Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals

Theorem

Let $M_1 = \left({I^\omega, d_2}\right)$ be the Hilbert cube:

$M_1 = \displaystyle \prod_{k \mathop \in \N} \left[{0 \,.\,.\, \dfrac 1 k}\right]$

under the same metric as that of the Hilbert sequence space:

$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in I^\omega: d_2 \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$

Let $M_2$ be the metric space defined as:

$M_2 = \displaystyle \prod_{k \mathop \in \N} \left[{0 \,.\,.\, 1}\right]$

under the Tychonoff topology.

Then $M_1$ and $M_2$ are homeomorphic.

Proof

Let $x = \left\langle{x_i}\right\rangle$ be an arbitrary element of $M_1$.

Let $f: M_1 \to M_2$ be the mapping defined as:

$\forall x \in M_1: f \left({x}\right) = \left({x_1, 2 x_2, 3 x_3, \ldots}\right)$

Then $f$ is seen to be a bijection.

It remains to be seen that an open set in $M_1$ is mapped to an open set in $M_2$ by $f$.

Sources

• 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology ... (previous) ... (next): $\text{II}: \ 38: \ 1$