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

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


    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

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

    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\}.$

    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

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

    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.

  5. 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.

Richard B. Melrose 2004-05-18