Metric Space

Important Equations

Lemma

Assume positive real number \(\alpha\), \(\beta\) satisfies conjugate condition \(\alpha+\beta=1\), then

\[ u^\alpha v^\beta\leq u\cdot\alpha+v\cdot\beta. \]

Proof

Since \(f(x)=x^\alpha (0<\alpha<1)\) is concave function, so tangent function at any point on the function curve lies above it. Choose \((1,1)\), we have tangent function

\[ y=\alpha(x-1)+1=\alpha x+1-\alpha=\alpha x+\beta \]

with \(\beta=1-\alpha\in (0,1)\). So

\[ x^\alpha\leq \alpha x + \beta. \]

Let \(x=\frac{u}{v}\), then

\[ \begin{align*} \left(\frac{u}{v}\right)^\alpha &\leq \alpha \frac{u}{v} +\beta\\ u^\alpha\cdot v^{-\alpha}&\leq \alpha u \cdot v^{-1}+\beta\\ u^\alpha\cdot v^{1-\alpha}&\leq \alpha u +\beta v\\ u^\alpha\cdot v^{\beta}&\leq \alpha u +\beta v. \end{align*} \]

If we let \(\alpha=\frac{1}{p}\), \(\beta=\frac{1}{q}\) in inequation, we have

\[ \begin{align} u^{\frac{1}{p}}\cdot v^{\frac{1}{q}}\leq \frac{1}{p}u+\frac{1}{q}v, \quad \frac{1}{p}+\frac{1}{q}=1.\label{conjugate-inequation} \end{align} \]

Hölder Inequation

Assume \(p\), \(q\) satisfies conjugate condition \(\frac{1}{p}+\frac{1}{q}=1\), then

(i) For \(\{\xi_n\}_{n\geq 1}\) and \(\{\eta_n\}_{n\geq 1}\), \(\xi_n,\eta_n\in \mathbb{C}\) we have

\[ \sum_{n=1}^\infty |\xi_n\eta_n|\leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\cdot \left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}. \]

(ii) For \(f\), \(g\) satisfies \(\int_E |f(x)|^pdx<\infty\), \(\int_E |g(x)|^qdx<\infty\), we have

\[ \int_E |f(x)g(x)|dx\leq \left(\int_E |f(x)|^pdx\right)^{\frac{1}{p}}\cdot \left(\int_E |g(x)|^qdx\right)^{\frac{1}{g}}. \]

Proof for (i)Proof for (ii)

In inequation \(\ref{conjugate-inequation}\), for \(n=1,2,\cdots\) let

\[ u=\frac{|\xi_n|^p}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)}, \quad v=\frac{|\eta_n|^q}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)}, \]

we have

\[ \frac{|\xi_n|}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}}\cdot \frac{|\eta_n|}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}}\leq \frac{1}{p}\frac{|\xi_n|^p}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)}+\frac{1}{q}\frac{|\eta_n|^q}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)} \]

summing up \(n=1,2,\cdots\), we have

\[ \begin{align*} \frac{\sum_{n=1}^\infty|\xi_n\eta_n|}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}} &\leq \frac{1}{p}\frac{\sum_{n=1}^\infty|\xi_n|^p}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)}+\frac{1}{q}\frac{\sum_{n=1}^\infty|\eta_n|^q}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)}\\ \frac{\sum_{n=1}^\infty|\xi_n\eta_n|}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}} &\leq \frac{1}{p}+\frac{1}{q}=1\\ \Rightarrow\quad \sum_{n=1}^\infty|\xi_n\eta_n|& \leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}. \end{align*} \]

In inequation \(\ref{conjugate-inequation}\), let

\[ u=\frac{|f(x)|^p}{\left(\int_E |f(x)|^p dx\right)}, \quad v=\frac{|g(x)|^q}{\left(\int_E |g(x)|^q dx\right)}, \]

we have

\[ \frac{|f(x)|}{\left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}}\cdot \frac{|g(x)|}{\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}}\leq \frac{1}{p}\frac{|f(x)|^p}{\left(\int_E |f(x)|^p dx\right)}+\frac{1}{q}\frac{|g(x)|^q}{\left(\int_E |g(x)|^q dx\right)} \]

integrate on \(E\) on both sides, we have

\[ \begin{align*} \frac{\int_E |f(x)g(x)|dx}{\left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}}&\leq \frac{1}{p}\frac{\int_E |f(x)|^pdx}{\left(\int_E |f(x)|^p dx\right)}+\frac{1}{q}\frac{\int_E |g(x)|^qdx}{\left(\int_E |g(x)|^q dx\right)}\\ \frac{\int_E |f(x)g(x)|dx}{\left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}}&\leq \frac{1}{p}+\frac{1}{q}=1\\ \Rightarrow \quad \int_E |f(x)g(x)|dx &\leq \left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}. \end{align*} \]

Minkowski Inequation

Assume \(p\), \(q\) satisfies conjugate condition \(\frac{1}{p}+\frac{1}{q}=1\), then

(i) For \(\{\xi_n\}_{n\geq 1}\) and \(\{\eta_n\}_{n\geq 1}\), \(\xi_n,\eta_n\in \mathbb{C}\), we have

\[ \left(\sum_{n=1}^\infty |\xi_n+\eta_n|^p\right)^{\frac{1}{p}} \leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}} + \left(\sum_{n=1}^\infty |\eta_n|^p\right)^{\frac{1}{p}} \]

(ii) For \(f\), \(g\) satisfies \(\int_E |f(x)|^pdx<\infty\), \(\int_E |g(x)|^qdx<\infty\), we have

\[ \left(\int_E |f(x)+g(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_E |f(x)|^pdx\right)^{\frac{1}{p}}+\left(\int_E |g(x)|^pdx\right)^{\frac{1}{p}} \]

Proof for (i)Proof for (ii)

\[ \begin{align*} \left(\sum_{n=1}^\infty |\xi_n+\eta_n|^p\right)&=\sum_{n=1}^\infty \left( |\xi_n+\eta_n|^{p-1}\cdot |\xi_n+\eta_n|\right)\\ &\leq \sum_{n=1}^\infty \left[|\xi_n+\eta_n|^{\frac{p}{q}}\cdot \left(|\xi_n|+|\eta_n|\right)\right]\\ &=\sum_{n=1}^\infty \left( |\xi_n+\eta_n|^{\frac{p}{q}}\cdot |\xi_n|\right) + \sum_{n=1}^\infty \left( |\xi_n+\eta_n|^{\frac{p}{q}}\cdot |\eta_n|\right)\\ &\leq \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{\frac{p}{q}\cdot q} \right)^\frac{1}{q}\cdot \left( \sum_{n=1}^\infty |\xi_n|^p\right)^\frac{1}{p} + \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{\frac{p}{q}\cdot q} \right)^\frac{1}{q}\cdot \left(\sum_{n=1}^\infty |\eta_n|^p\right)^\frac{1}{p}\\ \Rightarrow \quad \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{\frac{p}{q}\cdot q} \right)^{1-\frac{1}{q}} &\leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^\frac{1}{p}+\left(\sum_{n=1}^\infty |\eta_n|^p\right)^\frac{1}{p}\\ \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{p}\right)^{\frac{1}{p}} &\leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^\frac{1}{p}+\left(\sum_{n=1}^\infty |\eta_n|^p\right)^\frac{1}{p}\\ \end{align*} \]

\[ \begin{align*} \left(\int_E |f(x)+g(x)|^p dx\right)&=\int_E \left( |f(x)+g(x)|^{p-1}\cdot |f(x)+g(x)|\right) dx\\ &\leq \int_E \left[|f(x)+g(x)|^{\frac{p}{q}}\cdot \left(|f(x)|+|g(x)|\right)\right] dx\\ &=\int_E \left( |f(x)+g(x)|^{\frac{p}{q}}\cdot |f(x)|\right) dx + \int_E \left( |f(x)+g(x)|^{\frac{p}{q}}\cdot |g(x)|\right) dx\\ &\leq \left( \int_E |f(x)+g(x)|^{\frac{p}{q}\cdot q} dx\right)^\frac{1}{q}\cdot \left( \int_E |f(x)|^pdx\right)^\frac{1}{p} + \left( \int_E |f(x)+g(x)|^{\frac{p}{q}\cdot q} dx\right)^\frac{1}{q}\cdot \left(\int_E |g(x)|^pdx\right)^\frac{1}{p}\\ \Rightarrow \quad \left( \int_E |f(x)+g(x)|^{\frac{p}{q}\cdot q} dx\right)^{1-\frac{1}{q}} &\leq \left(\int_E |f(x)|^pdx\right)^\frac{1}{p}+\left(\int_E |g(x)|^pdx\right)^\frac{1}{p}\\ \left( \int_E |f(x)+g(x)|^{p}dx\right)^{\frac{1}{p}} &\leq \left(\int_E |f(x)|^pdx\right)^\frac{1}{p}+\left(\int_E |g(x)|^pdx\right)^\frac{1}{p}\\ \end{align*} \]

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

\[ \rho(x,y)>0,\quad x\neq y \]

• Definite

\[ \rho(x, y)=0 \Leftrightarrow x=y \]

• Symmetry

\[ \rho(x,y) = \rho(y,x) \]

• Triangle inequality

\[ \rho(x,z) \leq \rho(x,y) +\rho(y,z) \]

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

\[ \rho(x,y)=\left(\sum_{i=1}^n|\xi_i-\eta_i|^2\right)^\frac{1}{2} \]

(ii) \(M=C[a,b]\), \(\forall f,g\in C[a,b]\), their metric is defined by

\[ \rho(f,g)=\max_{t\in[a,b]}|f(t)-g(t)|. \]

Proof

(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.

\[ l^p=\left\{ \{\xi_n\}_{n\geq 1}: \sum_{n=1}^\infty |\xi_n|^p<\infty. \right\} \]

\(\forall x=\{\xi_n\}_{n\geq 1}, y=\{\eta_n\}_{n\geq 1}\in l^p\), their metric is defined by

\[ \rho(x,y)=\left(\sum_{n=1}^\infty |\xi_n-\eta_n|^p\right)^{\frac{1}{p}}. \]

(ii) \(M=l^\infty\), or

\[ l^\infty:=\{\{\xi_n\}_{n\geq 1}: \exists M>0, \text{ s.t } |\xi_n|\leq M,\forall n.\} \]

\(\forall x=\{\xi_n\}_{n\geq 1}, y=\{\eta_n\}_{n\geq 1}\in l^\infty\), their metric is defined by

\[ \rho(x,y)=\sup_{n}|\xi_n-\eta_n|. \]

Proof

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.

\[ L^p=\left\{ \text{measurable } f : \int_E |f(t)|^pdt<\infty. \right\} \]

\(\forall f, g \in L^p(E)\), their metric is defined by

\[ \rho(f,g)=\left( \int_E |f(t)-g(t)|^p dt\right)^{\frac{1}{p}}. \]

(ii) \(M=L^\infty(E)\), or

\[ L^\infty:=\{\text{measurable } f : \exists M>0, \exists E_0\subset E, \text{ s.t } m(E_0)=0, \text{ & } |f(t)|\leq M,\forall t\in E-E_0.\} \]

\(\forall f,g\in L^\infty(E)\), their metric is defined by

\[ \rho(f,g)=\inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|f(t)-g(t)|\right\}. \]

Proof

(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

\[ \lim_{n\rightarrow \infty}x_n=x_0 \]

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.

Proof

(i) Assume there are two distinct limits \(x_0,y_0\in M\),

\[ \rho(x_0,y_0)<\rho(x_0,x_n)+\rho(x_n,y_0) \rightarrow 0(n\rightarrow \infty). \]

(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)

\[ B(x_0, r)=\{x: \rho(x_0,x)<r\}, \]

and closed ball

\[ \overline{B}(x_0, r)=\{x: \rho(x_0,x)\leq r\}. \]

(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

\[ \forall x \in G, x \text{ is an interior point of } G. \]

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 \(\forall \varepsilon>0\),

\[ B(x_0, \varepsilon)\cap (G-\{x_0\})\neq \varnothing, \]

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:

\[ \overline{G} = G\cup G'. \]

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) \(\overline{\varnothing}=\varnothing\).

Density & Seperability

Definition of Density

Assume \(X\) is a metric space, \(A,B\subset X\). If \(\overline{B}\supset A\), then \(B\) is dense in \(A\).

Equivalent proposition of Density

Assume \(X\) is a metric space, \(A,B\subset X\). Then the following statements are equivalent.

(i) \(B\) is dense in \(A\),

(ii) \(\forall x\in A\), \(\forall \varepsilon>0\), \(\exists y\in B\), such that \(x\in B(y,\varepsilon)\) (or \(\rho(x,y)<\varepsilon\)).

(iii) \(\forall x\in A\), \(\exists \{x_n\}\in B\), such that \(\lim\limits_{n\rightarrow\infty}x_n=x\).

(iv) \(\forall \varepsilon>0\), \(A\subset \bigcup\limits_{x\in B}B(x, \varepsilon)\).

Proof

• (i) \(\Rightarrow\) (ii)

\(\forall x\in A\), since \(A\subset \overline{B}\), then \(x\in \overline{B}\), so \(\forall \varepsilon>0\), \(B(x,\varepsilon)\cap B\neq\varnothing\). So \(\exists y\in B(x,\varepsilon)\cap B\), which satisfies \(\rho(x,y)<\varepsilon\).

• (ii) \(\Rightarrow\) (iii)

\(\forall x\in A\), \(\forall \varepsilon_k=\frac{1}{k}>0\), \(\exists y_k\in B\), such that \(\rho(x,y_k)<\varepsilon_k\). So \(\lim\limits_{k\rightarrow \infty}y_k=x\).

• (iii) \(\Rightarrow\) (iv)

\(\forall x\in A\), \(\exists \{x_n\}\in B\), such that \(\lim\limits_{n\rightarrow\infty}x_n=x\). So \(\forall \varepsilon>0\), \(\exists N>0\), \(\rho(x,x_N)<\varepsilon\), which means \(x\in B(x_N,\varepsilon)\). So \(A\subset \bigcup\limits_{x\in B}B(x, \varepsilon)\).

• (iv) \(\Rightarrow\) (i)

\(\forall \varepsilon>0\), \(A\subset \bigcup\limits_{x\in B}B(x, \varepsilon)\), so \(\forall x\in A\), \(\exists y\in B\), \(x\in B(y,\varepsilon)\), which means \(B(x,\varepsilon)\cap B=\neq \varnothing\) (at least we have \(y\)), so \(x\in \overline{B}\).

Definition of Seperability

Assume \(X\) is a metric space. If there exists denumerable set \(B\subset X\), \(B\) is dense in \(X\), then \(X\) is seperable.

Continuous Mapping

Mapping & Continuous Mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If \(\forall x\in X\), \(\exists ! y\in Y\), such that \(Tx=y\), then we call \(T\) is a mapping from \(X\) to \(Y\).

If mapping \(T\) satisfies that for a fixed poinr \(x_0\in X\), \(\forall \varepsilon>0\), \(\exists \delta>0\), such that \(\rho_1(Tx,Tx_0)<\varepsilon\) whenever \(\rho(x,x_0)<\delta\), then we call \(T\) is a continuous mapping at \(x_0\).

If the above property holds for all \(x_0\in X\), then we call \(T\) continuous mapping on \(X\).

Usually we call a mapping from metric space \(X\) to \(\mathbb{R}\) functional.

Example. (i) Assume \(X\) is a metric space, \(x_0\in X\), mapping \(T(x)=\rho(x,x_0), (x\in X)\) is a continuous functional from \(X\) to \(\mathbb{R}\).

(ii) Assume \(X\) is a metric space, \(F\subset X\), mapping \(T(x)=\inf\limits_{y\in F}\rho(x,y), (x\in X)\) is continuous functional from \(X\) to \(\mathbb{R}\).

Proof for (i)Proof for (ii)

\(\square\)

\(\square\)

Equivalent definition for continuous mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively.

(i) Choose a point \(x_0\in X\). A mapping \(T\) from \(X\) to \(Y\) is continuous at \(x_0\), iff

\[ \forall \{x_n\}_{n\geq 1}\subset X, \lim\limits_{n\rightarrow \infty}x_n=x, s.t. \lim\limits_{n\rightarrow \infty}Tx_n=Tx. \]

(ii) A mapping \(T\) from \(X\) to \(Y\) is continuous, iff one of the following two condition holds

\[ \forall \text{ open set } G\in Y, T^{-1}(G) \text{ is an open set.} \]

\[ \forall \text{ closed set } F\in Y, T^{-1}(F) \text{ is a closed set.} \]

Proof

\(\square\)

Definition of inverse mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If a mapping \(T\) from \(X\) to \(Y\) is injective and surjective, or bijective, then \(\forall y\in Y\), \(\exists ! x\in X\), such that

\[ Tx=y. \]

here we have a new mapping, called Inverse Mapping, deonted as \(T^{-1}\).

Definition of Homeomorphic Mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If a mapping \(T\) from \(X\) to \(Y\) is bijective, and \(T\), \(T^{-1}\) are both continuous on \(X\), \(Y\), respectively, then we call \(T\) is a Homeomorphic Mapping.

If there exists a homeomorphic mapping \(T\) from \(X\) to \(Y\), then we call \(X\) and \(Y\) are homeomorphic.

Definition of Isometric Mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If a mapping \(T\) from \(X\) to \(Y\), satisfies that \(\forall x,y\in X\), \(\rho(x,y)=\rho_1(Tx,Ty)\), then we call \(T\) is an isometric mapping.

If there exists a bijective and isometric mapping \(T\) from \(X\) to \(Y\), then we call \(X\) and \(Y\) are isometric.

When Two metric space are homeomorphic or isometric, we view them as the same.

Completeness

Definition of Basic sequence (Cauchy Sequence)

Assume \(X\) is a metric space with metric \(\rho\). Sequence \(\{x_n\}_{n\geq 1}\subset X\) is said to be a Cauchy Sequence, if \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n,m>N\), such that

\[ \rho(x_n,x_m)<\varepsilon. \]

The core problem of functional analysis, lies in formulating a complete space, i.e. every Cauchy sequence of \(X\) converge to the point in the space \(X\) itself.

Example. Basic space \(\mathbb{R}\) and \(C[a,b]\) are complete.

Meager sets

Definition of nowhere dense sets

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is said to be Nowhere Dense Set, if \(\forall \text{ open set }G\subset X\), \(A\) is not dense in \(G\).

Equivalent definition of nowhere dense sets

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is said to be Nowhere Dense Set, iff

\[ \forall \text{ open ball }S(x_0,r), \exists \text{ open ball }S(y_0,r')\subset S(x_0,r), s.t. A\cap S(y_0,r')=\varnothing. \]

Proof

\(\square\)

Set of the first & second category

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is said to be a set of the first category, if there exists a sequence of nowhere dense sets \(\{F_n\}_{n\geq 1}\), such that

\[ A=\bigcup_{n=1}^\infty F_n. \]

If not, we call \(A\) a set of the second category.

Theorem of Nested Closed Balls

Metric Space \(X\) is complete, iff $\forall $ nested closed ball \(\{K_n=\overline{S}(x_n,r_n)\}_{n\geq 1}\), which satisfies

\[ \overline{S}(x_1,r_1)\supset \overline{S}(x_2,r_2)\supset \cdots, \]

and \(\lim\limits_{n\rightarrow \infty}r_n=0\), then there exists a unique point \(x_0\in X\), such that \(x_0\in \bigcap_{n=1}^\infty \overline{S}(x_n,r_n)\).

Proof

\(\square\)

Baire Theorem

Every complete space is a set of the second category.

Proof

\(\square\)

Compact Sets

Definition of Bounded set

Assume \(X\) is a metric space, and \(A\subset X\). \(A\) is said to be a bounded set, if there exists a open ball \(K=S(x,r)\) (or closed ball \(K=\overline{S}(x,r)\)) such that \(A\subset K\).

Definition for relative compact set & compact set

Assume \(X\) is a metric space, \(A\subset X\). If \(\forall \{x_n\}\subset A\), \(\exists \{n_k\}_{k\geq 1}\) and \(x\in X\), such that \(\lim\limits_{k\rightarrow \infty}x_{n_k}=x\), then we call \(A\) a relative compact set. If the limit point \(x\in A\), then \(A\) is called compact set.

If \(X\) itself is a compact set, then \(X\) is called compact metric space.

Properties for relative compact set & compact set

(i) Any set with finite number elements is compact set.

(ii) Assume \(A\subset X\) is a relative compact set, then \(\forall B\subset A\), \(B\) is also a relative compact set.

(iii) Assume \(A\subset X\) is a compact set, then \(\forall \text{ closed set }B\subset A\), \(B\) is also a compact set.

Proof

(ii) By definition. \(\forall \{x_n\}_{n\geq 1}\subset B\subset A\), so \(\exists \{n_k\}_{k\geq 1}\) and \(x\in X\), such that \(\lim\limits_{k\rightarrow \infty}x_{n_k}=x\), so \(B\) is a relative compact set.

(iii) \(\forall \{x_n\}_{n\geq 1}\subset B\subset A\), \(\exists \{n_k\}_{k\geq 1}\) and \(x\in A\), such that \(\lim\limits_{k\rightarrow \infty}x_{n_k}=x\). Since \(B\) is a closed set, the limit point should be in \(B\), i.e. \(x\in B\). So \(B\) is a compact set.

\(\square\)

Now we give a description for relative compact set.

Definition for \(\varepsilon\)-net

Assume \(X\) is a metric space with metric \(\rho\), \(A,B\subset X\). If for a given number \(\varepsilon>0\), \(\forall x\in A\), \(\exists y\in B\), such that \(\rho(x,y)<\varepsilon\), then we call \(B\) is an \(\varepsilon\)-net of \(A\).

Note we do not ask \(B\cap A\neq \varnothing\).

Definition for Totally Bounded Set

Assume \(X\) is a metric space, \(A\subset X\). If \(\forall \varepsilon>0\), \(\exists \varepsilon\)-net with finite number of elements for \(A\), then we call \(A\) is a Totally Bounded Set.

Properties for totally bounded set

(i) Any set with finite number of elements is a totally bounded set.

(ii) Assume \(A\subset X\) is a totally bounded set, then \(\forall B\subset A\), \(B\) is also a totally bounded set.

(iii) Assume \(A\subset X\) is a totally bounded set, then \(\forall\varepsilon>0\), \(\exists B\subset A\), s.t \(B\) is a \(\varepsilon\)-net of \(A\).

Proof

(i) Choose the set itself as an \(\varepsilon\)-net, which has finite number of elements.

(ii) \(\forall \varepsilon>0\), \(\exists C\) with finite number of elements, such that \(B\subset A\subset \bigcup\limits_{x\in C}S(x,\varepsilon)\), so \(C\) is also an \(\varepsilon\)-net of \(B\).

(iii) \(\forall \varepsilon>0\), \(\exists C=\{y_n\}_{1\leq n\leq N}\) is an \(\frac{\varepsilon}{2}\)-net of \(A\). That is, \(\forall x\in A\), \(\exists y\in C\), such that \(\rho(x,y)<\frac{\varepsilon}{2}\). By assumption, \(\exists \{n_k\}_{1\leq k\leq K} (K\leq N)\), such that \(S(y_{n_k},\frac{\varepsilon}{2})\cap A\neq \varnothing\), then choose \(z_k\in S(y_{n_k},\frac{\varepsilon}{2})\cap A\). Then define \(B=\bigcup_{k=1}^K \{z_k\}\), which satisfies

\[ \forall x\in A, \exists z_k\in B, \rho(x,z_k)<\varepsilon. \]

So \(B\) is a \(\varepsilon\)-net of \(A\).

Theorem for totally bounded set

Totally bounded set is bounded and seperable.

Proof

(i) easy to see it is bounded. Choose one element of an \(\varepsilon\)-net as center of ball, estimate the sufficient radious of the ball.

(ii) Leverage the \(\frac{1}{n}\)-net to formulate a dense denumerable subsequence.

\(\square\)

.

Theorem for relative compact description

Assume \(X\) is a metric space with metric \(\rho\), \(A\subset X\) is a relative compact set, then \(A\) is totally bounded set.

On the other hand, If \(X\) is a complete metric space, if \(A\subset X\) is a totally bounded set, then \(A\) is also a relative compact set.

Proof

The following theorem is useful in proof. Because it is hard to find an \(\varepsilon\)-net which proved to be finite number, but easy to show it is relative compact.

Theorem for relative compact set

Assume \(X\) is a complete metric space, \(A\subset X\) is a relative compact set, iff \(A\) has a relative compact \(\varepsilon\)-net.