Show that 1/x is NOT uniformly continuous

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I have to prove that the function $f(x)=frac1x$ on $(0,infty)$ is not uniformly continuous (for the definition of uniform continuity see here).



I negated the definition of uniform continuity getting:



$$ exists epsilon>0 texts.t. forall delta>0 textI can always find x,y in(0,infty) texts.t. mid x-ymid<delta textbut mid f(x)-f(y)mid geq epsilon. $$



So I chose $epsilon=1$. Set $y=fracx2$. Then I have to find an $x$ such that this holds: $mid x-ymid = fracx2 <delta$ & $mid f(x)-f(y)mid = frac1x geq 1$, which is equivalent to $ x<min [2delta,1] $.



Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $delta$. Thanks!




epsilon-delta uniform-continuity




share|cite|improve this question



















  • It's just fine.
    – José Carlos Santos
    Aug 2 at 14:52














up vote
0
down vote

favorite












I have to prove that the function $f(x)=frac1x$ on $(0,infty)$ is not uniformly continuous (for the definition of uniform continuity see here).



I negated the definition of uniform continuity getting:



$$ exists epsilon>0 texts.t. forall delta>0 textI can always find x,y in(0,infty) texts.t. mid x-ymid<delta textbut mid f(x)-f(y)mid geq epsilon. $$



So I chose $epsilon=1$. Set $y=fracx2$. Then I have to find an $x$ such that this holds: $mid x-ymid = fracx2 <delta$ & $mid f(x)-f(y)mid = frac1x geq 1$, which is equivalent to $ x<min [2delta,1] $.



Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $delta$. Thanks!




epsilon-delta uniform-continuity




share|cite|improve this question



















  • It's just fine.
    – José Carlos Santos
    Aug 2 at 14:52












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to prove that the function $f(x)=frac1x$ on $(0,infty)$ is not uniformly continuous (for the definition of uniform continuity see here).



I negated the definition of uniform continuity getting:



$$ exists epsilon>0 texts.t. forall delta>0 textI can always find x,y in(0,infty) texts.t. mid x-ymid<delta textbut mid f(x)-f(y)mid geq epsilon. $$



So I chose $epsilon=1$. Set $y=fracx2$. Then I have to find an $x$ such that this holds: $mid x-ymid = fracx2 <delta$ & $mid f(x)-f(y)mid = frac1x geq 1$, which is equivalent to $ x<min [2delta,1] $.



Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $delta$. Thanks!




epsilon-delta uniform-continuity




share|cite|improve this question











I have to prove that the function $f(x)=frac1x$ on $(0,infty)$ is not uniformly continuous (for the definition of uniform continuity see here).



I negated the definition of uniform continuity getting:



$$ exists epsilon>0 texts.t. forall delta>0 textI can always find x,y in(0,infty) texts.t. mid x-ymid<delta textbut mid f(x)-f(y)mid geq epsilon. $$



So I chose $epsilon=1$. Set $y=fracx2$. Then I have to find an $x$ such that this holds: $mid x-ymid = fracx2 <delta$ & $mid f(x)-f(y)mid = frac1x geq 1$, which is equivalent to $ x<min [2delta,1] $.



Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $delta$. Thanks!





epsilon-delta uniform-continuity





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 2 at 14:49









Alessandro

142




142











  • It's just fine.
    – José Carlos Santos
    Aug 2 at 14:52
















  • It's just fine.
    – José Carlos Santos
    Aug 2 at 14:52















It's just fine.
– José Carlos Santos
Aug 2 at 14:52




It's just fine.
– José Carlos Santos
Aug 2 at 14:52










1 Answer
1






active

oldest

votes

















up vote
0
down vote













It is correct. You can choose x depending on $delta$ because $x$ is quantified before $delta$.



Alternatively, we have $$1/n - 1/(n+1) to 0$$ but



$$f(1/n) -f(1/(n+1)) = n - (n+1) to -1 $$



And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.






share|cite|improve this answer





















    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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    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%2f2870145%2fshow-that-1-x-is-not-uniformly-continuous%23new-answer', 'question_page');

    );

    Post as a guest

















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    It is correct. You can choose x depending on $delta$ because $x$ is quantified before $delta$.



    Alternatively, we have $$1/n - 1/(n+1) to 0$$ but



    $$f(1/n) -f(1/(n+1)) = n - (n+1) to -1 $$



    And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.






    share|cite|improve this answer

























      up vote
      0
      down vote













      It is correct. You can choose x depending on $delta$ because $x$ is quantified before $delta$.



      Alternatively, we have $$1/n - 1/(n+1) to 0$$ but



      $$f(1/n) -f(1/(n+1)) = n - (n+1) to -1 $$



      And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        It is correct. You can choose x depending on $delta$ because $x$ is quantified before $delta$.



        Alternatively, we have $$1/n - 1/(n+1) to 0$$ but



        $$f(1/n) -f(1/(n+1)) = n - (n+1) to -1 $$



        And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.






        share|cite|improve this answer













        It is correct. You can choose x depending on $delta$ because $x$ is quantified before $delta$.



        Alternatively, we have $$1/n - 1/(n+1) to 0$$ but



        $$f(1/n) -f(1/(n+1)) = n - (n+1) to -1 $$



        And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 15:26









        Math_QED

        6,30831344




        6,30831344






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870145%2fshow-that-1-x-is-not-uniformly-continuous%23new-answer', 'question_page');

            );

            Post as a guest
































































            Comments

            Post a Comment

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
            Read more

            Color the edges and diagonals of a regular polygon

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
            Read more

            Relationship between determinant of matrix and determinant of adjoint?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
            Read more