Analysis – HW 1

This set is due Monday, September 16, at the beginning of lecture.

Recall that \mathbb N=\{0,1,2,\dots\}. Given a sequence \vec a=a_0,a_1,\dots of nonnegative real numbers, for F a finite subset of \mathbb N, the expression

\displaystyle \sum_{F}\vec a=\sum_{n\in F}a_n

has what is hopefully the obvious meaning: If n_0<n_1<\dots<n_k is the increasing enumeration of the elements of F, then

\sum_{F}\vec a=\sum_{i=0}^k a_{n_i}=a_{n_0}+a_{n_1}+\dots+a_{n_k},

with the (standard) convention that if F is empty, then \sum_{F}\vec a=0.

For S an arbitrary subset of \mathbb N (so S may be finite or infinite), define

\displaystyle \sum_{S}\vec a=\sup\{\sum_{F}\vec a\mid F\mbox{\ is a finite subset of\ }S\},

provided that the supremum exists. There is a small ambiguity here, in that if S is finite, we have defined \sum_{S}\vec a in two potentially conflicting ways.

1. Show that both definitions coincide if S is finite.

2. Give an example of a sequence \vec a and a set S such that \sum_{n\in S}a_n is not defined. Show that for any \vec a and any S, if \sum_{n\in S}a_n is not defined, then neither is \sum_{n\in\mathbb N}a_n.

3. Show that, if \sum_{\mathbb N}\vec a is defined, then

\sum_{\mathbb N}\vec a=\sup\{\sum_{k=0}^m a_k\mid m\in\mathbb N\}.

More generally, show that, as long as \sum_{S}\vec a is defined, then

\sum_{S}\vec a=\sup\{\sum_{S\cap[0,m]} \vec a\mid m\in\mathbb N\}

and that, if this supremum exists, then so does \sum_{S}\vec a, and the displayed equality holds.

4. Fix a positive integer k\ge 2. Show that if \vec a is such that, for every n, a_n has the form \displaystyle \frac{b_n}{k^{n+1}} where b_n\in\{0,1,\dots,k-1\} then, for any S, \sum_{n\in S}a_n is defined, and is a number in the interval [0,1].

5. Show that for every x\in[0,1] and every positive integer k\ge2 there is some \vec a as in item 4. such that \sum_{\mathbb N}\vec a=x. Describe as precisely as possible all the quadruples (k,x,\vec a,\vec a') such that k\ge 2 is an integer, x\in[0,1], \vec a\ne \vec a' are sequences as in 4., and yet

\sum_{n\in\mathbb N}a_n=x=\sum_{n\in \mathbb N}a'_n.

Hopefully it is clear that all we are describing is the base k representation of any number x\in[0,1].

6. Indicate how to extend the above so any real has a base k representation (for any k\ge2).

7. Given k\ge 2, let \vec a be the sequence with n-th term a_n=1/k^n for all n. Show that 2 is the only value of k such that there are S_1\ne S_2 with \sum_{S_1}\vec a=\sum_{S_2}\vec a. Describe all such pairs (S_1,S_2). Show that for all k\ge 2 there is some \vec a as in 4., with the same “failure of injectivity” property.

The above gives us that |\mathbb R|\ge|\mathcal P(\mathbb N)| in the sense that there is an injection \psi:\mathcal P(\mathbb N)\to \mathbb R.

8. Make this explicit, that is, give an example of such an injection \psi, hopefully related to these sums we are considering.

One can also show that |\mathcal P(\mathbb N)|\ge|\mathbb R| and in fact there is a bijection between these two sets, though you do not need to do this here.

As indicated in item 7., when \vec a=1,1/2,1/4,1/8,\dots the function \rho:\mathcal P(\mathbb N)\to\mathbb R given by \rho(S)=\sum_{n\in S}a_n is not an injection.

9. For this \vec a, show that the collection of sets S such that there is a set T\ne S with \sum_{S}\vec a=\sum_{T}\vec a is countable. Show that if \mathcal F\subset\mathcal P(\mathbb N) is countable, then there is a bijection between \mathcal P(\mathbb N) and \mathcal P(\mathbb N)\setminus \mathcal F so, in particular, even \rho allows us to verify that |\mathbb R|\ge|\mathcal P(\mathbb N)|.

TeX file.

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This entry was posted on Thursday, September 5th, 2013 at 11:21 am and is filed under 414/514: Analysis I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

5 Responses to Analysis – HW 1

  1. andrescaicedo says:
    September 6, 2013 at 7:03 pm

    I’ve posted the TeX file for the homework, in case it is useful.

    Reply
  2. Rustyn Yazdanpour says:
    September 8, 2013 at 8:35 pm

    Note that on problem 7, is it implied that b_n = 1, \forall n?.

    E.g.

    For S_1 = \{1\}, S_2 = \{ 2,3,4,\dots\}, if we say b_1 = 1, b_n = 2, \forall n\ge 2, we have \sum_{i\in S_1} \frac{b_i}{3^{i+1}} = \sum_{i\in S_2} \frac{b_i}{3^{i+1}} = \frac{1}{9}.

    Reply
  3. Rustyn Yazdanpour says:
    September 14, 2013 at 1:29 pm

    Reply

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