$\{0,1\}^\mathbb{N}$ is homeomorphic to which subset of $\mathbb{R}$.

Question

In an interview the interviewer asked me the following but I failed to give the answer.

$\{0,1\}^\mathbb{N}$ with product topology is homeomorphic to which subset of $\mathbb{R}$?

Can anyone give me the answer and explain me please? thanks for your kind help.

Answer 1

I am going to assume that the topology on $\{0,1\}$ is the discrete topology.

The space $\{0,1\}^\Bbb N$ is called the Cantor space. It is a zero-dimensional compact metric space which has no isolated points. From these properties also follow that it is totally disconnected as well.

But there is a nontrivial theorem (by Brouwer, I believe) stating the following:

Every completely metrizable, totally disconnected compact space without isolated points is homeomorphic to the Cantor space.

Since every subset of $\Bbb R$ is metrizable, and every compact subset is completely metrizable (because compact metric spaces are complete), it follows that $A\subseteq\Bbb R$ is homeomorphic to the Cantor space if and only if it is compact, totally disconnected and without isolated points.

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I just noticed that I read the question wrong, and it asks about a particular subset, rather than a characterization of all the subsets.

In this case the simplest answer would be the set of all real numbers in the interval $[0,1]$ that in their base $3$ digit expansion (taking finite expansion over infinite when we can, e.g. $0.2$ over $0.\bar 1$) the digit $1$ does not appear at all.

To see this, note that we can map a sequence $\langle a_n\mid n\in\Bbb n\rangle$ in the product to the real number $\sum_{n=1}^\infty2\frac{a_n}{3^n}$, which is a number in the interval $[0,1]$ that in its trenary expansion there is no digit $1$. It is not hard to show that this is a bijection.

To show that it is a homeomorphism, note that a basic open set in $\{0,1\}^\Bbb N$ is all the sequences which have some finite number of coordinates fixed. This defines some number in $[0,1]$, and we can find some $\varepsilon$ that every sequence in our basic open set is within $\varepsilon$ distance of this number; and vice versa, given a number in the range and some basic open interval around it we can define the basic open set in $\{0,1\}^\Bbb N$ whose image is contained in the interval.

Answer 2

This idea was dealt with by Roger Penrose in page 370 of his book "The Road to Reality: A Complete Guide to the Laws of the Universe".

The set is called the Cantor Set, and according to Wikipedia: As a topological space, the Cantor set is naturally homeomorphic to the product of countably many copies of the space $\{0, 1\}$, where each copy carries the discrete topology. This is the space of all sequences in two digits.