多元函数的极限与连续_NKU(答案)¶

练习10.1¶

Sol.

(1)考虑构造一个收敛的点列，\(X_{k}=\left\{ \left( x_{1},x_{2},\dots,x_{n}\right)|\lvert x_{i} \rvert< \frac{1}{k} \right\}\)

那么可以得到\(\bigcap\limits_{k=1}^{\infty}X_{k}=\left\{ 0 \right\}\)，单点集不为开集

(2)考虑构造收敛于区间端点的点列，\(X_{k}=\left\{ (x_{1},x_{2},\dots,x_{n})|d=\sqrt{ \sum\limits_{i=1}^{n}x_{i}^{2} }\leqslant 1- \frac{1}{k}\right\}\)

那么可以得到\(\bigcup\limits_{k=1}^{\infty}X_{k}=\left\{ (x_{1},x_{2},\dots,x_{n})|d<1 \right\}\)，是一个开球(开集)

Proof.

根据聚点的定义知道\(\forall\delta>0,\exists x'\in V(X_{0},\delta)\bigcap A\)，那么构造如下点列：

固定\(r_{1}=1,\exists X_{1}\in V(X_{0},r_{1}=1)\bigcap A,X_{1}\neq X_{0}\)，再固定\(r_{2}=\min\left\{ \frac{1}{2},\lvert X_{1}-X_{0} \rvert \right\},\exists X_{2}\in V(X_{0},r_{2})\bigcap A,X_{2}\neq X_{0}\)

以此类推，\(r_{n}=\min\left\{ \frac{1}{n},\lvert X_{n-1}-X_{0} \rvert \right\},\exists X_{n}\in V(X_{0},r_{n})\bigcap A,X_{n}\neq X_{0}\)

这样就得到了\(\left\{ X_{m} \right\},\lim\limits_{ m \to \infty }\lvert X_{m}-X_{0} \rvert =0,X_{m}\neq X_{0},\lim\limits_{ m \to \infty }X_{m}=X_{0}\)

Proof.

根据定义，若有\(x\in \overline{A}\)，讨论\(x\in A ,x\not\in A\)，如果\(x\in A \implies x\in A\bigcup A'\);\(x\not\in A,x\in A'\implies x\in A\bigcup A'\)，那么有\(\overline{A}\subset A\bigcup A'\)

反方向\(x\in A\)显然有\(x\in \overline{A}\)，若是\(x\in A'\)同样根据定义可得\(A'\subset \overline{A}\),综上可知\(A\bigcup A' \subset \overline{A}\)

由相互包含可知\(\overline{A}=A\bigcup A'\)

Proof.

(1)

若是\(A\subseteq B\implies \forall x\in A^{\circ},\exists V(x,\varepsilon)\subset A\subset B \implies x\in B^{\circ}\implies A^{\circ}\subseteq B^{\circ}\)

同样的\(\forall x\in \overline{A}\)，分类讨论\(x\in A\)显然\(x\in B\implies x\in \overline{B}\)，若\(x\not\in A,x\in A'\)，根据定义有\(V(x,\varepsilon)\setminus \left\{ x \right\}\bigcap A\neq \emptyset\implies V(x,\varepsilon)\setminus \left\{ x \right\}\bigcap B\neq \emptyset \implies x\in \overline{B}\)

所以可得\(\overline{A}\subseteq \overline{B}\)

(2)

\(\forall x\in \overline{A\bigcup B},\forall V(x,\varepsilon)\bigcap\left( A\bigcup B \right)\setminus \left\{ x \right\}\neq \emptyset\)

由此可得\(\left( V(x,\varepsilon)\bigcap A\setminus \left\{ x \right\}\neq \emptyset \right) \bigcup \left( V(x,\varepsilon)\bigcap B\setminus \left\{ x \right\}\neq \emptyset \right)\)，也就是属于A或者B的闭包(\(\overline{A}\bigcup \overline{B}\)),\(\overline{A\bigcup B}\subset \overline{A}\bigcup \overline{B}\)

\(V(x,\varepsilon)\bigcap A(B)\setminus \left\{ x \right\}\neq \emptyset\implies V(x,\varepsilon)\bigcap \left( A\bigcup B \right)\setminus \left\{ x \right\}\neq \emptyset\)，也就是属于\(A\bigcup B\)的闭包

综上可知：\(\overline{A\bigcup B}=\overline{A}\bigcup \overline{B}\)

同理\(\forall x\in \overline{A\bigcap B},\forall V(x,\varepsilon)\bigcap\left( A\bigcap B \right)\setminus \left\{ x \right\}\neq \emptyset\)

也就是说\(\left( V(x,\varepsilon)\bigcap A\setminus \left\{ x \right\}\neq \emptyset \right) \bigcap \left( V(x,\varepsilon)\bigcap B\setminus \left\{ x \right\}\neq \emptyset \right)\)

故\(\overline{A\bigcap B} \subseteq \overline{A} \bigcap \overline{B}\)

(3)

\(\forall x \in \left( A\bigcap B \right)^{\circ},\exists V(x,\varepsilon)\subset \left( A\bigcap B \right)\implies V(x,\varepsilon)\subset A(B)\implies x\in A^{\circ}\bigcap B^{\circ}\)，可知有\(A^{\circ}\bigcap B^{\circ}\supset \left( A\bigcap B \right)^{\circ}\)

若是\(x\in A^{\circ}\bigcap B^{\circ}\)，\(\exists V(x,\varepsilon_{1})\subset A,V(x,\varepsilon_{2})\subset B\),取\(\varepsilon=\min\left\{ \varepsilon_{1},\varepsilon_{2} \right\},V(x,\varepsilon)\subset A\bigcap B\),可知有\(A^{\circ}\bigcap B^{\circ}\subset \left( A\bigcap B \right)^{\circ}\)

综上可知,\(A^{\circ}\bigcap B^{\circ}=\left( A\bigcap B \right)^{\circ}\)$

\(\forall x\in A^{\circ}\bigcup B^{\circ},\exists V(x,\varepsilon)\subset A(B),V(x,\varepsilon)\subset \left( A\bigcup B \right)\implies x\in \left( A\bigcup B \right)^{\circ}\)

即证\(A^{\circ}\bigcup B^{\circ}\subset \left( A\bigcup B \right)^{\circ}\)

Proof.

(1)

根据定义证明\((A^{\circ})^{\circ}=A^{\circ}\)，利用距离函数可知集合内部是其上每一点的邻域:\(\forall x\in A^{\circ},V(x,\varepsilon)\)上任取一点\(x_{0}\)，再选择一个邻域\(V(x_{0},\varepsilon-d(x,x_{0}))\)上的点记为\(y\)，那么有三角不等式\(d(x,y)\leqslant d(x,x_{0})+d(x_{0},y)<\varepsilon\),证毕。

那么对于A内部上的任意点，选择比原本含于A的开邻域更小的邻域即可含于A的内部，也就是内部的内部仍然为内部，等价为\(A^{\circ}\)是开集

(2)

反证法，假设存在\(x\not\in \partial A,\forall V(x,\varepsilon),V(x,\varepsilon)\bigcap \partial A \neq \emptyset\)，目标是导出\(x\in \partial A\)

分类讨论，\(x\in A\)时，\(\forall\varepsilon,V(x,\varepsilon)\bigcap A\neq \emptyset\)，再根据假设条件\(V\left( x,\frac{\varepsilon}{2} \right)\bigcap \partial A\neq \emptyset,\exists x_{1}\in V\left( x, \frac{\varepsilon}{2} \right)\)且\(x_{1}\in \partial A\)，那么根据定义有\(V\left( x_{1}, \frac{\varepsilon}{2} \right)\bigcap A^{c}\neq \emptyset,\exists x_{2}\in V\left( x_{1}, \frac{\varepsilon}{2} \right)\bigcap A^{c},d(x,x_{2})\leqslant d(x,x_{1})+d(x_{1},x_{2})<\varepsilon\)，也就是\(x_{2}\in V(x,\varepsilon),V(x,\varepsilon)\bigcap A^{c}\neq \emptyset\)，再根据\(\varepsilon\)的任意性得到\(x\in \partial A\)

同理可得\(x\in A^{c}\)时也有\(x\in \partial A\)，综上所述矛盾，\(\partial A=\overline{\partial A}\)，\(\partial A\)是闭集

(3)

考虑证明\(\overline{A}^{c}\)是开的，\(\forall x\in \overline{A}^{c},x\not\in A,\exists V(x,\varepsilon)\bigcap A\neq \emptyset\)，还需证明\(V(x,\varepsilon)\bigcap \overline{A}=\emptyset\)，若是不然，与A的闭包有交，记交点为\(x_{1}\)，闭包与A交非空，
那么取\(d=\varepsilon-d(x,x_{1}),V(x_{1},d)\bigcap A\neq \emptyset\)，记交点为\(x_{2}\in A\)，\(d(x,x_{2})<\varepsilon,x\in A'\subset \overline{A}\),矛盾，所以\(\overline{A}\)是闭集

(4)

同样考虑\(\left( A' \right)^{c},\forall x\in \left( A' \right)^{c}\)，\(\exists V(x,\varepsilon)\)，\(x\in A\)时\(V(x,\varepsilon)\bigcap A=\left\{ x \right\}\)，\(x\not\in A\)时\(V(x,\varepsilon)\bigcap A=\emptyset\)，欲证\(\exists V(x,\varepsilon)\bigcap A'=\emptyset\)，反证不然，\(\forall \varepsilon,V(x,\varepsilon)\bigcap A'\neq \emptyset\)，不妨记为\(x_{1}\)，方便起见取\(\varepsilon'= \frac{\varepsilon}{2},x_{1}\in V(x,\varepsilon')\)，由于\(x_{1}\in A',V\left( x_{1},\varepsilon'= \frac{\varepsilon}{2} \right)\bigcap A\neq \emptyset\)，交点记为\(x_{2}\neq x_{1}\)，那么\(x_{2}\in V(x,\varepsilon)\)，\(V(x,\varepsilon)\)与A有异于\(x\)的交点，故矛盾，\(\left( A' \right)^{c}\)是开集，取补集知结论成立

Proof.

反证，有界，根据Bolzano-Weierstrass定理(聚点定理)可知有界点列必有聚点即收敛子列，可知与无收敛子列矛盾，所以\(\lim\limits_{ m \to \infty }\lvert X_{m} \rvert= +\infty\)

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