open set is the disjoint union of a countable collection of open intervals

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I don't understand that $I_x_xin O$ is disjoint. For example, $O = (1 , 2)$. Let $x = 1.5$. Then, $a_x = 1$, and $b_x = 2$. Therefore, $I_x = (1, 2)$. Let $y = 1.6$. Then, similarly, $I_y = (1, 2)$. That is, $I_x$ and $I_y$ are not disjoint, but equal.



Could you explain how $I_x_x in O$ can be disjoint?




real-analysis




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  • Royden Fitzpatrick?
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up vote
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down vote

favorite
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I don't understand that $I_x_xin O$ is disjoint. For example, $O = (1 , 2)$. Let $x = 1.5$. Then, $a_x = 1$, and $b_x = 2$. Therefore, $I_x = (1, 2)$. Let $y = 1.6$. Then, similarly, $I_y = (1, 2)$. That is, $I_x$ and $I_y$ are not disjoint, but equal.



Could you explain how $I_x_x in O$ can be disjoint?




real-analysis




share|cite|improve this question





















  • Royden Fitzpatrick?
    – BCLC
    Jul 28 at 9:34












up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1









I don't understand that $I_x_xin O$ is disjoint. For example, $O = (1 , 2)$. Let $x = 1.5$. Then, $a_x = 1$, and $b_x = 2$. Therefore, $I_x = (1, 2)$. Let $y = 1.6$. Then, similarly, $I_y = (1, 2)$. That is, $I_x$ and $I_y$ are not disjoint, but equal.



Could you explain how $I_x_x in O$ can be disjoint?




real-analysis




share|cite|improve this question

















I don't understand that $I_x_xin O$ is disjoint. For example, $O = (1 , 2)$. Let $x = 1.5$. Then, $a_x = 1$, and $b_x = 2$. Therefore, $I_x = (1, 2)$. Let $y = 1.6$. Then, similarly, $I_y = (1, 2)$. That is, $I_x$ and $I_y$ are not disjoint, but equal.



Could you explain how $I_x_x in O$ can be disjoint?





real-analysis





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edited Jul 28 at 10:31









Asaf Karagila

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asked Jul 28 at 9:17









Sihyun Kim

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  • Royden Fitzpatrick?
    – BCLC
    Jul 28 at 9:34
















  • Royden Fitzpatrick?
    – BCLC
    Jul 28 at 9:34















Royden Fitzpatrick?
– BCLC
Jul 28 at 9:34




Royden Fitzpatrick?
– BCLC
Jul 28 at 9:34










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The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x cap I_y not = emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)






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    1 Answer
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    active

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    active

    oldest

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    up vote
    7
    down vote



    accepted










    The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x cap I_y not = emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)






    share|cite|improve this answer

























      up vote
      7
      down vote



      accepted










      The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x cap I_y not = emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)






      share|cite|improve this answer























        up vote
        7
        down vote



        accepted







        up vote
        7
        down vote



        accepted






        The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x cap I_y not = emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)






        share|cite|improve this answer













        The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x cap I_y not = emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)







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        share|cite|improve this answer











        answered Jul 28 at 9:21









        Patrick Stevens

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