Definition: Suppose $\mathcal A$ is the Boolean algebra of all Jordan measurable sets in $I=[0,1]$ (i.e $\mathcal A=\{A\subseteq I: \mu(\partial(A))=0\}$, where $\mu$ is the Lebesgue measure and $\partial$ means the topological boundary).

Definition: "Interval algebra" suppose $\mathcal B$ on $I=[0,1]$ is the algebra generated by the sets of the form $([a_{0},b_{0})\cup[a_{1},b_{1})\cup...\cup[a_{n},b_{n}))\cap I$ where $0\leq a_{0}< b_{0}<a_{1}<b_{1}<...<a_{n}<b_{n}$.

(1) Prove or disprove: Every (nontrivial) uncountable atomless subalgebra of $\mathcal A$ is isomorphic to $\mathcal B$.

(2) If (1) is false, I would say: Every (nontrivial) uncountable atomless subalgebra of $\mathcal A/N$ ($N$ is the family of all null sets) is isomorphic to $\mathcal B$.

By atomless I mean a Boolean algebra which has no atom.

  • $\begingroup$ Do you mean $\mu(\partial(A))=0$ in the definition of Jordan measurable? $\endgroup$ Feb 3, 2014 at 17:00
  • $\begingroup$ @JoelDavidHamkins Oh yes, exactly. Thanks for letting me know. I will correct it. $\endgroup$
    – Ameen
    Feb 3, 2014 at 17:02
  • $\begingroup$ As for (2), $\mathcal A/N$ is itself an atomless algebra. I see no reason why it should be isomorphic to $\mathcal B$. Certainly the natural inclusion of $\mathcal B$ in $\mathcal A/N$ is not onto. $\endgroup$ Feb 3, 2014 at 20:38
  • $\begingroup$ @EmilJeřábek Yes, the inclusion map of $\mathcal B$ is not onto but this is not enough to say they are not isomorphic, because inclusion map of odd numbers in $\Bbb N$ is also not onto but they are isomorphic. $\endgroup$
    – Ameen
    Feb 8, 2014 at 0:20

2 Answers 2


It seems to me that the cardinalities are off, so (1) fails, since $\cal A$ contains an atomless Boolean algebra of cardinality $2^{\frak c}$, and so many distinct atomless Boolean subalgebra of that size, but $\cal B$ has size continuum. So in these instances, they cannot be isomorphic.

Similarly, a cardinality argument shows at least that (2) will fail if the continuum hypothesis is false, since in that case ${\cal A}/N$ will have size continuum, and therefore by the downward Lowenheim-Skolem theorem will have atomless subalgebras of size $\aleph_1$, which will again be too small to be isomorphic to $\cal B$. So (2) fails when CH fails.

  • $\begingroup$ Thanks for the answer. I suspect that (1) might be false, but I could not bring any example (would be more helpful if you have an example in mind). But (2) might be true under the continuum hypothesis. What do you think? $\endgroup$
    – Ameen
    Feb 3, 2014 at 17:21
  • 2
    $\begingroup$ I expect that (2) will fail regardless of CH. $\endgroup$ Feb 3, 2014 at 17:27
  • $\begingroup$ Appreciate your ideas $\endgroup$
    – Ameen
    Feb 3, 2014 at 17:48
  • $\begingroup$ One more thing I would like to ask you if you do not mind. what is the relation between Cohen algebra and interval algebra. $\endgroup$
    – Ameen
    Feb 3, 2014 at 23:04
  • $\begingroup$ Your interval algebra, if you mean just the Boolean algebra generated by those intervals, is not a complete Boolean algebra, but the Cohen algebra is usually taken as the complete Boolean algebra. But otherwise, they are related in that the Boolean completion of the interval algebra is isomorphic to the Cohen algebra. A condition in the Cohen algebra specifies a finite binary string, and each such string corresponds to a tiny interval in your interval algebra. $\endgroup$ Feb 3, 2014 at 23:32

An answer to (2) in ZFC: The algebra $\mathcal B$ is not only uncountable, it has the property that below any positive element there are uncountably many elements.

Let $\mathcal A'$ be the algebra of all sets of the form $([a_{0},b_{0})\cup[a_{1},b_{1})\cup...\cup[a_{n},b_{n}))\cap I$ where all $a_i,b_i$ are in $([0,\frac12]\cap \mathbb Q) \cup [\frac12,1]$. This is a subalgebra of $\mathcal A$. So $\mathcal A'/N$ is a subalgebra of $\mathcal A/N$, it is uncountable and atomless.

Now consider the element $[0,\frac12]/N$ in $\mathcal A'/N$. It is positive, but there are only countably many elements below it.

So $\mathcal A'/N$ is not isomorphic to $\mathcal B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.