Finite Product Space is Connected iff Factors are Connected

Theorem
Let $T_1 = \struct {S_1, \tau_1}, T_2 = \struct {S_2, \tau_2}, \dotsc, T_n = \struct {S_n, \tau_n}$ be topological spaces.

Let $T = \ds \prod_{i \mathop = 1}^n T_i$ be the product space of $T_1, T_2, \ldots, T_n$.

Then $T$ is connected each of $T_1, T_2, \ldots, T_n$ are connected.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
 * $T$ is connected each of $T_1, T_2, \ldots, T_n$ are connected.

Basis for the Induction
$\map P 2$ is the case:
 * The product space $T_1 \times T_2$ is connected $T_1$ and $T_2$ are connected.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \prod_{i \mathop = 1}^k T_i$ is connected each of $T_1, T_2, \ldots, T_k$ are connected.

from which it is to be shown that:
 * $\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is connected each of $T_1, T_2, \ldots, T_{k + 1}$ are connected.

Induction Step
This is the induction step:

By definition of product space:
 * $\ds \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$

But from the basis for the induction:
 * $\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is connected $\ds \prod_{i \mathop = 1}^k T_i$ is connected and $T_{k + 1}$ is connected

and from the induction hypothesis:
 * $\ds \prod_{i \mathop = 1}^k T_i$ is connected each of $T_1, T_2, \ldots, T_k$ are connected.

Hence:


 * $\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is connected each of $T_1, T_2, \ldots, T_{k + 1}$ are connected.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction:
 * For all $n \in \Z_{\ge 2}$, $T$ is connected each of $T_1, T_2, \ldots, T_n$ are connected.

Also see

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