The existence of left-hand limits for a strictly increasing function












0












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My question is : "Assume $f: mathbb{R} to mathbb{R}$, and $f$ is strictly increasing; that is, $f(a) < f(b)$ whenever $a<b$. Show that for any $c in mathbb{R}$, $lim_{x to c^{-}} f(x)$ exists.
" I attempt to construct two increasing sequences $(x_n)$ and $(y_n)$ both approach to the same $c$, and then I argue that limit of $f(x_n) = f(y_n)$ by using contradiction argument. Any other approach or problems with my sketch-proof?






real-analysis sequences-and-series limits







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  • $begingroup$
    Your idea is fine, but it would help if you can show some details to see whether your intended implementation of it is accurate.
    $endgroup$
    – DanielWainfleet
    Dec 6 '18 at 1:13
















0












$begingroup$


My question is : "Assume $f: mathbb{R} to mathbb{R}$, and $f$ is strictly increasing; that is, $f(a) < f(b)$ whenever $a<b$. Show that for any $c in mathbb{R}$, $lim_{x to c^{-}} f(x)$ exists.
" I attempt to construct two increasing sequences $(x_n)$ and $(y_n)$ both approach to the same $c$, and then I argue that limit of $f(x_n) = f(y_n)$ by using contradiction argument. Any other approach or problems with my sketch-proof?






real-analysis sequences-and-series limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your idea is fine, but it would help if you can show some details to see whether your intended implementation of it is accurate.
    $endgroup$
    – DanielWainfleet
    Dec 6 '18 at 1:13














0












0








0





$begingroup$


My question is : "Assume $f: mathbb{R} to mathbb{R}$, and $f$ is strictly increasing; that is, $f(a) < f(b)$ whenever $a<b$. Show that for any $c in mathbb{R}$, $lim_{x to c^{-}} f(x)$ exists.
" I attempt to construct two increasing sequences $(x_n)$ and $(y_n)$ both approach to the same $c$, and then I argue that limit of $f(x_n) = f(y_n)$ by using contradiction argument. Any other approach or problems with my sketch-proof?






real-analysis sequences-and-series limits







share|cite|improve this question









$endgroup$




My question is : "Assume $f: mathbb{R} to mathbb{R}$, and $f$ is strictly increasing; that is, $f(a) < f(b)$ whenever $a<b$. Show that for any $c in mathbb{R}$, $lim_{x to c^{-}} f(x)$ exists.
" I attempt to construct two increasing sequences $(x_n)$ and $(y_n)$ both approach to the same $c$, and then I argue that limit of $f(x_n) = f(y_n)$ by using contradiction argument. Any other approach or problems with my sketch-proof?






real-analysis sequences-and-series limits




real-analysis sequences-and-series limits






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asked Dec 5 '18 at 23:01









Dong LeDong Le

517




517












  • $begingroup$
    Your idea is fine, but it would help if you can show some details to see whether your intended implementation of it is accurate.
    $endgroup$
    – DanielWainfleet
    Dec 6 '18 at 1:13


















  • $begingroup$
    Your idea is fine, but it would help if you can show some details to see whether your intended implementation of it is accurate.
    $endgroup$
    – DanielWainfleet
    Dec 6 '18 at 1:13
















$begingroup$
Your idea is fine, but it would help if you can show some details to see whether your intended implementation of it is accurate.
$endgroup$
– DanielWainfleet
Dec 6 '18 at 1:13




$begingroup$
Your idea is fine, but it would help if you can show some details to see whether your intended implementation of it is accurate.
$endgroup$
– DanielWainfleet
Dec 6 '18 at 1:13










2 Answers
2






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Hint: $xi_c = sup { f(x) mid x < c }$ exists. Show that $lim_{xto c-} f(x) = xi_c$.



By the way, it suffices to assume that $f$ is increasing ($f(a) le f(b)$ whenever $a le b$).






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    A different method. Let $(x_n)_n$ and $(y_n)_n$ be strictly increasing sequences, each converging to $c.$ Since $(f(x_n))_n$ and $(f(y_n))_n$ are increasing and each is bounded above by $f(c),$ the limits $X=lim_{nto infty}f(x_n)$ and $Y=lim_{nto infty}f(y_n)$ exist.



    Consider the set $A={x_n:ninBbb N}cup {y_n:nin Bbb N}.$ We can enumerate $A={a_n: nin Bbb N}$ with $a_n<a_{n+1}$ for each $n$ because (i) each of ${x_n:nin Bbb N}$ and ${y_n: nin Bbb N}$ is countably infinite, so $A$ is countably infinite, and (ii) $Asubset (-infty,c)$ and for any $r<c$ the set ${ain A:a<r}$ is finite.



    Now the sequence $(f(a_n))_n$ is increasing and bounded above by $f(c)$ so it has a limit $L.$ Any sub-sequence of a convergent sequence converges to the same limit. Both $(f(x_n))_n$ and $(f(y_n))_n$ are sub-sequences of $(f(a_n))_n,$ so $X=L$ and $Y=L.$



    It is sufficient that $f$ is increasing but not necessarily strictly increasing. BTW, for $nin Bbb N$ there is a unique $g(n)$ such that $x_n=a_{g(n)},$ and $g:Bbb Nto Bbb N$ is strictly increasing. So $(f(x_n))_n=(f(a_{g(n)}))_n$ is a sub-sequence of $(f(a_n))_n.$ And similarly so is $(f(y_n))_n.$






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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      Hint: $xi_c = sup { f(x) mid x < c }$ exists. Show that $lim_{xto c-} f(x) = xi_c$.



      By the way, it suffices to assume that $f$ is increasing ($f(a) le f(b)$ whenever $a le b$).






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Hint: $xi_c = sup { f(x) mid x < c }$ exists. Show that $lim_{xto c-} f(x) = xi_c$.



        By the way, it suffices to assume that $f$ is increasing ($f(a) le f(b)$ whenever $a le b$).






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: $xi_c = sup { f(x) mid x < c }$ exists. Show that $lim_{xto c-} f(x) = xi_c$.



          By the way, it suffices to assume that $f$ is increasing ($f(a) le f(b)$ whenever $a le b$).






          share|cite|improve this answer











          $endgroup$



          Hint: $xi_c = sup { f(x) mid x < c }$ exists. Show that $lim_{xto c-} f(x) = xi_c$.



          By the way, it suffices to assume that $f$ is increasing ($f(a) le f(b)$ whenever $a le b$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 23:20

























          answered Dec 5 '18 at 23:09









          Paul FrostPaul Frost

          9,8453732




          9,8453732























              1












              $begingroup$

              A different method. Let $(x_n)_n$ and $(y_n)_n$ be strictly increasing sequences, each converging to $c.$ Since $(f(x_n))_n$ and $(f(y_n))_n$ are increasing and each is bounded above by $f(c),$ the limits $X=lim_{nto infty}f(x_n)$ and $Y=lim_{nto infty}f(y_n)$ exist.



              Consider the set $A={x_n:ninBbb N}cup {y_n:nin Bbb N}.$ We can enumerate $A={a_n: nin Bbb N}$ with $a_n<a_{n+1}$ for each $n$ because (i) each of ${x_n:nin Bbb N}$ and ${y_n: nin Bbb N}$ is countably infinite, so $A$ is countably infinite, and (ii) $Asubset (-infty,c)$ and for any $r<c$ the set ${ain A:a<r}$ is finite.



              Now the sequence $(f(a_n))_n$ is increasing and bounded above by $f(c)$ so it has a limit $L.$ Any sub-sequence of a convergent sequence converges to the same limit. Both $(f(x_n))_n$ and $(f(y_n))_n$ are sub-sequences of $(f(a_n))_n,$ so $X=L$ and $Y=L.$



              It is sufficient that $f$ is increasing but not necessarily strictly increasing. BTW, for $nin Bbb N$ there is a unique $g(n)$ such that $x_n=a_{g(n)},$ and $g:Bbb Nto Bbb N$ is strictly increasing. So $(f(x_n))_n=(f(a_{g(n)}))_n$ is a sub-sequence of $(f(a_n))_n.$ And similarly so is $(f(y_n))_n.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                A different method. Let $(x_n)_n$ and $(y_n)_n$ be strictly increasing sequences, each converging to $c.$ Since $(f(x_n))_n$ and $(f(y_n))_n$ are increasing and each is bounded above by $f(c),$ the limits $X=lim_{nto infty}f(x_n)$ and $Y=lim_{nto infty}f(y_n)$ exist.



                Consider the set $A={x_n:ninBbb N}cup {y_n:nin Bbb N}.$ We can enumerate $A={a_n: nin Bbb N}$ with $a_n<a_{n+1}$ for each $n$ because (i) each of ${x_n:nin Bbb N}$ and ${y_n: nin Bbb N}$ is countably infinite, so $A$ is countably infinite, and (ii) $Asubset (-infty,c)$ and for any $r<c$ the set ${ain A:a<r}$ is finite.



                Now the sequence $(f(a_n))_n$ is increasing and bounded above by $f(c)$ so it has a limit $L.$ Any sub-sequence of a convergent sequence converges to the same limit. Both $(f(x_n))_n$ and $(f(y_n))_n$ are sub-sequences of $(f(a_n))_n,$ so $X=L$ and $Y=L.$



                It is sufficient that $f$ is increasing but not necessarily strictly increasing. BTW, for $nin Bbb N$ there is a unique $g(n)$ such that $x_n=a_{g(n)},$ and $g:Bbb Nto Bbb N$ is strictly increasing. So $(f(x_n))_n=(f(a_{g(n)}))_n$ is a sub-sequence of $(f(a_n))_n.$ And similarly so is $(f(y_n))_n.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A different method. Let $(x_n)_n$ and $(y_n)_n$ be strictly increasing sequences, each converging to $c.$ Since $(f(x_n))_n$ and $(f(y_n))_n$ are increasing and each is bounded above by $f(c),$ the limits $X=lim_{nto infty}f(x_n)$ and $Y=lim_{nto infty}f(y_n)$ exist.



                  Consider the set $A={x_n:ninBbb N}cup {y_n:nin Bbb N}.$ We can enumerate $A={a_n: nin Bbb N}$ with $a_n<a_{n+1}$ for each $n$ because (i) each of ${x_n:nin Bbb N}$ and ${y_n: nin Bbb N}$ is countably infinite, so $A$ is countably infinite, and (ii) $Asubset (-infty,c)$ and for any $r<c$ the set ${ain A:a<r}$ is finite.



                  Now the sequence $(f(a_n))_n$ is increasing and bounded above by $f(c)$ so it has a limit $L.$ Any sub-sequence of a convergent sequence converges to the same limit. Both $(f(x_n))_n$ and $(f(y_n))_n$ are sub-sequences of $(f(a_n))_n,$ so $X=L$ and $Y=L.$



                  It is sufficient that $f$ is increasing but not necessarily strictly increasing. BTW, for $nin Bbb N$ there is a unique $g(n)$ such that $x_n=a_{g(n)},$ and $g:Bbb Nto Bbb N$ is strictly increasing. So $(f(x_n))_n=(f(a_{g(n)}))_n$ is a sub-sequence of $(f(a_n))_n.$ And similarly so is $(f(y_n))_n.$






                  share|cite|improve this answer









                  $endgroup$



                  A different method. Let $(x_n)_n$ and $(y_n)_n$ be strictly increasing sequences, each converging to $c.$ Since $(f(x_n))_n$ and $(f(y_n))_n$ are increasing and each is bounded above by $f(c),$ the limits $X=lim_{nto infty}f(x_n)$ and $Y=lim_{nto infty}f(y_n)$ exist.



                  Consider the set $A={x_n:ninBbb N}cup {y_n:nin Bbb N}.$ We can enumerate $A={a_n: nin Bbb N}$ with $a_n<a_{n+1}$ for each $n$ because (i) each of ${x_n:nin Bbb N}$ and ${y_n: nin Bbb N}$ is countably infinite, so $A$ is countably infinite, and (ii) $Asubset (-infty,c)$ and for any $r<c$ the set ${ain A:a<r}$ is finite.



                  Now the sequence $(f(a_n))_n$ is increasing and bounded above by $f(c)$ so it has a limit $L.$ Any sub-sequence of a convergent sequence converges to the same limit. Both $(f(x_n))_n$ and $(f(y_n))_n$ are sub-sequences of $(f(a_n))_n,$ so $X=L$ and $Y=L.$



                  It is sufficient that $f$ is increasing but not necessarily strictly increasing. BTW, for $nin Bbb N$ there is a unique $g(n)$ such that $x_n=a_{g(n)},$ and $g:Bbb Nto Bbb N$ is strictly increasing. So $(f(x_n))_n=(f(a_{g(n)}))_n$ is a sub-sequence of $(f(a_n))_n.$ And similarly so is $(f(y_n))_n.$







                  share|cite|improve this answer












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                  answered Dec 6 '18 at 0:58









                  DanielWainfleetDanielWainfleet

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