Hilbert Cube is Compact

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

Then $M$ is a compact space.


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 $M$.


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 $\struct {\ell^2, \norm {\, \cdot \,}_2}$.