Hilbert Cube is Arc-Connected

Theorem

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

Then $M$ is an arc-connected space.

Proof

Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$.

Consider the mapping $f: \closedint 0 1 \to I^\omega$ defined as:

$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$

\(\ds \sum_{i \mathop \ge 0} \paren {t x_i + \paren {1 - t} y_i}^2\)

\(=\)

\(\ds \sum_{i \mathop \ge 0} \paren {t^2 x_i^2 + \paren {1 - t}^2 y_i^2 + 2 t \paren {1 - t} x_i y_i}\)

\(\ds \)

\(=\)

\(\ds t^2 \sum_{i \mathop \ge 0} x_i^2 + \paren {1 - t}^2 \sum_{i \mathop \ge 0} y_i^2 + 2 t \paren {1 - t} \sum_{i \mathop \ge 0} x_i y_i\)

which is convergent.

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Then $f$ is an injective path joining $x$ to $y$.

Hence the result.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $5$