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In our Topology course, we have been studying the anti-equivalance of categories between zero-dimensional compact Hausdorff spaces and Boolean algebras. The Cantor set has come up a lot, and I have a question with this set.

Why is the Boolean algebra of all clopen subsets of the Cantor set isomorphic to its own square?

Also, are there other sets that satisfy this property?

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Hint: The Cantor set is a fractal. It is homeomorphic to two disjoint copies of itself. –  Brett Frankel Feb 28 '12 at 3:08
    
It's even homeomorphic to its square! –  Patrick Feb 28 '12 at 3:57
    
By the square of an algebra do you mean the set of all $u\times u$ or $u\times v$ where $u,v$ are in the algebra? .. or something else. –  Patrick Feb 28 '12 at 4:07
    
The square of a Boolean algebra $B$ is the direct product of two copies of $B$. –  GEdgar Feb 28 '12 at 4:13
    
Probably just $A\times A$. –  Patrick Feb 28 '12 at 4:14
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2 Answers 2

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Let $C$ be the Cantor set, and let $\mathscr{B}$ be its clopen algebra.

  • If you’re thinking of the Cantor set $C$ as the middle-thirds Cantor set, let $C_1=C\cap\left[0,\frac13\right]$, and let $C_2=C\cap\left[\frac23,1\right]$.

  • If you’re thinking of the Cantor set as $\{0,1\}^\omega$, let $C_1=\{x\in C:x_0=0\}$, and let $C_2=\{x\in C:x_0=1\}$.

In either case note that $C$ is the disjoint union of $C_1$ and $C_2$. Now let $\mathscr{B}_1$ and $\mathscr{B}_2$ be the clopen algebras of $C_1$ and $C_2$, respectively, and show the following:

  1. $C$ is homeomorphic to $C_1$ and $C_2$, so $\mathscr{B},\mathscr{B}_1$, and $\mathscr{B}_2$ are isomorphic.

  2. The map $$h:\mathscr{B}\to\mathscr{B}_1\times\mathscr{B}_2:U\mapsto\langle U\cap C_1,U\cap C_2\rangle$$ is an isomorphism.

It follows immediately from (1) and (2) that $\mathscr{B}\cong\mathscr{B}\times\mathscr{B}$.

The same thing (in the second version) works for the clopen algebra of the product space $\{0,1\}^\kappa$ for any infinite cardinal $\kappa$, not just $\omega$; this is the free Boolean algebra on $\kappa$ generators.

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Let $p_1,p_2,p_3,\ldots$ be primitive propositions, and consider the Boolean algebra made up of all propositions you can form from finitely many of them connected by "and", "or", and "not". Each such proposition corresponds to a clopen subset of the Cantor set. If $p_n$ is true, that corresponds to the $n$th ternary digit being $1$; if false then $0$. So a proposition specifies something about finitely many binary digits. Thus a clopen set.

Now let $q_1,q_2,q_3,\ldots$ be more primitive propositions.

The whole set $p_1,p_2,p_3,\ldots,q_1,q_2,q_3,\ldots$ is still countable, so the Boolean algebra you get is isomorphic to the one you get with just the "$p$"s and not the "$q$"s. And it's the product space.

Notice that the union of the two sets of propositions corresponds to the product of two topological spaces. That's one aspect of the "anti-equivalence" of the category of Boolean algebras and Boolean homomorphisms, and the category of totally disconnected compact Hausdorf spaces and continuous functions.

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