and, importantly why it simplifies so drastically for monotone sequences of sets.

Let $(A_n)_n$ be any sequence of sets. Then we can define its limit whenever the following expressions are equal: $$ \liminf_{n\rightarrow \infty} A_n=\bigcup_{n=1}^\infty\bigcap_{k\geq n}A_k,$$ and $$ \limsup_{n\rightarrow \infty} A_n=\bigcap_{n=1}^\infty \bigcup_{k\geq n}A_k.$$ In that case we write $$\lim_{n\rightarrow\infty}A_n= \limsup_{n\rightarrow \infty} A_n = \liminf_{n\rightarrow \infty}A_n.$$ Now, suppose that $(A_n)_n$ is nondecreasing, then the limit of the sets always exists. Note that $\bigcap_{k\geq n}A_k=A_n$, and $\bigcup_{k\geq n}A_k=\bigcup_{n=1}^\infty A_n$, for each $n$. This implies $$ \liminf_{n\rightarrow \infty} A_n=\bigcup_{n=1}^\infty A_n ,$$ and $$\limsup_{n\rightarrow \infty} A_n=\bigcap_{n=1}^\infty \bigcup_{k=1}^\infty A_k = \bigcup_{k=1}^\infty A_k,$$ so $\lim_{n\rightarrow \infty} A_n = \bigcup_{n=1}^\infty A_n$. A similar result is true for nonincreasing sets, and it’s a good exercise to prove this result.

Jan
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    Sorry, I had to down vote. The meaning of lim sup and lim inf for sets is not obvious, and readers of this answer will likely come away with misconceptions. Maybe at least link to the Wikipedia article, and emphasize a bit more strongly how much we're relying on the special case of nondecreasing sets to be able to use the simpler formula? – JonathanZ Nov 01 '24 at 14:37
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    I understand your concern and provided an addendum. If you find any mistakes or gaps, let me know! – Jan Nov 01 '24 at 15:03
  • @JonathanZ The concept of limits of sequences of sets is well known and often used in measure theory. See, for example, Measure theory from Paul Halmos. – jjagmath Nov 01 '24 at 15:19
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    @Jan - Thanks, I think that really improves the answer. Always happy to switch a down-vote to an up-vote. – JonathanZ Nov 01 '24 at 15:57
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    @jjagmath - It's important to understand what context/level the questioner is coming from. It was clear to me that they were not referencing that definition, and probably not even familiar with it. Clarification was needed. – JonathanZ Nov 01 '24 at 15:59
  • Related: https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Hans Lundmark Nov 01 '24 at 21:39
  • It's not clear why $1 \notin I_n$ for each $n$ implies $1 \notin \bigcup_{n=1}^{\infty}I_n$. Limits commonly do not exist anywhere in their generating sequences. Are you claiming $\lim_{n\rightarrow\infty}{1/n}$ does not exist? One could argue it "obviously" should be ${0}$, and the answer to the original question is "obviously" [0, 1]. – JounceCracklePop Nov 02 '24 at 09:11
  • @JounceCracklePop That implication has nothing to do with limits, as it is purely set theoretic. It is just the contraposition of $1 \in \bigcup_{n=1}^\infty I_n \Rightarrow \exists n \in \mathbb{N}^+ : 1\in I_n$, which is obviously true. And, $\lim_{n\rightarrow \infty} {1/n}=\varnothing$. Also, I find it pointless to argue with a definition. You can always define your own types of limits for sets. – Jan Nov 02 '24 at 10:48
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Three almost trivial but interesting cases true for arbitrary topological spaces:

$\bullet$ A nondecreasing sequence of open sets has limit and the limit is open.

$\bullet$ A nonincreasing sequence of closed sets has limit and the limit is closed.

$\bullet$ A nonincreasing sequence of compact nonempty sets has limit and the limit is compact and nonempty (Is the decreasing sequence of non empty compact sets non empty and compact?).

Martín-Blas Pérez Pinilla
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