Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements from $B$ always gives a sum of $2$ or less. Show that $B$ must be finite or at most countable.

$B$ = {$x \in R:x>0\}$, $x_1,x_2...x_n \in B$ such that $x_1+x_2+...+x_n \le 2$.

Question: for any $a,b$ $(a,b)$~$R$, but $B$ is $(0,+\infty)$ so why $B$ is not uncountable (taking as $a = 0$, and letting $b$->$\infty$)?

And why for $B$ being countable doesn't contradict: for any $a,b$ $(a,b)$~$R$?

P.S. I read Showing a set is finite or countable and understood it.

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    $\begingroup$ It sounds like you are confused about what the question is asking. It is not saying that B is the set of positive real numbers; it is a set of positive real numbers. This means that B is a subset of the set of all positive real numbers. $\endgroup$ Aug 8, 2018 at 17:23
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    $\begingroup$ No. B is some unspecified subset of the positive real numbers such that if you add up a finite number of the elements of B the sum is always less than 2. For example, B could be {0.1, 0.2, 0.5, 1.1} or B could be the infinite set {1, 1/2, 1/4, 1/8, 1/16, ...}. Note that B cannot be {1, 1.4, 1.8} and B cannot be {1, 1/2, 1/3, 1/4, ...} $\endgroup$ Aug 8, 2018 at 17:47
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    $\begingroup$ B does not have to be an interval (0,a). In fact, it cannot be such an interval. Your job is to prove that B must be finite or countable. Intervals are uncountable. $\endgroup$ Aug 8, 2018 at 17:49
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    $\begingroup$ Well, I'm just trying to help clarify the question. I haven't given a solution. I think the hints and discussion below do that for you. But the point is you cannot assume anything about B except: 1. its elements are all positive real numbers and 2. if you take a finite number of elements from B and add them up then you get a sum less than two. Given any such B you have to prove that B is either finite or countable. $\endgroup$ Aug 8, 2018 at 17:59
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    $\begingroup$ Here is a proof that B cannot be the open interval (0,1). If it were then these numbers would be in B: 0.5, 0.51, 0.52, and 0.53. These numbers add up to more than 2. Do you see? You could do something similar to show that B cannot be (0,a) for any positive a. But you have to show more than that. You have to show B is finite or countable. $\endgroup$ Aug 8, 2018 at 18:00

3 Answers 3


Hint 1: How many elements of $B$ can be in the set $[2,\infty)$?

Hint 2: How many elements of $B$ can be in the set $[1,2)$?

Hint 3: How many elements of $B$ can be in the set $[0.5,1)$?

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    $\begingroup$ At this rate you are going to need countably many hints :) $\endgroup$ Aug 8, 2018 at 17:04
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    $\begingroup$ @ArnaudMortier Can't he give a hint schema? $\endgroup$
    – saulspatz
    Aug 8, 2018 at 17:07
  • $\begingroup$ 1) One $x = 2$, 2) set $x_1=1$ so one for sure, then I don't know how many $x_2,x_3,...x_n$ to choose so that they sum up to $1$. 3) The same, I don't know. $\endgroup$
    – user13
    Aug 8, 2018 at 17:10
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    $\begingroup$ Obviously $B\cap[1,\infty)$ has at most two elements, because if $x,y,z\ge1$ then $x+y+z>2$. Similarly $B\cap[2/n,\infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...) $\endgroup$ Aug 8, 2018 at 17:17
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    $\begingroup$ No, we certainly can't write $B$ as $(0,a)$. If $B=(0,a)$ then there are finite subsets of $B$ with sum larger than $2$. $\endgroup$ Aug 8, 2018 at 18:56

Not only can we show that it's countable, it's not too difficult to construct an enumeration: given an element $b$, just count how many other elements of $B$ are larger. This has to be a finite integer, since if there were an infinite number of elements greater than $b$, then the sum of $n$ such elements would be greater than $nb$, so picking an $n>2/b$ would give a sum greater than 2.


For any $a\in B$, define $B_a\equiv\{b\in B:a<b\}$. The cardinality of $B_a$ must be less than $\lceil2/a\rceil$, otherwise the sum of any $\lceil2/a\rceil$ elements of $B_a$ would exceed 2. That is, $\sum_{i=1}^{\lceil2/a\rceil}x_i>\sum_{i=1}^{\lceil2/a\rceil}a=a\lceil2/a\rceil\ge2$ for any sequence $x_i$ in $B_a$, contradicting the requirements on $B$.

Define the function $f:B\rightarrow\mathbb{N}$ as $f(a)=\left|B_a\right|$.

Homework: If you can show that $f$ is one-to-one then you can conclude that $|B|\le|\mathbb{N}|$. That is, $B$ is either finite or countable.

Credit goes to Acccumulation for the outline of this proof.

  • $\begingroup$ What if $a =$ sup$B$? Then $B_a $ is the empty set. If $B_a \equiv \{b \in B: a \leq b \}$, then the sum of any $\lceil r/a \rceil$ for $r > 2$ elements of $B_a$ would exceed 2. $\endgroup$
    – hiroshin
    Jul 18, 2019 at 23:08

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