Show that a certain set of positive real numbers must be finite or countable

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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.







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  • 4




    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.
    – Leonard Blackburn
    Aug 8 at 17:23






  • 2




    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, ...
    – Leonard Blackburn
    Aug 8 at 17:47






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:49






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:59






  • 1




    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.
    – Leonard Blackburn
    Aug 8 at 18:00














up vote
3
down vote

favorite
1













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.







share|cite|improve this question


















  • 4




    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.
    – Leonard Blackburn
    Aug 8 at 17:23






  • 2




    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, ...
    – Leonard Blackburn
    Aug 8 at 17:47






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:49






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:59






  • 1




    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.
    – Leonard Blackburn
    Aug 8 at 18:00












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






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.







share|cite|improve this question















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.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 8 at 18:23









dmtri

347213




347213










asked Aug 8 at 16:57









Sargis Iskandaryan

48212




48212







  • 4




    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.
    – Leonard Blackburn
    Aug 8 at 17:23






  • 2




    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, ...
    – Leonard Blackburn
    Aug 8 at 17:47






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:49






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:59






  • 1




    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.
    – Leonard Blackburn
    Aug 8 at 18:00












  • 4




    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.
    – Leonard Blackburn
    Aug 8 at 17:23






  • 2




    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, ...
    – Leonard Blackburn
    Aug 8 at 17:47






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:49






  • 2




    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.
    – Leonard Blackburn
    Aug 8 at 17:59






  • 1




    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.
    – Leonard Blackburn
    Aug 8 at 18:00







4




4




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.
– Leonard Blackburn
Aug 8 at 17:23




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.
– Leonard Blackburn
Aug 8 at 17:23




2




2




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, ...
– Leonard Blackburn
Aug 8 at 17:47




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, ...
– Leonard Blackburn
Aug 8 at 17:47




2




2




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.
– Leonard Blackburn
Aug 8 at 17:49




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.
– Leonard Blackburn
Aug 8 at 17:49




2




2




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.
– Leonard Blackburn
Aug 8 at 17:59




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.
– Leonard Blackburn
Aug 8 at 17:59




1




1




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.
– Leonard Blackburn
Aug 8 at 18:00




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.
– Leonard Blackburn
Aug 8 at 18:00










3 Answers
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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.






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    up vote
    17
    down vote













    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 $[0.5,1)$?






    share|cite|improve this answer
















    • 24




      At this rate you are going to need countably many hints :)
      – Arnaud Mortier
      Aug 8 at 17:04






    • 9




      @ArnaudMortier Can't he give a hint schema?
      – saulspatz
      Aug 8 at 17:07










    • 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.
      – Sargis Iskandaryan
      Aug 8 at 17:10







    • 2




      Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
      – David C. Ullrich
      Aug 8 at 17:17






    • 1




      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 $2$.
      – David C. Ullrich
      Aug 8 at 18:56

















    up vote
    1
    down vote













    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.






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      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.






      share|cite|improve this answer
























        up vote
        3
        down vote



        accepted










        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.






        share|cite|improve this answer






















          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 8 at 18:25









          Acccumulation

          4,5382314




          4,5382314




















              up vote
              17
              down vote













              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 $[0.5,1)$?






              share|cite|improve this answer
















              • 24




                At this rate you are going to need countably many hints :)
                – Arnaud Mortier
                Aug 8 at 17:04






              • 9




                @ArnaudMortier Can't he give a hint schema?
                – saulspatz
                Aug 8 at 17:07










              • 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.
                – Sargis Iskandaryan
                Aug 8 at 17:10







              • 2




                Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
                – David C. Ullrich
                Aug 8 at 17:17






              • 1




                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 $2$.
                – David C. Ullrich
                Aug 8 at 18:56














              up vote
              17
              down vote













              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 $[0.5,1)$?






              share|cite|improve this answer
















              • 24




                At this rate you are going to need countably many hints :)
                – Arnaud Mortier
                Aug 8 at 17:04






              • 9




                @ArnaudMortier Can't he give a hint schema?
                – saulspatz
                Aug 8 at 17:07










              • 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.
                – Sargis Iskandaryan
                Aug 8 at 17:10







              • 2




                Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
                – David C. Ullrich
                Aug 8 at 17:17






              • 1




                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 $2$.
                – David C. Ullrich
                Aug 8 at 18:56












              up vote
              17
              down vote










              up vote
              17
              down vote









              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 $[0.5,1)$?






              share|cite|improve this answer












              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 $[0.5,1)$?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 8 at 17:04









              vadim123

              73.8k895184




              73.8k895184







              • 24




                At this rate you are going to need countably many hints :)
                – Arnaud Mortier
                Aug 8 at 17:04






              • 9




                @ArnaudMortier Can't he give a hint schema?
                – saulspatz
                Aug 8 at 17:07










              • 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.
                – Sargis Iskandaryan
                Aug 8 at 17:10







              • 2




                Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
                – David C. Ullrich
                Aug 8 at 17:17






              • 1




                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 $2$.
                – David C. Ullrich
                Aug 8 at 18:56












              • 24




                At this rate you are going to need countably many hints :)
                – Arnaud Mortier
                Aug 8 at 17:04






              • 9




                @ArnaudMortier Can't he give a hint schema?
                – saulspatz
                Aug 8 at 17:07










              • 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.
                – Sargis Iskandaryan
                Aug 8 at 17:10







              • 2




                Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
                – David C. Ullrich
                Aug 8 at 17:17






              • 1




                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 $2$.
                – David C. Ullrich
                Aug 8 at 18:56







              24




              24




              At this rate you are going to need countably many hints :)
              – Arnaud Mortier
              Aug 8 at 17:04




              At this rate you are going to need countably many hints :)
              – Arnaud Mortier
              Aug 8 at 17:04




              9




              9




              @ArnaudMortier Can't he give a hint schema?
              – saulspatz
              Aug 8 at 17:07




              @ArnaudMortier Can't he give a hint schema?
              – saulspatz
              Aug 8 at 17:07












              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.
              – Sargis Iskandaryan
              Aug 8 at 17:10





              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.
              – Sargis Iskandaryan
              Aug 8 at 17:10





              2




              2




              Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
              – David C. Ullrich
              Aug 8 at 17:17




              Obviously $Bcap[1,infty)$ has at most two elements, because if $x,y,zge1$ then $x+y+z>2$. Similarly $Bcap[2/n,infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...)
              – David C. Ullrich
              Aug 8 at 17:17




              1




              1




              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 $2$.
              – David C. Ullrich
              Aug 8 at 18:56




              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 $2$.
              – David C. Ullrich
              Aug 8 at 18:56










              up vote
              1
              down vote













              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.






              share|cite|improve this answer


























                up vote
                1
                down vote













                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.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  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.






                  share|cite|improve this answer














                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 9 at 11:06

























                  answered Aug 9 at 3:13









                  Matt

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