# Hilbert Cube is Compact

## Theorem

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

Then $M$ is a compact space.

## Proof

Let $M'$ be the metric space defined as:

- $M' = \ds \prod_{k \mathop \in \N} \closedint 0 1$

under the product topology.

By definition, $\closedint 0 1$ is the closed unit interval under the usual (Euclidean) topology.

From Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals, $M$ is homeomorphic to $M'$.

From Closed Real Interval is Compact in Metric Space, $\closedint 0 1$ is a compact space.

But $M'$ is also a compact space by Tychonoff's Theorem.

From Compactness is Preserved under Continuous Surjection, if $M'$ is compact then so is $M$.

$\blacksquare$

This article is complete as far as it goes, but it could do with expansion.In particular: Add the direct first-principles proof from S&S as Proof 2.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Expand}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

### Hilbert Cube is Compact in 2-Sequence Space with 2-Norm

Let $\ell^2$ be the 2-sequence space.

Let $\norm {\, \cdot \,}_2$ be the 2-norm.

Let $C$ be the Hilbert cube.

Then $C$ is compact in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $4$