837: Map from Topological Space into Finite Product Topological Space Is Continuous iff All Component Maps Are Continuous

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description/proof of that map from topological space into finite product topological space is continuous iff all component maps are continuous

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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: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of product topology.
• The reader knows a definition of continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$\{T_1,...,T_n\}$: $T_j\in \{\text{ the topological spaces }\}$
$T_1\times ...\times T_n$: $=\text{ the product topological space }$
$f$: $:T\to T_1\times ...\times T_n$
$\{{\pi }_1,...,{\pi }_n\}$: ${\pi }_j:T_1\times ...\times T_n\to T_j=\text{ the projection }$
$\{{\pi }_1\circ f,...,{\pi }_n\circ f\}$:
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Statements:
$f\in \{\text{ the continuous maps }\}$
$⟺$
$\mathrm{\forall }j\in \{1,...,n\}({\pi }_j\circ f\in \text{ the continuous maps })$
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2: Note

The product's being finite is crucial for this logic.

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3: Proof

Whole Strategy: Step 1: suppose that $f$ is continuous, and see that each ${\pi }_j\circ f$ is continuous at each $t\in T$; Step 2: suppose that each ${\pi }_j\circ f$ is continuous, and see that $f$ is continuous at each $t\in T$.

Step 1:

Let us suppose that $f$ is continuous.

Let us see that ${\pi }_j\circ f$ is continuous at each $t\in T$.

Let $U_{{\pi }_j\circ f(t)}\subseteq T_j$ be any open neighborhood of ${\pi }_j\circ f(t)$.

For each $k\in \{1,...,n\}\setminus \{j\}$, let us take any open neighborhood of ${\pi }_k\circ f(t)\in T_k$, $U_{{\pi }_k\circ f(t)}\subseteq T_k$.

$U_{{\pi }_1\circ f(t)}\times ...\times U_{{\pi }_n\circ f(t)}\subseteq T_1\times ...\times T_n$ is an open neighborhood of $f(t)$.

As $f$ is continuous, there is an open neighborhood of $t$, $U_t\subseteq T$, such that $f(U_t)\subseteq U_{{\pi }_1\circ f(t)}\times ...\times U_{{\pi }_n\circ f(t)}$.

Then, ${\pi }_j\circ f(U_t)\subseteq {\pi }_j(U_{{\pi }_1\circ f(t)}\times ...\times U_{{\pi }_n\circ f(t)})=U_{{\pi }_j\circ f(t)}$.

That means that ${\pi }_j\circ f$ is continuous at $t$.

As $t\in T$ is arbitrary, ${\pi }_j\circ f$ is continuous all over $T$, which means that ${\pi }_j\circ f$ is continuous.

Step 2:

Let us suppose that each ${\pi }_j\circ f$ is continuous.

Let $t\in T$ be any.

For each open neighborhood of $f(t)\in T_1\times ...\times T_n$, there is a $U_{{\pi }_1\circ f(t)}\times ...\times U_{{\pi }_n\circ f(t)}\subseteq T_1\times ...\times T_n$ contained in the neighborhood, where $U_{{\pi }_j\circ f(t)}\subseteq T_j$ is an open neighborhood of ${\pi }_j\circ f(t)$.

As ${\pi }_j\circ f$ is continuous, there is an open neighborhood of $t$, $U_{t,j}\subseteq T$, such that ${\pi }_j\circ f(U_{t,j})\subseteq U_{{\pi }_j\circ f(t)}$.

Let us think of ${\cap }_l{U}_{t,l}\subseteq T$, which is an open neighborhood of $t$ as the finite intersection of open neighborhoods.

$f({\cap }_l{U}_{t,l})\subseteq U_{{\pi }_1\circ f(t)}\times ...\times U_{{\pi }_n\circ f(t)}$, because ${\pi }_j\circ f({\cap }_l{U}_{t,l})\subseteq {\pi }_j\circ f(U_{t,j})\subseteq U_{{\pi }_j\circ f(t)}$, which means that $f$ is continuous at $t$.

As $t\in T$ is arbitrary, $f$ is continuous all over $T$, which means that $f$ is continuous.

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References

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