Solutions to first Test

Total Marks possible: $10\times 6=60$

Average Mark: 43

Median: 40

You are permitted to bring the book `Rudin: Principles of Mathematical Analysis' with you - just the book, nothing else is permitted (and no notes in your book!) You may use theorems, lemmas and propositions from the book. Note that where $\mathbb{R}^k$ is mentioned below the standard metric is assumed.

1. Suppose that $\{p_n\}$ is a sequence in a metric space, $X,$ and $p\in X.$ Assuming that every subsequence of $\{p_n\}$ itself has a subsequence which converges to $p$ show that $p_n\to p.$

   Solution:- To say that $\{p_n\}$ converges to $p$ is to say that for every $\epsilon >0$ the set $\{n\in\mathbb{N};d(p,p_n)>\epsilon \}$ is finite. Thus if $\{p_n\}$ were not to converge to $p$ then for some $\epsilon >0$ this set would be infinite. Then we can take the subsequence $\{p_{n_k}\}$ where $\{n_k\}$ is the unique increasing sequence with range $\{n\in\mathbb{N};d(p,p_n)>\epsilon \}.$ Thus sequence cannot have any subsequence converge to $p$ since for any susbsequence (of the subsequence) $\{p_{n_{k_l}}\}$ all points lie outside $B(p,\epsilon ).$ This proves the result by contradiction.

2. Let $x_n,$ $n=1,2,\dots,$ be a sequence of real numbers with $x_n\to0.$ Show that there is a subsequence $x_{n(k)},$ $k=1,2,\dots,$ such that $\sum\limits_{k}\vert x_{n(k)}\vert<\infty.$

   Solution:- By definition of convergence, give for every $k$ there exists $N$ such that $n\ge N$ implies $\vert x_n\vert<2^{-k}.$ Thus we can choose a subsequence $\{x_{n_k}\}$ with $\vert x_{n_k}\vert<2^{-k}$ for all $k.$ Then $\sum\limits_{k=1}^N\vert x_{n_k}\vert<\sum\limits_{k=1}^N2^{-k}<1$ so the sequence of partial sums is increasing and bounded above, hence convergent.

3. Give examples of:
   1. A countable subset of $\mathbb{R}^2$ which is infinite and closed.
   2. A subset of the real interval $[-2,2]$ which contains $[0,1]$ but is not compact.
   3. A metric space in which all subsets are compact.
   4. A cover of $[0,1]$ as a subset of $\mathbb{R}$ which has no finite subcover.

   Solution:- Note that I did not ask you to prove this, but they do need to be right. Not just discrete in the third case, etc.

   1. The subset $\{x=(1/n,0);n\in\mathbb{N}\}\cup\{0\}\subset\mathbb{R}^2$ is closed since the only limit point is $0.$ It is certainly infinite and countable.
   2. For example $(-2,1]$ contains $[0,1]$ but is not compact since it is not closed.
   3. Any finite metric space has the property that all subsets are finite, hence compact.
   4. Taking the collection of single-point subsets $V_x=\{x\},$ $x\in[0,1]$ is a cover of $[0,1]$ but has no finite subcover. (I did not say open).

4. Let $K$ be a compact set in a metric space $X$ and suppose $p\in X\setminus K.$ Show that there exists a point $q\in K$ such that

   $$d(p,q)=\inf\{d(p,x); x\in K\}.$$

   
   

   Solution:-

   Method A

   By definition of the infimum, there exists a sequence $q_n$ in $K$ such that $d(p,q_n)\to I=\inf\{d(p,x); x\in K\}.$ Since $K$ us compact, this has a convergent subsequence. Replacing the original sequence by this subsequence we may assume that $q_n\to q$ in $K.$ Now, by the triangle inquality

   $$\vert d(p,q)-d(p,q_n)\vert\le d(q,q_n)\to0$$

   
   

   by the definition of convergence. Thus $d(p,q)$ must be the limit of the sequence $d(p,q_n)$ in $\mathbb{R},$ so $d(p,q)=I$ as desired.

   Method B

   Let $I=\inf\{d(p,x); x\in K\}.$ If there is no point $q\in K$ with $d(p,q)=I$ then the open sets $E_{\epsilon }=\{x\in X;d(p,x)>I+\epsilon \},$ $\epsilon >0$ cover $K.$ By compactness there is a finite subcover, so $K\subset E_\epsilon$ for some $\epsilon >0$ which contradicts the definition of $I.$

   Method C

   (really uses later stuff) Since $f:K\ni x\longmapsto d(p,x)$ is continuous (prove using triangle inequality) and $K$ is compact, $f(K)$ is compact in $\mathbb{R},$ so contains its infimum. Thus there exists $q\in K$ with $d(p,q)=\inf\{d(p,x); x\in K\}.$

5. 1. Let $X$ be a (non-empty) metric space with metric $d$ and let $q\notin X$ be an external point. Show that there is a unique metric $d_Y$ one $Y=X\cup\{q\}$ satisfying

      $$d_Y(x,x')=d(x,x'),\ \forall\ x,x'\in X,\ d_Y(q,x)=1,\ \forall x\in X.$$

      
      

   2. Show that with this metric $Y$ is not connected.

   Solution:- Unfortunately I got carried away here and this is not true! I should have said that $X$ is a metric space with $d(x,x')\le 2$ for all $x,x'\in X;$ then it works fine. I hope I did not confuse anyone too much by this. I gave everyone full marks for the whole question.

6. Let $X$ be a metric space and $A\subset X.$ Let $A^{\circ}$ be the union of all those open sets in $X$ which are subsets of $A.$ Show that the complement of $A^\circ$ is the closure of the complement of $A.$

Solution:- A set contained in $A$ is exactly one with complement containing the complement of $A.$ Thus, from the definition, the complement of $A^{\circ}$ is the intersection of all closed sets which contain the complement of $A.$ This we know to be its closure.

Or:- $A^\circ=A\setminus (A^{\complement})'$ - since a point in $A$ is either an interior point (lies in an open subset of $A$ or else is a limit point of the complement). Thus $(A^\circ)^{\complement}=A^{\complement}\cup(A^{\complement})'$ which is the closure of the complement.