A little question on rationals and open sets in $mathbbR$.

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Let $A$ be an open set of real numbers, and $D$ be the set of rationals in $A$. For every $din D$, let $J(d)$ be an (arbitrary) open interval such that $d in J(d) subseteq A$. Is it true that $bigcup J (d) = A$ ?






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    Let $A$ be an open set of real numbers, and $D$ be the set of rationals in $A$. For every $din D$, let $J(d)$ be an (arbitrary) open interval such that $d in J(d) subseteq A$. Is it true that $bigcup J (d) = A$ ?






    general-topology real-numbers rational-numbers







    share|cite|improve this question

























      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      Let $A$ be an open set of real numbers, and $D$ be the set of rationals in $A$. For every $din D$, let $J(d)$ be an (arbitrary) open interval such that $d in J(d) subseteq A$. Is it true that $bigcup J (d) = A$ ?






      general-topology real-numbers rational-numbers







      share|cite|improve this question















      Let $A$ be an open set of real numbers, and $D$ be the set of rationals in $A$. For every $din D$, let $J(d)$ be an (arbitrary) open interval such that $d in J(d) subseteq A$. Is it true that $bigcup J (d) = A$ ?






      general-topology real-numbers rational-numbers




      general-topology real-numbers rational-numbers






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      edited Sep 1 at 3:44









      Cornman

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      asked Sep 1 at 3:04









      Alex123

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          No, it isn't true.



          Consider $A=(0,1)$ and $J(d)$ being whichever one of $(0,frac1pi)$ or $(frac1pi,1)$ which contains $d$.






          share|cite|improve this answer




















          • And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
            – Alex123
            Sep 1 at 3:22










          • @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
            – Arthur
            Sep 1 at 3:40











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          No, it isn't true.



          Consider $A=(0,1)$ and $J(d)$ being whichever one of $(0,frac1pi)$ or $(frac1pi,1)$ which contains $d$.






          share|cite|improve this answer




















          • And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
            – Alex123
            Sep 1 at 3:22










          • @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
            – Arthur
            Sep 1 at 3:40















          up vote
          4
          down vote



          accepted










          No, it isn't true.



          Consider $A=(0,1)$ and $J(d)$ being whichever one of $(0,frac1pi)$ or $(frac1pi,1)$ which contains $d$.






          share|cite|improve this answer




















          • And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
            – Alex123
            Sep 1 at 3:22










          • @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
            – Arthur
            Sep 1 at 3:40













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          No, it isn't true.



          Consider $A=(0,1)$ and $J(d)$ being whichever one of $(0,frac1pi)$ or $(frac1pi,1)$ which contains $d$.






          share|cite|improve this answer












          No, it isn't true.



          Consider $A=(0,1)$ and $J(d)$ being whichever one of $(0,frac1pi)$ or $(frac1pi,1)$ which contains $d$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 1 at 3:07









          Arthur

          102k795176




          102k795176











          • And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
            – Alex123
            Sep 1 at 3:22










          • @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
            – Arthur
            Sep 1 at 3:40

















          • And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
            – Alex123
            Sep 1 at 3:22










          • @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
            – Arthur
            Sep 1 at 3:40
















          And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
          – Alex123
          Sep 1 at 3:22




          And if we were to impose J(d) = (d-1/n,d+1/n) for some n = 1,2,... ?
          – Alex123
          Sep 1 at 3:22












          @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
          – Arthur
          Sep 1 at 3:40





          @Alex still no. List the rationals in $(0,1)$ as $d_1,d_2,ldots$, and let $J(d_k)=(d_k-frac12^k+2,d_k+frac12^k+2)$. Then the combined width of all the $J(d_k)$'s is $frac12$, so even with no overlap, (and no "overspill", as in $J(d_k)notsubseteq (0,1)$), half of $A$ (as measured by, well, measure) is still uncovered. In fact, this still works for $A=Bbb R$, meaning by far most of the number line is uncovered. We can also adjust the exponent to make the sum of the widths as small as we want.
          – Arthur
          Sep 1 at 3:40


















           

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