Finite Product Space is Connected iff Factors are Connected/Basis for the Induction

Theorem

Let $T_1$ and $T_2$ be topological spaces.

Then the product space $T_1 \times T_2$ is connected if and only if $T_1$ and $T_2$ are connected.

Proof

Necessary Condition

Let $T_1 \times T_2$ be connected.

By Projection from Product Topology is Continuous, $T_1$ and $T_2$ are continuous images under the projections $\pr_1$ and $\pr_2$.

Hence by Continuous Image of Connected Space is Connected, $T_1$ and $T_2$ are connected.

$\Box$

Sufficient Condition

Suppose that $T_1$ and $T_2$ are connected.

Define:

$C_y = T_1 \times \set y$ for each $y \in T_2$

$B = \set {x_0} \times T_2$ for some fixed $x_0 \in T_1$.

Each $C_y$ is homeomorphic to $T_1$ by Topological Product with Singleton.

By Connectedness is a Topological Property, each $C_y$ is therefore connected.

By the same argument, $B$ is also connected.

Also:

$C_y \cap B = \set {\tuple {x_0, y} }$ and hence is non-empty

$\ds T_1 \times T_2 = B \cup \bigcup_{y \mathop \in T_2} C_y$.

So by the corollary to Union of Connected Sets with Non-Empty Intersections is Connected, it follows that $T_1 \times T_2$ is connected.

$\blacksquare$

Also see

• Product Space is Path-connected iff Factor Spaces are Path-connected

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $6.2$: Connectedness: Proposition $6.2.17$