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$\begingroup$

For $a

Let us consider the following class:

$\mathscr{F}(I):=\{E \subseteq \mathbb{R}: E=\bigcup_{k=1}^nI_k, I_k\in I, \text{and } I_k\cap I_l=\varnothing \text{ for }1\leq k\neq l\leq n, n\in\mathbb{N} \}$

I have read that this is an algebra, but where is the empty set in this class? Even considering $a\leq b$, I cannot undesrtand the claim. Am I missing something?

Any help will be appreciated.

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    If we take $n = 0$, don't we get the empty set? – 2017-02-28
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    Isn't $(5, 5]$ empty? Isn't it the set of all numbers simultaneously greater than $5$ and less than or equal to $5$? – 2017-02-28
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    @KennyWong: I suspect $0 \not\in \mathbb{N}$ for the OP. That's the more common interpretation. – 2017-02-28
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    But we need $a < b$ in $(a,b]$. – 2017-02-28
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    What does the first paragraph of this question have to do with the rest? It asks us to think of a certain collection, but it doesn't give that collection a name, and it does not seem to be mentioned ever after. – 2017-02-28
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    But @BrianTung, a – 2017-02-28
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    I have edited a bit the question @HenningMakholm. Sorry I am not an expert mathematician. – 2017-02-28
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    @JoséAntonioDíazNavas: You either have to allow $a \le b$, or allow $n$ to be $0$ (implying that $\cup _{n=1} ^0$ is the empty union) - without any of these, there is no $\emptyset$ in $\mathscr F (I)$. – 2017-02-28
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    I have read this [link](http://math.stackexchange.com/questions/938880/question-about-measure-theory-the-least-sigma-algebra-generated-by-certain) and I can see the differences, maybe the initial condition a – 2017-02-28
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    Yes @AlexM., those are my suspicions. – 2017-02-28
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    @JoséAntonioDíazNavas: Sorry, I did not see $a < b$. – 2017-02-28

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