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10.16 Topology, Compactness

  1. Prove \(\forall x,y\in\R^{n},||x|-|y||\leq |x-y|\)

    Proof

    \(|x|=|x-y+y|\leq |x-y|+|y|\implies |x|-|y|\leq |x-y|\)

    \(|y|=|y-x+x|\leq |y-x|+|x|\implies |y|-|x|\leq |y-x|\)

    Then \(||x|-|y||\leq |x-y|\)

  2. Suppose \(x\in(x_1,x_2,...,x_n)\in \R^n\). Prove that \(|x|\leq \sum_{i=1}|x_i|\)

    Proof

    Let \(x=\sum^n_{i=1}x_ie_i\), then \(|x|=\left|\sum^n_{x=1}x_ie_i\right|\leq \sum^n_{i=1}|x_ie_i|=\sum^n_{x=1}|x_i||e_i|=\sum^n_{i=1}|x_i|\)

  3. Let \(A=\{(x,y)\in \R^{n}:1<x^{2}+y^{2}\leq 4\}\). Find \(\text{int}(A),\text{ext}(A),\partial A,\bar A=A\cup \partial A\)

    Solution

    \(\text{int}(A)=\{(x,y)\in \R^{n}:1<x^{2}+y^{2}< 4\}\)

    \(\text{ext}(A)=\{(x,y)\in \R^{n}:x^{2}+y^{2}>4\text{ or }0\leq x^{2}+y^{2}< 1\}\)

    \(\partial(A)=\{(x,y)\in \R^{n}:x^{2}+y^{2}=4\text{ or }x^{2}+y^{2}= 1\}\)

    \(\bar A=\{(x,y)\in\R^{n}:1\leq x^{2}+y^{2}\leq 4\}\)

  4. A set \(F\) is closed iff it contains its boundary

    Proof

    \(\Rightarrow\)) Suppose \(F\) is closed, we need to show that \(\partial F\subset F\).

    Take \(\forall x\in\partial F\), NTP: \(x\in F\). Suppose \(x\notin F\), then \(x\in \R^n\setminus F\).

    Since \(x\in\partial F\), then \(x\notin F\wedge x\notin \R^n\setminus F\). Thus contradiction. Thus \(x\in F\)

    \(\Leftarrow\)) Suppose \(F\) contains its boundary, NTP: set \(F\) is closed.

    We know \(\partial F\subset F\), NTP: \(\R^n\setminus F\) is open. Take any \(x\in \R^n\setminus F\), NTP: \(\exists \varepsilon:U^{\varepsilon}\subset \R^{n}\setminus F\)

    Since \(x\notin F\), then \(x\notin \partial F\). Then there is a neighborhood \(U\) of \(x\) s.t. \(U\subset F\) or \(U\subset \R^n\setminus F\)

    But \(x\notin F\), thus \(\exists \varepsilon:U^{\varepsilon}\subset \R^{n}\setminus F\)

  5. For a set \(A\), \(x\) is a limit point of \(A\) if every deleted neighborhood of \(x\) intersects \(A\)

    A set is closed iff it contains all its limit points.

    An example of a limit point that is not in the set \(A\): \(A=\{\frac{1}{n}:n\in\N\}\) and \(x=0\)

  6. \(A=\{(\frac{1}{n},\frac{1}{m})\in \R^2:n,m\in \N\}\)

    Not open: There are points in the neighborhood which doesn't have the given form

    Not closed: \((0,0)\notin A\) which is a limit point

    Neither open nor closed

  7. Find \(\text{int}(\mathbb{Q}),\text{ext}(\mathbb{Q}),\partial \mathbb{Q},\bar{\mathbb{Q}}\)

    Remark: The set of all limit points of \(A\) is closure of \(A\)

    \(\text{int}(\mathbb{Q})=\empty\), \(\text{ext}(\mathbb{Q})=\empty,\bar{\mathbb{Q}}=\R,\partial\mathbb{Q}=\bar{\mathbb{Q}} \setminus\text{int}(\mathbb{Q})=\R\), since \(x-\frac{1}{n}<a_n<x+\frac{1}{n}\)

  8. Is the set \(\{\frac{1}{n}:n\in \N\}\) compact in \(\R\)?

    It is bounded but not closed since \(0\notin\) set. Thus by theorem, it is not compact

  9. Give an example of a closed set that is not compact in \(\R\)

    \([1,\infty)\). It is closed in \(\R\) but not compact since it's not bounded

  10. Suppose \(K\subset \R^n\) is compact and \(F\subset K\) is closed. Prove that \(F\) is compact

    Proposition: A closed subset of a compact set is compact

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