# Cantor space: example of an open set that's not closed?

Very basic question: I can't get my head around the Cantor space. It has a basis of clopen sets. Finite unions of closed sets are closed, and unions of open sets are open, so a finite union of basis elements of a Cantor space is clopen. The only open sets in a Cantor space that are not closed, if there are any, are infinite unions of basis elements. An example of an open set that's not closed is ... what?

Not a homework question; I managed to stump myself. Thanks for your help.

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## 2 Answers

Hint: Look at the complement of a point—the Cantor space is not discrete.

Later: Since the Cantor set is homeomorphic to a countable product $\prod_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ of the cyclic group of order two (use the identification of the cantor set with the points in $[0,1]$ whose infinite ternary expansion contains no $1$), it is homogeneous. This means in particular that the Cantor set has no isolated points and hence it has no open points.

Now note that a basic open set is of the form $\prod_{n=1}^{\infty} X_n$ with all but finitely many $X_n = \mathbb{Z}/2\mathbb{Z}$. But this means that a basic open set contains a space homeomorphic to the entire Cantor space, hence non-empty open sets are uncountable (in fact of cardinality $\mathfrak{c}$). In particular we see that a convergent sequence (which is of course closed) can't be open. Passing to complements we get an open set that's not closed, as required.

[Meta: Thanks to ccc for pointing this out and to Brian for making me think again.]

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Hold it horizontally (landscape view, I mean), use both your thumbs :) – t.b. Aug 1 '11 at 5:06
How about convergent sequences? These are compact subsets of a Hausdorff space, and therefore closed, right? – gary Aug 1 '11 at 7:07
@Theo: The observation that a basic open set in the product space $2^\omega$ is homeomorphic to $2^\omega$ is pretty elementary; the problem is that too often students see only the middle-thirds Cantor set in elementary courses. (And of course that means that every non-empty open set is not just uncountable, but of cardinality $\mathfrak{c}$.) – Brian M. Scott Aug 2 '11 at 10:08
@Theo: A good student familiar only with the middle-thirds Cantor set might still come up with an example like $[0.1/9]\cup\bigcup_{n>0}\left[\sum_{k=1}^n 2/3^k,\sum_{k=1}^n 2/3^k+1/3^{n+2} \right]$, which is clearly open, being a union of clopen pieces of the construction. but which has $1$ as a fairly obvious limit point in its (non-trivial) complement. – Brian M. Scott Aug 2 '11 at 10:37
@Theo: Das Brett vorm Kopf ist ein Berufsrisiko! :-) – Brian M. Scott Aug 2 '11 at 10:43

Just make a sort of fishbone, and remove the spine: $$\bigcup_{i=1}^{\infty} [10^i].$$

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