279: Product of Connected Topological Spaces Is Connected

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that product of connected topological spaces is connected

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of product topology.

---

Target Context

• The reader will have a description and a proof of the proposition that the product of any connected topological spaces is connected.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any product topological space, $T={\times }_{\alpha \in A}{T}_{\alpha }$ where $A$ is any possibly uncountable indices set, such that each $T_{\alpha }$ is connected, $T$ is connected.

---

2: Proof

Let us suppose that $T$ was not connected. $T=U_1\cup U_2,U_1\cap U_2=\mathrm{\varnothing }$, $U_1\ne \mathrm{\varnothing }$ and $U_2\ne \mathrm{\varnothing }$ where $U_i$ would be an open set. $U_i={\cup }_{\beta \in B_i}{\times }_{\alpha \in A}{U}_{i-\beta -\alpha }$ where $B_i$ would be any possibly uncountable indices set, $U_{i-\beta -\alpha }$ would be an open set such that each of only a finite number of $U_{i-\beta -\alpha }$s would not be $T_{\alpha }$ for each $i$ and $\beta$, by the definition of product topology.

Let us take any point, $p_{\underset{―}{\gamma }}\in {\pi }_{\underset{―}{\gamma }}T$, where ${\pi }_{\underset{―}{\gamma }}$ would be the projection of $T$ to $T_{\underset{―}{\gamma }}:={\times }_{\alpha \in A\setminus \{\gamma \}}{T}_{\alpha }$. Let us define $B_{i-\underset{―}{\gamma }}:=\{\beta \in B_i|p_{\underset{―}{\gamma }}\in {\times }_{\alpha \in A\setminus \{\gamma \}}{U}_{i-\beta -\alpha }\}$ and $U_{i-\gamma }:={\cup }_{\beta \in B_{i-\underset{―}{\gamma }}}{U}_{i-\beta -\gamma }$. $U_{1-\gamma }\cup U_{2-\gamma }=T_{\gamma }$, because for any $p_{\gamma }\in T_{\gamma }$, the corresponding point, $p\in T$, such that ${\pi }_{\underset{―}{\gamma }}p=p_{\underset{―}{\gamma }}$ and ${\pi }_{\gamma }p=p_{\gamma }$ would be in ${\times }_{\alpha \in A}{U}_{i-\beta -\alpha }$ for a $\beta \in B_{i-\underset{―}{\gamma }}$, as $p\in U_1$ or $p\in U_2$. $U_{1-\gamma }\cap U_{2-\gamma }=\mathrm{\varnothing }$, because if $p_{\gamma }\in U_{1-\gamma }\cap U_{2-\gamma }$, the corresponding point, $p\in T$, such that ${\pi }_{\underset{―}{\gamma }}p=p_{\underset{―}{\gamma }}$ and ${\pi }_{\gamma }p=p_{\gamma }$ would be in $U_1\cap U_2$.

As $T_{\gamma }$ is connected, $U_{1-\gamma }=\mathrm{\varnothing }$ or $U_{2-\gamma }=\mathrm{\varnothing }$, which would mean that for any fixed $p_{\underset{―}{\gamma }}$, all the $p_{\gamma }\in T_{\gamma }$s would be entirely in $U_1$ or in $U_2$. Let us suppose that $U_{2-\gamma }=\mathrm{\varnothing }$ and so, inevitably $U_{1-\gamma }=T_{\gamma }$ without loss of generality. For any $p_{\gamma }\in U_{1-\gamma }$, $p_{\gamma }\in U_{1-\beta -\gamma }$ for a $\beta \in B_{1-\underset{―}{\gamma }}$, and the corresponding point, $p\in T$, such that ${\pi }_{\underset{―}{\gamma }}p=p_{\underset{―}{\gamma }}$ and ${\pi }_{\gamma }p=p_{\gamma }$ would be in a ${\times }_{\alpha \in A}{U}_{1-\beta -\alpha }$ for a $\beta \in B_{1-\underset{―}{\gamma }}$.

But there are only finite number of $U_{1-\beta -\alpha }$s such that $U_{1-\beta -\alpha }\ne T_{\alpha }$ for the $\beta$. Let us define $A_{\beta }$ as the set of such $\alpha$s, $\{{\alpha }_1,{\alpha }_2,...,{\alpha }_n\}$. Then, every point, $p^{\prime }\in T$, such that ${\pi }_{\alpha \in A_{\beta }}{p}^{\prime }=p_{\alpha }\in U_{1-\beta -\alpha }$ for every $\alpha \in A_{\beta }$ would be in ${\times }_{\alpha \in A}{U}_{1-\beta -\alpha }\subseteq U_1$. ${\alpha }_1$ could be taken to be the $\gamma$ of the above procedure, then, every point, $p^{\prime }\in T$, such that ${\pi }_{\alpha \in A_{\beta }\setminus \{{\alpha }_1\}}{p}^{\prime }=p_{\alpha }\in U_{1-\beta -\alpha }$ for every $\alpha \in A_{\beta }\setminus \{{\alpha }_1\}$ would be in $U_1$, because as the 1 $p_{{\alpha }_1}$ is known to be in $U_1$, the whole $T_{{\alpha }_1}$ have to be in $U_1$. Then, ${\alpha }_2$ could be taken to be the $\gamma$ of the above procedure, then, every point, $p^{\prime }\in T$, such that ${\pi }_{\alpha \in A_{\beta }\setminus \{{\alpha }_1,{\alpha }_2\}}{p}^{\prime }=p_{\alpha }\in U_{1-\beta -\alpha }$ for every $\alpha \in A_{\beta }\setminus \{{\alpha }_1,{\alpha }_2\}$ would be in $U_1$, likewise, and so on. After all, every point, $p^{\prime }\in T$ would be in $U_1$, a contradiction.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>