description/proof of that map from topological space into finite product topological space is continuous iff all component maps are continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of continuous map.
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.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\{T_1, ..., T_n\}\): \(T_j \in \{\text{ the topological spaces }\}\)
\(T_1 \times ... \times T_n\): \(= \text{ the product topological space }\)
\(f\): \(: T \to T_1 \times ... \times T_n\)
\(\{\pi_1, ..., \pi_n\}\): \(\pi_j: T_1 \times ... \times T_n \to T_j = \text{ the projection }\)
\(\{\pi_1 \circ f, ..., \pi_n \circ f\}\):
//
Statements:
\(f \in \{\text{ the continuous maps }\}\)
\(\iff\)
\(\forall j \in \{1, ..., n\} (\pi_j \circ f \in \text{ the continuous maps })\)
//
2: Note
The product's being finite is crucial for this logic.
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.
