Metric Space¶
Basic Concepts¶
度量空间的定义 | Definition of Metric Space
Assume \(M\) is a set. If there is a function \(\rho: M\times M\mapsto \mathbb{R}\) defined on it, satisfies the following axioms. That is, \(\forall x,y,z \in M\)
- Positive
- Definite
- Symmetry
- Triangle inequality
then \(M\) is called metric space with metrc \(\rho\), denoted as \((M,\rho)\).
Example. Cases for \((M,\rho)\) with background of Mathematical analysis, show that the following \(\rho\) satisfies axiom of triangle inequation.
(i) \(M=\mathbb{R}^n\), \(\forall x=(\xi_1,\cdots,\xi_2), y=(\eta_1,\cdots,\eta_n)\in \mathbb{R}\), their metric is defined by
(ii) \(M=C[a,b]\), \(\forall f,g\in C[a,b]\), their metric is defined by
(i) Use Cauchy inequation.
Example. Cases for \((M,\rho)\) regarding sequence space, show that the following \(\rho\) satisfies axiom of triangle inequation.
(i) \(M=l^p(1\leq p<\infty)\), i.e.
\(\forall x=\{\xi_n\}_{n\geq 1}, y=\{\eta_n\}_{n\geq 1}\in l^p\), their metric is defined by
(ii) \(M=l^\infty\), or
\(\forall x=\{\xi_n\}_{n\geq 1}, y=\{\eta_n\}_{n\geq 1}\in l^\infty\), their metric is defined by
Example. Cases for \((M,\rho)\) with background of Real analysis, to be more specific, \(L^p\) Space, show that the following \(\rho\) satisfies axiom of triangle inequation. Note that we use "\(=\)" to denote equation almost everywhere.
(i) \(M=L^p(E), (1\leq p<\infty, E\subset \mathbb{R} \text{ is measurable})\), i.e.
\(\forall f, g \in L^p(E)\), their metric is defined by
(ii) \(M=L^\infty(E)\), or
\(\forall f,g\in L^\infty(E)\), their metric is defined by
(i)
Convergence¶
Readers could compare this definition to convergence in \(L^p\) Space.
Definitions of Convergence
Assume \(\{x_n\}_{n\geq 1}\subset M\). If there exists \(x_0\in M\), such that \(\rho(x_n,x_0)\rightarrow 0(n\rightarrow \infty)\), then we call \(\{x_n\}\) converges to \(x_0\), denoted by
and \(x_0\) is a limit of \(\{x_n\}\).
Properties of convergence
Assume \(\{x_n\}_{n\geq 1}\subset M\) is a convergent sequence, then
(i) Limit of \(\{x_n\}_{n\geq 1}\) exists uniquely.
(ii) \(\forall y_0\in M\), \(\{\rho(x_n,y_0)\}_{n\geq 1}\) is bounded.
(iii) Subsequence. for all subsequence \(\{x_{n_j}\}\) converges to the same limit. Conversely, if for all subsequence \(\{x_{n_j}\}\) converges, then \(\{x_n\}\) also converges.
(i) Assume there are two distinct limits \(x_0,y_0\in M\),
(ii) easy to see.
(iii) The forward direction is easy. But reversely, if \(\{x_n\}\) does not converge, then there exist at least two distinct subsequential limits \(L_1\) and \(L_2\), \(L_1\neq L_2\), so which contradicts the assumption!
Topology¶
Definitions
Assume metric space \((M,\rho)\).
(i) If \(x_0\in M\), \(r>0\), Define open ball (neighborhood)
and closed ball
(ii) Assume \(G\subset M\), \(x\in G\), \(x\) is an interior point of \(G\), if \(\exists\) neighborhood \(B(x,r)\subset G\). The kernel of a set \( G \) is a set made up of all its interior points, denoted as \(G^o\). \(G\) is an open set if
Define \(\varnothing\) is an open set.
Properties of open sets
(i) \(M\) is an open set.
(ii) \(\{X_\lambda\}_{\lambda\in \Lambda}\), \(\bigcup_{\lambda\in \Lambda} X_\lambda\) is an open set.
(iii) \(\{X_n\}_{1\leq n\leq N}\), \(\bigcap_{n=1}^N X_n\) is an open set.
Definitions 2
Assume metric space \((M,\rho)\). \(G\subset M\), \(x_0\in M\).
(i) If \(\forall \varepsilon>0\),
then \(G\) is an accumulation point.
(ii) The Derived Set of \( G \) is defined as the set composed of all accumulation points of \( G \), denoted as \( G' \).
(iii) The Closure of \( G \) is the union of \( G \) and its derived set, denoted as:
Properties of Closure
Assume metric space \((M,\rho)\), \(A,B\subset X\), then
(i) \(A\subset \overline{A}\), (ii) \(\overline{\overline{A}}=\overline{A}\), (iii) \(\overline{A\cup B}=\overline{A}\cup \overline{B}\), (iv) \(\overline{\varnothing}=\varnothing\).