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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.

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.

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 $[0.5,1)$?

For any $ain B$, define $B_aequivbin B:a<b$. The cardinality of $B_a$ must be less than $lceil2/arceil$, otherwise the sum of any $lceil2/arceil$ elements of $B_a$ would exceed 2. That is, $sum_i=1^lceil2/arceilx_i>sum_i=1^lceil2/arceila=alceil2/arceilge2$ for any sequence $x_i$ in $B_a$, contradicting the requirements on $B$.

Define the function $f:BrightarrowmathbbN$ as $f(a)=left|B_aright|$.

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

Credit goes to Acccumulation for the outline of this proof.