- #1

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I was able to prove that A is an algebra, since for any C,D ε A, C[itex]\cap[/itex]D and C[itex]^{c}[/itex] belong to A.

I'm not able to understand why A isn't a σ-algebra. Can anyone please outline a proof or give me a counter-argument.

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- Thread starter Old Monk
- Start date

- #1

- 8

- 0

I was able to prove that A is an algebra, since for any C,D ε A, C[itex]\cap[/itex]D and C[itex]^{c}[/itex] belong to A.

I'm not able to understand why A isn't a σ-algebra. Can anyone please outline a proof or give me a counter-argument.

- #2

- 22,089

- 3,297

Are singletons in A?? Is [itex]\mathbb{Q}[/itex] in A?

- #3

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- #4

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Yeah singletons are in A. Since singletons of the form (a,a) exist in J, the single union of such elements should exist in A.

Isn't [itex](a,a)=\emptyset[/itex]?? Singletons would rather come from [itex][a,a]=\{a\}[/itex]. But ok, let's assume that the singletons are in A.

That I assume implies, all the rationals in [0,1] are also in A.

Why??

- #5

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We have [0,r)[itex]\cup[/itex](r,1] in A, if r is any rational in [0,1]. Therefore the element {r} ε A. So I think we could generalize this to all r ε Q[itex]\cap[/itex][0,1].

- #6

- 22,089

- 3,297

We have [0,r)[itex]\cup[/itex](r,1] in A, if r is any rational in [0,1]. Therefore the element {r} ε A. So I think we could generalize this to all r ε Q[itex]\cap[/itex][0,1].

Yeah, I agree that every {r} is in A. But why does that imply that [itex]\mathbb{Q}\in A[/itex]. Is it even true??

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