Is the set of rational numbers in [0,1] a set in the algebra?

[tunaaa]

Hello, first I’d like to clarify that the only difference between an algebra and a sigma-algebra, is that we have

[tex]A,B \in \mathcal{A} \Rightarrow A \cup B \in \mathcal{A} \text{ (1) for } \mathcal{A} \text{ algebra}[/tex][tex]A_1, A_2, A_3, \ldots\in\mathcal{A} \Rightarrow \bigcup_{i=1}^{\infty}A_i \in \mathcal{A}\text{ (2) for } \mathcal{A} \text{ sigma-algebra}[/tex]If this understanding is correct, then what I am confused about is that surely (1) implies (2), since, starting with (1),

[tex]\text{if } A_1, A_2 \in\mathcal{A} \text{ then } A_1 \cup A_2 \in\mathcal{A} \text{ by definition}[/tex][tex]\text{now if also } A_3 \in\mathcal{A} \text{ then } \underbrace{(A_1 \cup A_2)}_{A} \cup \underbrace{A_3}_{B} \in\mathcal{A}[/tex][tex]\text{now if also } A_4 \in\mathcal{A} \ldots[/tex]and so do we not arrive at (2)?


Thanks for your help.

 

[Stephen Tashi]

surely (1) implies (2),

No, the fact that a statement is true "for any finite N" does not imply the statement is true "for an infinite number". In fact, what the statement means in the case of "an infinite number" may require its own definition (for example, finite series, vs infinite series).

I think what you mean by "algebra" is "the algebra of sets". I think you are correct that the essential additional feature of a sigma algebra is that it is closed under countable unions.

 

[tunaaa]

Thanks. So could you give an example of a set that is an algebra, but not a sigma algebra? Thanks

 

[Stephen Tashi]

Look at a post on the forum by pivoxa15:
Physics Forums > Mathematics > Calculus & Analysis
Sigma algebra?

It has the example:
Universal set = real numbers in [0,1]
Sets in the algebra: all subsets of [0,1] of the form [a,b],(a,b],[a,b) or (a,b) and all finite unions of such sets.

A subset of [0,1] that is expressible as a countably infinite intersections and unions of such sets
(but not as a finite number of operations on them) is all rational numbers in [0,1].

 

[tunaaa]

Thanks - but to me that just suggests the rationals should in fact be in the algebra, rather than proving a contradiction. Why are the rationals not in the algebra to begin with?

 

[micromass]

A very elementary example is the following set

[tex]\{A\subseteq \mathbb{N}~\vert~A~\text{is finite or}~\mathbb{N}\setminus A~\text{is finite}\}[/tex]

this is an algebra that is not a sigma-algebra. In fact, this set is the algebra generated bby all the singletons. Note that the sigma-algebra generated by the singletons is [itex]\mathcal{P}(\mathbb{N})[/itex].

 

[Stephen Tashi]

Thanks - but to me that just suggests the rationals should in fact be in the algebra

There might be a moral or aesthetic sense in which the rationals "should" be in the algebra, but people who give counterexamples aren't bound by such considerations.

 

[tunaaa]

There might be a moral or aesthetic sense in which the rationals "should" be in the algebra, but people who give counterexamples aren't bound by such considerations.

OK fair enough, but that still doesn't explain why the rationals aren't in the algebra to begin with. I apologize for being slow to understand your argument. Thanks

 

[Stephen Tashi]

Your question would be clearer without the phrase "to begin with". If you are asking for a proof that the set of rational numbers in [0,1] is not a set in the algebra then this is a good question. My intuitive argument would be that a set in the algebra can have only a finite number of singleton points in it. The only way to get a singleton point is to take the intersection of two intervals of the form (a,b] and [b,c). Since there are more than a finite number of rational numbers in [0,1], for any set in the algebra. we can find a rational number q that is either not in the set or is in an interval that is a subset of the set. If is in an interval then the interval would also contain irrational numbers.