{"id":518,"date":"2014-02-15T01:00:25","date_gmt":"2014-02-15T01:00:25","guid":{"rendered":"http:\/\/blogs.ams.org\/visualinsight\/?p=518"},"modified":"2015-07-29T00:53:05","modified_gmt":"2015-07-29T00:53:05","slug":"cantors-cube","status":"publish","type":"post","link":"https:\/\/blogs.ams.org\/visualinsight\/2014\/02\/15\/cantors-cube\/","title":{"rendered":"Cantor’s Cube"},"content":{"rendered":"
\"Cantor's<\/a>

Cantor’s Cube<\/p><\/div>\n

<\/p>\n

This is Cantor’s cube. To make it, start with a cube. Chop it into 3\u00d73\u00d73 smaller cubes, and remove all of them except the 8 at the corners. Then do the same thing for each of these 8 smaller cubes, and so on, forever. The stuff that’s left is Cantor’s cube.<\/p>\n

What’s the volume of Cantor’s cube? It’s zero! At each stage of the procedure above you multiply the volume by 8\/27, since there are 3\u00d73\u00d73 little cubes, but you keep only 8. So, in the end there’s no volume left.<\/p>\n

You can do the same trick in any dimension. In dimension 1, you get the famous Cantor set by starting with an interval and repeatedly removing the middle 1\/3. The Cantor set has length zero, since at each stage you multiply its length by 2\/3.<\/p>\n

If we denote the Cantor set by $K$, then Cantor’s cube is $K^3$. But in fact, Cantor’s cube is homeomorphic the Cantor set! Indeed, $K^n$ is homeomorphic to $K$ for any $n \\ge 1$.<\/p>\n

A space homeomorphic to the Cantor set is called a Cantor space<\/a> In 1910, Brouwer showed that a topological space is a Cantor space if and only if it is:<\/p>\n

1) non-empty,<\/p>\n

2) compact<\/a>,<\/p>\n

3) totally disconnected<\/a>,<\/p>\n

4) has no isolated points<\/a>,<\/p>\n

5) metrizable<\/a>.<\/p>\n

A compact Hausdorff space is metrizable if and only if it is second countable<\/a>, by a spinoff of the Urysohn Metrization Theorem<\/a>. So, a Hausdorff space obeying 1)-4) but not 5) must be ‘big’: that is, not second countable. <\/p>\n

We can get such a space by putting the product topology on $2^\\kappa$ for an uncountable cardinal $\\kappa$. On the other hand, if we take $\\kappa = \\aleph_0$ we get the Cantor set, $K$. Indeed, this is one way to see that $K^n \\cong K$ for any finite $n \\ge 1$:<\/p>\n

$$ (2^{\\aleph_0})^n \\; \\cong \\; 2^{n \\aleph_0} \\; \\cong \\; 2^{\\aleph_0} $$<\/p>\n

Another approach is to use Stone’s representation theorem<\/a>. The set of homomorphisms from any Boolean algebra<\/a> $A$ to the Boolean algebra $2 = \\{T,F\\}$ has a natural topology making it a compact totally disconnected Hausdorff space. This is called the Stone space<\/b> of $A$. The Stone space lacks isolated points iff $A$ is atomless<\/a>, and it is metrizable iff $A$ is countable. <\/p>\n

There is a unique countable atomless Boolean algebra, and its Stone space is homeomorphic to the Cantor set. If we take $A$ to be an uncountable atomless Boolean algebra, its Stone space obeys 1)-4) but not 5). <\/p>\n

The picture here was created by Solkoll<\/a>. It is in the public domain, and can be found on Wikicommons<\/a>.<\/p>\n

\n

Visual Insight<\/i> is a place to share striking images that help explain advanced topics in mathematics. I\u2019m always looking for truly beautiful images, so if you know about one, please drop a comment here<\/a> and let me know!<\/p>\n

<\/div>","protected":false},"excerpt":{"rendered":"

To make this shape, start with a cube. Chop it into 3\u00d73\u00d73 smaller cubes, and remove all of them except the 8 at the corners. Then do the same thing for each of these 8 smaller cubes, and so on, forever. The stuff that’s left is called ‘Cantor’s cube’.<\/p>\n

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