description/proof of that for product topological space, projection is continuous
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description 1
- 2: Proof 1
- 3: Structured Description 2
- 4: Proof 2
Starting Context
- The reader knows a definition of product topological space.
- The reader knows a definition of continuous map.
Target Context
- The reader will have a description and a proof of the proposition that for any product topological space, any projection 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 1
Here is the rules of Structured Description.
Entities:
\(J'\): \(\in \{\text{ the possibly uncountable infinite index sets }\}\)
\(\{T_j \vert j \in J'\}\): \(T_j \in \{\text{ the topological spaces }\}\)
\(T'\): \(= \times_{j \in J'} T_j\) with the product topology
\(J\): \(\subseteq J'\)
\(T\): \(= \times_{j \in J} T_j\) with the product topology
\(f\): \(: T' \to T, t' \mapsto (j \mapsto t' (j))\), \(= \text{ the projection }\)
//
Statements:
\(f \in \{\text{ the continuous maps }\}\)
//
2: Proof 1
Whole Strategy: Step 1: let \(t' \in T'\) be any and take any neighborhood of \(f (t')\), \(N_{f (t')}\); Step 2: take any basic open neighborhood of \(f (t')\), \(\times_{j \in J} U_j\), such that \(f (t') \in \times_{j \in J} U_j \subseteq N_{f (t')}\); Step 3: for each \(j' \in J' \setminus J\), take \(U_{j'} = T_{j'}\), and take \(U_{t'} := \times_{j' \in J'} U_{j'}\), and see that \(U_{t'}\) is an open neighborhood of \(t'\) on \(T'\) and \(f (U_{t'}) \subseteq N_{f (t')}\).
Step 1:
Let \(t' \in T'\) be any.
Let any neighborhood of \(f (t')\) on \(T\) be \(N_{f (t')} \subseteq T\).
Step 2:
There is a basic open neighborhood of \(f (t')\), \(\times_{j \in J} U_j\), such that \(f (t') \in \times_{j \in J} U_j \subseteq N_{f (t')}\), where \(U_j\) is an open neighborhood of \(f (t') (j)\) on \(T_j\) where only some finite number of \(U_j\) s are not \(T_j\), by the definition of product topology.
Step 3:
For each \(j' \in J' \setminus J\), let us take \(U_{j'} = T_{j'}\).
Let us take \(U_{t'} := \times_{j' \in J'} U_{j'} \subseteq T'\).
\(t' \in U_{t'}\), because for each \(j \in J\), \(t' (j) = f (t') (j) \in U_j\), and for each \(j' \in J' \setminus J\), \(t' (j') \in T_{j'} = U_{j'}\).
\(U_{t'}\) is open on \(T'\), by the definition of product topology: only some finite number of \(U_{j'}\) s are not \(T_{j'}\) s.
So, \(U_{t'}\) is an open neighborhood of \(t'\) on \(T'\).
\(f (U_{t'}) \subseteq \times_{j \in J} U_j\) (in fact, \(=\) holds, but \(\subseteq\) is enough for our purpose), because for each \(t'' \in U_{t'}\), for each \(j \in J\), \(f (t'') (j) = t'' (j) \subseteq U_j\).
So, \(f (U_{t'}) \subseteq \times_{j \in J} U_j \subseteq N_{f (t')}\).
3: Structured Description 2
Here is the rules of Structured Description.
Entities:
\(J'\): \(= \{1, ..., n\}\)
\(\{T_1, ..., T_n\}\): \(T_j \in \{\text{ the topological spaces }\}\)
\(T'\): \(= T_1 \times ... \times T_n\) with the product topology
\(J\): \(\subseteq J'\), \(= \{j_1, ..., j_m\}\)
\(T\): \(= T_{j_1} \times ... \times T_{j_m}\) with the product topology
\(f\): \(: T' \to T, (t'^1, ..., t'^n) \mapsto (t'^{j_1}, ..., t'^{j_m})\), \(= \text{ the projection }\)
//
Statements:
\(f \in \{\text{ the continuous maps }\}\)
//
4: Proof 2
Whole Strategy: Step 1: let \(t' \in T'\) be any and take any neighborhood of \(f (t')\), \(N_{f (t')}\); Step 2: take any basic open neighborhood of \(f (t')\), \(U_{j_1} \times ... \times U_{j_m}\), such that \(f (t') \in U_{j_1} \times ... \times U_{j_m} \subseteq N_{f (t')}\); Step 3: for each \(j' \in J' \setminus J\), take \(U_{j'} = T_{j'}\), and take \(U_{t'} := U_1 \times ... \times U_n\), and see that \(U_{t'}\) is an open neighborhood of \(t'\) on \(T'\) and \(f (U_{t'}) \subseteq N_{f (t')}\).
Step 1:
Let \(t' \in T'\) be any.
Let any neighborhood of \(f (t')\) on \(T\) be \(N_{f (t')} \subseteq T\).
Step 2:
There is a basic open neighborhood of \(f (t')\), \(U_{j_1} \times ... \times U_{j_m}\), such that \(f (t') \in U_{j_1} \times ... \times U_{j_m} \subseteq N_{f (t')}\), where \(U_{j_l}\) is an open neighborhood of \((f (t'))^{j_l}\) on \(T_{j_l}\), by the definition of product topology.
Step 3:
For each \(j' \in J' \setminus J\), let us take \(U_{j'} = T_{j'}\).
Let us take \(U_{t'} := U_1 \times ... \times U_n \subseteq T'\).
\(t' \in U_{t'}\), because for each \(j \in J\), \(t'^j = (f (t'))^j \in U_j\), and for each \(j' \in J' \setminus J\), \(t'^{j'} \in T_{j'} = U_{j'}\).
\(U_{t'}\) is open on \(T'\), by the definition of product topology.
So, \(U_{t'}\) is an open neighborhood of \(t'\) on \(T'\).
\(f (U_{t'}) \subseteq U_{j_1} \times ... \times U_{j_m}\) (in fact, \(=\) holds, but \(\subseteq\) is enough for our purpose), because for each \(t'' \in U_{t'}\), for each \(j \in J\), \((f (t''))^j = t''^j \subseteq U_j\).
So, \(f (U_{t'}) \subseteq U_{j_1} \times ... \times U_{j_m} \subseteq N_{f (t')}\).
