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












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












$begingroup$


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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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






active

oldest

votes


















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












    $begingroup$

    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027780%2fthe-existence-of-left-hand-limits-for-a-strictly-increasing-function%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 0:58









                  DanielWainfleetDanielWainfleet

                  34.6k31648




                  34.6k31648






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027780%2fthe-existence-of-left-hand-limits-for-a-strictly-increasing-function%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Berounka

                      Image
                      Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                      Read more

                      Sphinx de Gizeh

                      Image
                      Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                      Read more

                      Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                      Image
                      3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                      Read more