$A=\{a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}| k\in\mathbb N, \lambda_1,...,\lambda_k\in\mathbb R \;\text{are all distinct}\}$ is not bounded For any real number $\lambda\in\mathbb R$ choose a positive real number $a_\lambda>0$.
Show that the subset $A$ below is unbounded.
$$A=\{a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}| k\in\mathbb N, \lambda_1,...,\lambda_k\in\mathbb R \;\text{are all distinct}\}$$
And consider the case when $\lambda$s are from integers.
Here is an answer A boundedness question.

By hypothesis,


$$\begin{array}{l|rcl}
b : & \mathbb R & \longrightarrow & \mathbb R \\
    & \lambda & \longmapsto & b_\lambda \end{array}$$ is supposed to be a one-to-one map. Without loss of generality by changing $b$ into $-b$, we can suppose that $b(\mathbb R) \cap (0,\infty)$ is uncountable. Therefore, one of the sets $B_n = b(\mathbb R) \cap (1/n, \infty)$ is uncountable  (a union of countably many countable sets is countable). Taking $k$ distinct elements of $B_n$ we have $\sum_{i=1}^k b_{\lambda_i} \ge k/n$ proving that $B$ in unbounded.


The result doesn't hold if the $\lambda$s belong to $\mathbb N$. Take for example $b_n = 1/n^2$. As $\sum 1/n^2$ converges, every partial sum is also finite.

but I am not satisfied.

*

*Why we write -b into b? What is the point?


*It is obvious that $b(\mathbb R)\cap (1/n,\infty)$ is uncountable but $a_{\lambda_i}s$ ($a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}$) are not supposed to be inside $(1/n,\infty)$
I understand why we take $(1/n,\infty)$ because we want an lower bound that can be go to the infinity. But it is not good with me.


*And for the second consideration with $\lambda$s from $\mathbb Z$ I didnot quite understand why the $1/n^2,1/n^3$ sums are working?
 A: There are not any new ideas in my answer, but I thought the below might help. I will write $a_\lambda$ as $a(\lambda)$ and use standard function notation.
There is nothing in the hypotheses that implies the map $a:\mathbb R \to (0,\infty)$ is $1-1.$ For example, you could simply define $a(\lambda) = 1$ for all $\lambda.$ (It's the $\lambda$'s that are distinct, not the $a(\lambda)$'s.) So the "proof" in 1. is a nonproof as far as I can see.
All we need to assume is that $a:\mathbb R \to (0,\infty).$ The desired result will follow.
Proof: For $n=1,2,\dots$ let $E_n=a(\mathbb R)\cap (1/n,\infty).$ Then $\mathbb R = \cup_n a^{-1}(E_n).$ Since $\mathbb R$ is uncountable, $a^{-1}(E_n)$ must be infinite for some $n,$ say $n_0.$ Now every infinite set contains a countably infinite set, so there exist countably many distinct $y_1,y_2,\dots \in a^{-1}(E_{n_0}).$ We then have, for each $k,$
$$\tag 1 \sum_{j=1}^{k}a(y_j) > k\cdot \frac{1}{n_0}$$
As $k\to \infty,$ the right side of $(1)\to \infty,$ showing the set $A$ is unbounded.
To show the result need not hold for $a:\mathbb Z\to (0,\infty), $  define $a(n) =2^{-n}$ for $n\in \mathbb N,$ $a(n)=0$ for $n<1.$ Suppose $n_1,\dots, n_k\in \mathbb Z$ are distict. Then
$$\sum_{j=1}^{k}a(n_j) \le \sum_{n=1}^{\infty}2^{-n} =1.$$
Thus the set $A$ in this case is bounded above by $1.$
A: For each $n\in\mathbb{N}$, let $B_{n}=\{\lambda\in\mathbb{R}\mid a_{\lambda}>\frac{1}{n}\}$.
Clearly $\cup_{n}B_{n}=\mathbb{R}.$ Since $\mathbb{R}$ is uncountable,
there exists $n$ such that $B_{n}$ is uncountable. For any $k\in\mathbb{N}$,
we can choose pairwisely distinct $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in B_{n}$.
Then $\sum_{i=1}^{k}a_{\lambda_{i}}>\frac{k}{n}.$ Note that $\sum_{i=1}^{k}a_{\lambda_{i}}\in A.$
This shows that the set $A$ is unbounded above.

If $\lambda$ is only allowed to take values in $\mathbb{Z},$ the
set $A$ may be bounded or unbounded, depending on $a_{\lambda}$.
For example, if $a_{n}=\frac{1}{2^{|n|}},$ then $A$ is bounded. If
$a_{n}=|n|$, then $A$ is unbounded.
