$$ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Self-defined math definitions %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Math symbol commands \newcommand{\intd}{\,{\rm d}} % Symbol 'd' used in integration, such as 'dx' \newcommand{\diff}{{\rm d}} % Symbol 'd' used in differentiation \newcommand{\Diff}{{\rm D}} % Symbol 'D' used in differentiation \newcommand{\pdiff}{\partial} % Partial derivative \newcommand{DD}[2]{\frac{\diff}{\diff #2}\left( #1 \right)} \newcommand{Dd}[2]{\frac{\diff #1}{\diff #2}} \newcommand{PD}[2]{\frac{\pdiff}{\pdiff #2}\left( #1 \right)} \newcommand{Pd}[2]{\frac{\pdiff #1}{\pdiff #2}} \newcommand{\rme}{{\rm e}} % Exponential e \newcommand{\rmi}{{\rm i}} % Imaginary unit i \newcommand{\rmj}{{\rm j}} % Imaginary unit j \newcommand{\vect}[1]{\boldsymbol{#1}} % Vector typeset in bold and italic \newcommand{\phs}[1]{\dot{#1}} % Scalar phasor \newcommand{\phsvect}[1]{\boldsymbol{\dot{#1}}} % Vector phasor \newcommand{\normvect}{\vect{n}} % Normal vector: n \newcommand{\dform}[1]{\overset{\rightharpoonup}{\boldsymbol{#1}}} % Vector for differential form \newcommand{\cochain}[1]{\overset{\rightharpoonup}{#1}} % Vector for cochain \newcommand{\bigabs}[1]{\bigg\lvert#1\bigg\rvert} % Absolute value (single big vertical bar) \newcommand{\Abs}[1]{\big\lvert#1\big\rvert} % Absolute value (single big vertical bar) \newcommand{\abs}[1]{\lvert#1\rvert} % Absolute value (single vertical bar) \newcommand{\bignorm}[1]{\bigg\lVert#1\bigg\rVert} % Norm (double big vertical bar) \newcommand{\Norm}[1]{\big\lVert#1\big\rVert} % Norm (double big vertical bar) \newcommand{\norm}[1]{\lVert#1\rVert} % Norm (double vertical bar) \newcommand{\ouset}[3]{\overset{#3}{\underset{#2}{#1}}} % over and under set % Super/subscript for column index of a matrix, which is used in tensor analysis. \newcommand{\cscript}[1]{\;\; #1} % Star symbol used as prefix in front of a paragraph with no indent \newcommand{\prefstar}{\noindent$\ast$ } % Big vertical line restricting the function. % Example: $u(x)\restrict_{\Omega_0}$ \newcommand{\restrict}{\big\vert} % Math operators which are typeset in Roman font \DeclareMathOperator{\sgn}{sgn} % Sign function \DeclareMathOperator{\erf}{erf} % Error function \DeclareMathOperator{\Bd}{Bd} % Boundary of a set, used in topology \DeclareMathOperator{\Int}{Int} % Interior of a set, used in topology \DeclareMathOperator{\rank}{rank} % Rank of a matrix \DeclareMathOperator{\divergence}{div} % Curl \DeclareMathOperator{\curl}{curl} % Curl \DeclareMathOperator{\grad}{grad} % Gradient \DeclareMathOperator{\tr}{tr} % Trace \DeclareMathOperator{\span}{span} % Span $$

止于至善

As regards numerical analysis and mathematical electromagnetism

James Munkres Topology: Theorem 16.3

Theorem 16.3 If \(A\) is a subspace of \(X\) and \(B\) is a subspace of \(Y\), then the product topology on \(A \times B\) is the same as the topology \(A \times B\) inherits as a subspace of \(X \times Y\).

Comment: To prove the identity of two topologies, we can either show they mutually contain each other or prove the equivalence of their bases. Because a topological basis has smaller number of elements or cardinality than the corresponding topology, proof via basis is more efficient.

Proof: Let \(\mathcal{C}\) be the topological basis of \(X\) and \(\mathcal{D}\) be the basis of \(Y\). Because \(A \subset X\) and \(B \subset Y\), the subspace topological bases of them are \(\mathcal{B}_A = \{C \cap A \vert \forall C \in \mathcal{C} \}\) and \(\mathcal{B}_B = \{D \cap B \vert \forall D \in \mathcal{D} \}\) respectively according to Lemma 16.1.

Due to Lemma 15.1, the basis of the product topology on \(A \times B\) is

\[
\mathcal{B}_{A \times B} = \{ (C \cap A) \times (D \cap B) \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \}.
\]

Meanwhile, the basis of the product topology on \(X \times Y\) is

\[
\mathcal{B}_{X \times Y} = \{ C \times D \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \}.
\]

Restricting \(\mathcal{B}_{X \times Y}\) to the subset \(A \times B\), the basis of the subspace topology on \(A \times B\) is

\[
\begin{aligned}
\tilde{\mathcal{B}}_{A \times B} &= \{ (C \times D) \cap (A \times B) \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \} \\
&= \{ (C \cap A) \times (D \cap B) \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \},
\end{aligned}
\]

which is the same as that of the product topology on \(A \times B\). Hence, this theorem is proved.

The above process of proof can be illustrated as below.

Remark: The above two routes for generating topology on \(A \times B\) must lead to the same result, otherwise, the theory itself is inappropriately proposed. A theory must be at least self-consistent before its debut in reality.

 

posted @ 2018-12-13 23:18  皮波迪博士  阅读(352)  评论(0编辑  收藏  举报