Finite Product Space is Connected iff Factors are Connected

From ProofWiki
Jump to navigation Jump to search

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 if and only if each of $T_1, T_2, \ldots, T_n$ are connected.


General Case

Let $I$ be an indexing set.

Let $\family {T_\alpha}_{\alpha \mathop \in I}$ be an indexed family of topological spaces.

Let $T = \ds \prod_{\alpha \mathop \in I} T_\alpha$ be the Cartesian space of $\family {T_\alpha}_{\alpha \mathop \in I}$.

Let $T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$.


This article, or a section of it, needs explaining.
In particular: Notation in the above -- explain what the overline means in this context.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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 {{Explain}} from the code.


Let $\tau$ be a topology on $T$ such that the subsets ${S'}_\alpha \subseteq \ds \prod T_\alpha$ where ${S'}_\alpha = \set {\family {y_\beta} \in T: y_\beta = x \beta \text { for all } \beta \ge \alpha}$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.


Then $T$ is connected if and only if each of $T_\alpha: \alpha \in I$ 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 if and only if 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 if and only if $T_1$ and $T_2$ are connected.


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.

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 if and only if 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 if and only if 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 if and only if $\ds \prod_{i \mathop = 1}^k T_i$ is connected and $T_{k + 1}$ is connected

and from the induction hypothesis:

$\ds \prod_{i \mathop = 1}^k T_i$ is connected if and only if each of $T_1, T_2, \ldots, T_k$ are connected.

Hence:

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

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Finite_Product_Space_is_Connected_iff_Factors_are_Connected&oldid=672081"