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

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## Theorem

Let $M_1 = \struct {I^\omega, d_2}$ be the Hilbert cube:

- $M_1 = \displaystyle \prod_{k \mathop \in \N} \closedint 0 {\dfrac 1 k}$

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

- $\displaystyle \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$

Let $M_2$ be the metric space defined as:

- $M_2 = \displaystyle \prod_{k \mathop \in \N} \closedint 0 1$

under the Tychonoff topology.

Then $M_1$ and $M_2$ are homeomorphic.

## Proof

Let $x = \sequence {x_i}$ be an arbitrary element of $M_1$.

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

- $\forall x \in M_1: \map f x = \tuple {x_1, 2 x_2, 3 x_3, \ldots}$

Then $f$ is seen to be a bijection.

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

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $1$