2
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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". $\endgroup$ Jan 15 '16 at 21:39
3
$\begingroup$

Yes, your proof is correct.

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 $$\overline{\prod_{i\in I} A_i} = \prod_{i\in I} \overline{A_i}.$$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.