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This is probably a pretty dumb question, but I am confused by set theory again. The question is whether $$\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right]$$ equals $[0,1]$ or $[0,1)$. However, I am looking for some explanation and not only the result, since I'd like to understand why it's the one or the other.

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The element $1$ is in the union iff it is in one of the sets. –  Andrew Jul 16 '12 at 18:31
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But $1 \in \lim_{n \rightarrow \infty} [0,1-\frac{1}{n}]$ isn't it? Or does the limit make no sence here anyway? –  Haatschii Jul 16 '12 at 18:34
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In a way, the union above is the way of defining such a limit. But your notation does not make sense. You just need to treat the union as you treat finite unions. –  Andrew Jul 16 '12 at 18:38
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There will be some numbers in the union that are within (say) $0.1$ of $1$, also some numbers in the union which are within $0.00001$ of $1$, and so on. But there is no number in your union which is exactly equal to $1$. –  André Nicolas Jul 16 '12 at 18:39
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@AsafKaragila: Okay, then "any reasonable notion of limit". Where "reasonable" requires, among other things, that limits commute with any bijection with a sufficiently large domain, and that a constant sequence is its own limit. –  Henning Makholm Jul 16 '12 at 18:42
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4 Answers

up vote 19 down vote accepted

Remember that $x\in\bigcup\limits_{i\in I} A_i$ if and only if for some $i\in I$, $x\in A_i$.

So $1$ is in the union if and only if it appears in at least on of those intervals, so is it? No. It is not.


To the comment, don't think of $\bigcup_{i=1}^\infty$ as a limit in the calculus-sense of the word. Think of it as a logical operation which tells you that the index set is $\mathbb N$ (or some other set which is clear from context) and then use the above formula.

If you wish to think about it as $f(n)=\bigcup\limits_{k=1}^n [0,1-\frac1k]$, and think about the infinite union as $\lim\limits_{n\to\infty} f(n)$, then there are several caveats:

  1. Limits will usually require some sort of topology, some underlying structure which tells us about convergence. How would you define the limit here? For every $\varepsilon>0$...? It makes no sense, since subsets of $\mathbb R$ do not have a natural metric function.

  2. We can consider the following definition: $A$ is the limit of the sequence of $f(n)$ if and only if for every $x\in A$, there exists $n_0$ such that for all $n>n_0$, $x\in A_n$.

    Observe, however, that this coincides with the definition above, that $x$ is in the union if and only if it appears in at least one of the functions. This definition, however, coincides with the above only because this sequence of sets is increasing.

  3. Luckily, we can always think of an infinite union as an increasing sequence, but we would expect a definition for a limit to work for any sequence of sets, not just increasing unions.

  4. We can, however, think of it as a limit of a sequence of characteristic functions, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}=\chi_{[0,1)}$$ even as such limit, though, it is not "continuous" in the way you would like it to be, that is, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}\neq\chi_{\left[0,1-\lim\limits_{n\to\infty}\frac1n\right]}$$

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I will have to think about it for a while, but I hope I got the right idea of what is going on. Thanks for your help. –  Haatschii Jul 16 '12 at 21:20
    
@Haatschii: Glad I could help. Let me know if you have further questions. –  Asaf Karagila Jul 16 '12 at 21:26
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1 is not in any of the sets, so it can't be in their union.

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The whole point of a limit is that you go "up to but not including" the number. The limit of the upper bound of the union you gave is $1$, but that doesn't mean $1$ is in the union. Just like saying $\lim_{x\to 2}f(x)=3$ doesn't imply $f(2)=3$. To be an element of the union, it must be in at least one set. If you can't pick any single set (pre-union) that contains $1$, then it isn't in the union. It is a limit point, which is where the confusion lies, but that's a different thing entirely.

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+1: An excellent analogy! And considering $f:\Bbb N\to\mathcal{P}(\Bbb R)$ given by $$f(n)=\bigcup_{k=1}^{n+1}\left[0,1-\frac 1 k\right],$$ we're looking at just such a situation. –  Cameron Buie Jul 16 '12 at 19:38
    
@CameronBuie: So to get things rights. Am I right that using this $f(n)$, follows $$\lim_{n \rightarrow \infty} = [0,1] \neq [0,1) = \bigcup_{n=1}^\infty \left[ 0, 1- \frac{1}{k} \right]$$? Is this what you mean with this $\bigcup _{n=1}^\infty$ being different than a "normal" limit? –  Haatschii Jul 16 '12 at 19:47
    
@Haatschii: Limits talk about continuity. This simply tells you that the union is not a continuous function. Similarly you can consider a cardinal function, $n\mapsto|\mathcal P(\{0,\ldots,n\})|$. The limit is the cardinality of finite subsets of $\mathbb N$ which is $\aleph_0$, but $f("\infty")$ is uncountable. –  Asaf Karagila Jul 16 '12 at 19:52
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@Haatschii that sort of union is a normal limit. Or rather, it can be construed as one under the right circumstances (feel free to think of it as one, basically). The point we're making is that "$x$ is a limit point of a set" isn't the same as "$x$ is in the set". It is a normal limit, but like normal limits it doesn't (have to) actually take the value it approaches. –  Robert Mastragostino Jul 16 '12 at 20:02
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Note that, for this case, (with the "definition" of $\infty$ expanded)

$$\begin{aligned} \bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right] & \stackrel{\operatorname{def}}{=} \lim_{k\to\infty} \bigcup_{n=1}^k \left[0,1-\frac{1}{n}\right] \\ & = \lim_{k\to\infty} \left[0,1-\frac{1}{k}\right] \\ & = \left[0,1\right) \end{aligned}$$

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What do you mean by $\lim_{n\to\infty} \left[0,1-\frac{1}{n}\right]$? Without further explanation, the notation $\lim_{n \rightarrow \infty}$ is reserved, at least for me, for limits of sequences of real or complex numbers. –  Nils Matthes Jul 17 '12 at 9:01
    
In this case, it just means the union is equal to the last member. –  Mark Hurd Jul 17 '12 at 18:01
    
The expression "the last member" is a bit unfortunate, since this terminology suggests that the union is only over a finite, totally ordered set. Also your answer seems to hide away the essential points which explain why $\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right] = \left[0,1\right)$. For example, why is it not $[0,1]$? –  Nils Matthes Jul 18 '12 at 8:58
    
Courtesy of \stackrel{}{}. –  Asaf Karagila Jul 18 '12 at 12:34
    
I've expanded out the $\infty$. That shows my understanding of "last". As far as [0,1) instead of [0,1], I understand the former to be the only logical answer; other answers have gone into it further. I suppose my simple answer is like the top answer, the interval never actually contains 1, and because we have a symbol for it [0,1) is the answer. Note it is the fact that it is a one-sided limit: the simple $\lim_{n\to1}[0,n]=[0,1]$, but $\lim_{n\to1 \text{from below}}[0,n]=[0,1)$ –  Mark Hurd Jul 18 '12 at 12:44
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