Showing that monotone functions have at most countable discontinuities.












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$begingroup$


I want to show that a map $F: mathbf{R} to mathbf{R}$ has at most countable discountinuities, if $F(x) leq F(y)$ whenever $x leq y$.



Here's the idea. Let's use standard notation $F(x^+), F(x^-)$ for upper and lower limits of $F$ around $x$. And let $D(F) = {x : F(x^+) neq F(x^-)}$, which is the set of discountinuities for a monotone function.
Now suppose that $D(F)$ is uncountable. Suppose $x, y in D(F)$ are distinct. Then $(F(x^-), F(x^+)) cap (F(y^-), F(y^+)) = emptyset$. This is because with $r = d(x, y)$, and assuming $x < y$, if $F(x^+) > F(y^-)$, this means that $inf_{x leq s < x + r/2} F(x) > sup_{y - r/2 < t leq y} F(t)$, which happens if and only if for some particular $x leq s < x + r/2$ and $y - r/2 < t leq y$, we have $F(s)> F(t)$. But this is a contradiction since as stated, $s leq t$. Hence we have $F(x^+) leq F(y^-)$, which means the intervals are disjoint as required.



We have shown that $D(F)$ uncountable implies the existence of uncountably many disjoint intervals $(F(x^-), F(x^+))$, where $x$ ranges over $D(F)$. But this can't happen because each open interval contains a distinct rational, which is a countable set.






real-analysis proof-verification







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  • $begingroup$
    Your proof looks good. :)
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 20:59










  • $begingroup$
    Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right.
    $endgroup$
    – user25959
    Dec 8 '18 at 20:59
















3












$begingroup$


I want to show that a map $F: mathbf{R} to mathbf{R}$ has at most countable discountinuities, if $F(x) leq F(y)$ whenever $x leq y$.



Here's the idea. Let's use standard notation $F(x^+), F(x^-)$ for upper and lower limits of $F$ around $x$. And let $D(F) = {x : F(x^+) neq F(x^-)}$, which is the set of discountinuities for a monotone function.
Now suppose that $D(F)$ is uncountable. Suppose $x, y in D(F)$ are distinct. Then $(F(x^-), F(x^+)) cap (F(y^-), F(y^+)) = emptyset$. This is because with $r = d(x, y)$, and assuming $x < y$, if $F(x^+) > F(y^-)$, this means that $inf_{x leq s < x + r/2} F(x) > sup_{y - r/2 < t leq y} F(t)$, which happens if and only if for some particular $x leq s < x + r/2$ and $y - r/2 < t leq y$, we have $F(s)> F(t)$. But this is a contradiction since as stated, $s leq t$. Hence we have $F(x^+) leq F(y^-)$, which means the intervals are disjoint as required.



We have shown that $D(F)$ uncountable implies the existence of uncountably many disjoint intervals $(F(x^-), F(x^+))$, where $x$ ranges over $D(F)$. But this can't happen because each open interval contains a distinct rational, which is a countable set.






real-analysis proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your proof looks good. :)
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 20:59










  • $begingroup$
    Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right.
    $endgroup$
    – user25959
    Dec 8 '18 at 20:59














3












3








3


1



$begingroup$


I want to show that a map $F: mathbf{R} to mathbf{R}$ has at most countable discountinuities, if $F(x) leq F(y)$ whenever $x leq y$.



Here's the idea. Let's use standard notation $F(x^+), F(x^-)$ for upper and lower limits of $F$ around $x$. And let $D(F) = {x : F(x^+) neq F(x^-)}$, which is the set of discountinuities for a monotone function.
Now suppose that $D(F)$ is uncountable. Suppose $x, y in D(F)$ are distinct. Then $(F(x^-), F(x^+)) cap (F(y^-), F(y^+)) = emptyset$. This is because with $r = d(x, y)$, and assuming $x < y$, if $F(x^+) > F(y^-)$, this means that $inf_{x leq s < x + r/2} F(x) > sup_{y - r/2 < t leq y} F(t)$, which happens if and only if for some particular $x leq s < x + r/2$ and $y - r/2 < t leq y$, we have $F(s)> F(t)$. But this is a contradiction since as stated, $s leq t$. Hence we have $F(x^+) leq F(y^-)$, which means the intervals are disjoint as required.



We have shown that $D(F)$ uncountable implies the existence of uncountably many disjoint intervals $(F(x^-), F(x^+))$, where $x$ ranges over $D(F)$. But this can't happen because each open interval contains a distinct rational, which is a countable set.






real-analysis proof-verification







share|cite|improve this question









$endgroup$




I want to show that a map $F: mathbf{R} to mathbf{R}$ has at most countable discountinuities, if $F(x) leq F(y)$ whenever $x leq y$.



Here's the idea. Let's use standard notation $F(x^+), F(x^-)$ for upper and lower limits of $F$ around $x$. And let $D(F) = {x : F(x^+) neq F(x^-)}$, which is the set of discountinuities for a monotone function.
Now suppose that $D(F)$ is uncountable. Suppose $x, y in D(F)$ are distinct. Then $(F(x^-), F(x^+)) cap (F(y^-), F(y^+)) = emptyset$. This is because with $r = d(x, y)$, and assuming $x < y$, if $F(x^+) > F(y^-)$, this means that $inf_{x leq s < x + r/2} F(x) > sup_{y - r/2 < t leq y} F(t)$, which happens if and only if for some particular $x leq s < x + r/2$ and $y - r/2 < t leq y$, we have $F(s)> F(t)$. But this is a contradiction since as stated, $s leq t$. Hence we have $F(x^+) leq F(y^-)$, which means the intervals are disjoint as required.



We have shown that $D(F)$ uncountable implies the existence of uncountably many disjoint intervals $(F(x^-), F(x^+))$, where $x$ ranges over $D(F)$. But this can't happen because each open interval contains a distinct rational, which is a countable set.






real-analysis proof-verification




real-analysis proof-verification






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asked Dec 8 '18 at 20:45









Drew BradyDrew Brady

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721315












  • $begingroup$
    Your proof looks good. :)
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 20:59










  • $begingroup$
    Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right.
    $endgroup$
    – user25959
    Dec 8 '18 at 20:59


















  • $begingroup$
    Your proof looks good. :)
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 20:59










  • $begingroup$
    Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right.
    $endgroup$
    – user25959
    Dec 8 '18 at 20:59
















$begingroup$
Your proof looks good. :)
$endgroup$
– Tki Deneb
Dec 8 '18 at 20:59




$begingroup$
Your proof looks good. :)
$endgroup$
– Tki Deneb
Dec 8 '18 at 20:59












$begingroup$
Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right.
$endgroup$
– user25959
Dec 8 '18 at 20:59




$begingroup$
Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right.
$endgroup$
– user25959
Dec 8 '18 at 20:59










1 Answer
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$begingroup$

I assume your question is, is this proof valid? The answer is yes!



To be clearer, you should probably point out that $x + r/2 = y - r/2$, which is how you get $s < x + r/2 = y - r/2 < t$ and therefore $s le t$ (you say this is true "as stated", but I don't see you stating it).



You might also want to say more about why a discontinuity in a monotone function must have $F(x^+) ne F(x^-)$.






share|cite|improve this answer









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    $begingroup$

    I assume your question is, is this proof valid? The answer is yes!



    To be clearer, you should probably point out that $x + r/2 = y - r/2$, which is how you get $s < x + r/2 = y - r/2 < t$ and therefore $s le t$ (you say this is true "as stated", but I don't see you stating it).



    You might also want to say more about why a discontinuity in a monotone function must have $F(x^+) ne F(x^-)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I assume your question is, is this proof valid? The answer is yes!



      To be clearer, you should probably point out that $x + r/2 = y - r/2$, which is how you get $s < x + r/2 = y - r/2 < t$ and therefore $s le t$ (you say this is true "as stated", but I don't see you stating it).



      You might also want to say more about why a discontinuity in a monotone function must have $F(x^+) ne F(x^-)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I assume your question is, is this proof valid? The answer is yes!



        To be clearer, you should probably point out that $x + r/2 = y - r/2$, which is how you get $s < x + r/2 = y - r/2 < t$ and therefore $s le t$ (you say this is true "as stated", but I don't see you stating it).



        You might also want to say more about why a discontinuity in a monotone function must have $F(x^+) ne F(x^-)$.






        share|cite|improve this answer









        $endgroup$



        I assume your question is, is this proof valid? The answer is yes!



        To be clearer, you should probably point out that $x + r/2 = y - r/2$, which is how you get $s < x + r/2 = y - r/2 < t$ and therefore $s le t$ (you say this is true "as stated", but I don't see you stating it).



        You might also want to say more about why a discontinuity in a monotone function must have $F(x^+) ne F(x^-)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 21:02









        Hew WolffHew Wolff

        2,260716




        2,260716






























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