218: Product of Finite Number of Connected Topological Spaces Is Connected

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

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Topics

About: topological space

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The table of contents of this article

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

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Starting Context

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

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Target Context

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

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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.

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Main Body

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1: Description

For any finite number of topological spaces, $T_1,T_2,...,T_n$, $T_1\times T_2,\times ...,\times T_n$ is connected.

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2: Proof

1st, let us think of the case, $n=2$. Suppose that $T_1\times T_2$ was not connected. $T_1\times T_2=U_1\cup U_2$, $U_1\cap U_2=\mathrm{\varnothing }$ where $U_i$ would be a non-empty open set on $T_1\times T_2$. By the definition of product topology, $U_1={\cup }_{\alpha }{U}_{1-\alpha }\times U_{2-\alpha }$ where $\{\alpha \}$ would be a possibly uncountable index set and $U_{i-\alpha }$ would be non-empty open on $T_i$; $U_2={\cup }_{\beta }{U}_{1-\beta }\times U_{2-\beta }$ where $\{\beta \}$ would be a possibly uncountable index set and $U_{i-\beta }$ would be non-empty open on $T_i$. For any point, $p_1\in T_1$, $\{p_1\}\times T_2$ would have to be covered by $U_1\cup U_2$. Let us take the subset of $\{\alpha \}$, $\{{\alpha }^{\prime }\}=\{{\alpha }^{\prime }\in \{\alpha \}|p_1\in U_{1-{\alpha }^{\prime }}\}$ and the subset of $\{\beta \}$, $\{{\beta }^{\prime }\}=\{{\beta }^{\prime }\in \{\beta \}|p_1\in U_{1-{\beta }^{\prime }}\}$. $({\cup }_{{\alpha }^{\prime }}{U}_{2-{\alpha }^{\prime }})\cup ({\cup }_{{\beta }^{\prime }}{U}_{2-{\beta }^{\prime }})=T_2$, because for each $p_2\in T_2$, $(p_1,p_2)$ would have to be in a $U_{1-{\alpha }^{\prime }}\times U_{2-{\alpha }^{\prime }}$ or in a $U_{1-{\beta }^{\prime }}\times U_{2-{\beta }^{\prime }}$. ${\cup }_{{\alpha }^{\prime }}{U}_{2-{\alpha }^{\prime }}$ and ${\cup }_{{\beta }^{\prime }}{U}_{2-{\beta }^{\prime }}$ would not share any point, because if $p_2\in {\cup }_{{\alpha }^{\prime }}{U}_{2-{\alpha }^{\prime }},{\cup }_{{\beta }^{\prime }}{U}_{2-{\beta }^{\prime }}$, $(p_1,p_2)\in U_1,U_2$. ${\cup }_{{\alpha }^{\prime }}{U}_{2-{\alpha }^{\prime }}$ or ${\cup }_{{\beta }^{\prime }}{U}_{2-{\beta }^{\prime }}$ would be empty, because otherwise, $T_2$ would not be connected, being a union of disjoint non-empty open sets. Suppose that ${\cup }_{{\alpha }^{\prime }}{U}_{2-{\alpha }^{\prime }}=T_2$ without loss of generality. $p_1\in {\cup }_{\alpha }{U}_{1-\alpha }$, because otherwise, there would be no $(p_1,?)$ on $U_1\cup U_2$, and $(p_1,?)\notin U_2$, because otherwise, a $p_1,p_2$ would be shared by $U_1$ and $U_2$. But not all the points of $T_1$ would be on $U_1$, because otherwise, $U_2$ would be empty. So, while $({\cup }_{\alpha }{U}_{1-\alpha })\cup ({\cup }_{\beta }{U}_{1-\beta })=T_1$, because each $p_1\in T_1$ would have to be in a $U_{1-\alpha }$ or in a $U_{1-\beta }$, any point would not be shared by ${\cup }_{\alpha }{U}_{1-\alpha }$ and ${\cup }_{\beta }{U}_{1-\beta }$, neither of which would be empty, so, $T_1$ would not be connected, as being a union of disjoint non-empty open sets, a contradiction. So, $T_1\times T_2$ is connected.

$T_1\times T_2,\times ...,\times T_n$ can be proved to be connected inductively, as $T_1\times T_2$ is connected, $T_1\times T_2\times T_3=(T_1\times T_2)\times T_3$ is connected, and so on.

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References

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