description/proof of that map from topological space into product topological space is continuous iff each component map is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of projection from product set onto subproduct set.
- The reader admits the proposition that for any product topological space, any projection is continuous.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
Target Context
- The reader will have a description and a proof of the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_{2, j} \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_{2, j}\): \(= \text{ the product topological space }\)
\(f\): \(: T_1 \to \times_{j \in J} T_{2, j}\)
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Statements:
\(f \in \{\text{ the continuous maps }\}\)
\(\iff\)
\(\forall j \in J (\pi^j \circ f: T_1 \to T_j \in \{\text{ the continuous maps }\})\), where \(\pi^j: \times_{l \in J} T_{2, l} \to T_{2, j}\) is the projection
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2: Note
While 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 has already been proved, in fact, the product does not need to be finite, by this proposition.
3: Proof
Whole Strategy: Step 1: suppose that \(f\) is continuous; Step 2: see that \(\pi^j \circ f\) is continuous as a composition of continuous maps; Step 3: suppose that \(\pi^j \circ f\) is continuous; Step 4: see that \(f\) is continuous by seeing that \(f\) is continuous at each \(t \in T_1\).
Step 1:
Let us suppose that \(f\) is continuous.
Step 2:
For each \(j \in J\), \(\pi^j\) is continuous, by the proposition that for any product topological space, any projection is continuous.
\(\pi^j \circ f\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Step 3:
Let us suppose that for each \(j \in J\), \(\pi^j \circ f\) is continuous.
Step 4:
Let \(t \in T_1\) be any.
Let us see that \(f\) is continuous at \(t\).
Let \(U_{f (t)} \subseteq \times_{j \in J} T_{2, j}\) be any open neighborhood of \(f (t)\).
There are a finite \(J^` \subseteq J\) and an open neighborhood of \(f (t)^j\), \(U_{f (t)^j} \subseteq T_{2, j}\), for each \(j \in J^`\) and \(U_{f (t)^j} := T_{2, j}\) for each \(j \in J \setminus J^`\) such that \(\times_{j \in J} U_{f (t)^j} \subseteq \times_{j \in J} T_{2, j}\) is an open neighborhood of \(f (t)\) such that \(\times_{j \in J} U_{f (t)^j} \subseteq U_{f (t)}\), by the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
For each \(j \in J^`\), there is an open neighborhood of \(t\), \(U_{t, j} \subseteq T_1\), such that \(\pi^j \circ f (U_{t, j}) \subseteq U_{f (t)^j}\), because \(\pi^j \circ f\) is continuous.
Let us take \(U_t := \cap_{j \in J^`} U_{t, j}\), which is an open neighborhood of \(t\) as a finite intersection of open neighborhoods.
Then, \(f (U_t) \subseteq \times_{j \in J} U_{f (t)^j} \subseteq U_{f (t)}\), because for each \(t' \in U_t\), for each \(j \in J^`\), \(f (t')^j = \pi^j \circ f (t') \in U_{f (t)^j}\), because \(t' \in U_t \subseteq U_{t, j}\), and for each \(j \in J \setminus J^`\), \(f (t')^j = \pi^j \circ f (t') \in T_{2, j} = U_{f (t)^j}\).
So, \(f\) is continuous at \(t\).
So, \(f\) is continuous.
