## 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 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 $|\mathbb R|\ge|\mathcal P(\mathbb N)|$.

### 5 Responses to Analysis – HW 1

1. andrescaicedo says:

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

2. Rustyn Yazdanpour says:

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 $\sum_{i\in S_1} \frac{b_i}{3^{i+1}} = \sum_{i\in S_2} \frac{b_i}{3^{i+1}} = \frac{1}{9}.$

• andrescaicedo says:

Oh, yes, that question came up not as intended. Thanks for noticing it. I’ve fixed the text.

3. Rustyn Yazdanpour says: • andrescaicedo says:

Thank you!