Is the Cantor cube a closed subset of the Hilbert cube?

I have found this statement:

The cantor cube $C:= \{0,1\}^I$ is homeomorphic to a closed subset of the Hilbert cube $H:= [0,1]^I$.

I have an idea for a proof. I want to show $\overline C \subseteq C$, considering $C$ as a subspace of $H$. Suppose $H$ comes with euclidean topology.

Let $x \in \overline C$. Then, every open neighbourhood of $x$ has a non-empty intersection with $C$. Suppose $x \not \in C$. This implies there exists $i_0 \in I$ such that $x_{i_0} \not\in \{0,1\}$. Define $$X_{i_0} := U_\varepsilon(x_{i_0}) \subseteq [0,1]$$ with $\varepsilon := \frac{1}{2}\min\{x_{i_0}, 1-x_{i_0}\}$ and let $$U := \prod_{i \in I} X_i$$ be an open neighbourhood of $x$ with $X_i := [0,1]$ for $i \neq i_0$. By construction $X_{i_0} \cap \{0,1\} = \emptyset$ which implies $U \cap C = \emptyset$. This contradicts $x \in \overline C$.

Is this correct?

• Looks fine to me. I would add that if you remove the initial hypothesis (let $x\in\overline{C}$, then instead of getting a contradiction at the end, you have a direct proof of the contrapositive. In many ways, a such proofs are "preferred". Jan 15 '16 at 21:39

If you look at it, you may notice that by de-concretising it, the same argument works to show that in every product space, the product of closed subsets of the factors is a closed subset of the product:

Let $I$ be an index set, and for every $i\in I$ let $X_i$ be a topological space, and $F_i$ a closed subset of $X_i$. Then $F := \prod_{i\in I} F_i$ is a closed subset of $\prod_{i\in I} X_i$.

Slightly stronger even:

Let $I$ be an index set, and for every $i\in I$ let $X_i$ be a topological space, and $A_i \subset X_i$. Then $$\overline{\prod_{i\in I} A_i} = \prod_{i\in I} \overline{A_i}.$$

It is a fine proof.

In general, if $A_i \subseteq X_i$ is closed, then $\prod_{i \in I} A_i$ is closed in $\prod_{i \in I} X_i$, which we can apply here with $\{0,1\}$ being closed in $[0,1]$ for $|I|$ many such copies.

The prood of the latter is similar: suppose $(x_i)_{i \in I}$ is not a point of $\prod_{i \in I} A_i$, for some $i_0$ we have that $x_{i_0} \notin A_{i_0}$. Then $\prod_{i \in I} O_i$ is the basic open set with $O_{i_0} = X_{i_0} \setminus A_{i_0}$ (which is open by closedness of $A_{i_0}$) and all other $O_i = X_i$, also misses $\prod_{i \in I} A_i$. So every point of the complement of $\prod_{i \in I} A_i$ is an interior point of it, so the complement is open, etc.