Accumulation points: an example of a sequence that has exactly two accumulation points












5














The definition in my text defines A.P. as: A point p is an A.P. of a set S if it is the limit of a sequence of points of S-{p}.



I am confused by my classnotes and information about A.P. in my textbook is not enough for me to understand the topic.






real-analysis sequences-and-series







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  • So: "accumulation point" is for a set not for a sequence ??
    – GEdgar
    Oct 24 '12 at 21:24
















5














The definition in my text defines A.P. as: A point p is an A.P. of a set S if it is the limit of a sequence of points of S-{p}.



I am confused by my classnotes and information about A.P. in my textbook is not enough for me to understand the topic.






real-analysis sequences-and-series







share|cite|improve this question






















  • So: "accumulation point" is for a set not for a sequence ??
    – GEdgar
    Oct 24 '12 at 21:24














5












5








5


2





The definition in my text defines A.P. as: A point p is an A.P. of a set S if it is the limit of a sequence of points of S-{p}.



I am confused by my classnotes and information about A.P. in my textbook is not enough for me to understand the topic.






real-analysis sequences-and-series







share|cite|improve this question













The definition in my text defines A.P. as: A point p is an A.P. of a set S if it is the limit of a sequence of points of S-{p}.



I am confused by my classnotes and information about A.P. in my textbook is not enough for me to understand the topic.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question













share|cite|improve this question











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asked Oct 24 '12 at 20:48









Maximiliano

43021225




43021225












  • So: "accumulation point" is for a set not for a sequence ??
    – GEdgar
    Oct 24 '12 at 21:24


















  • So: "accumulation point" is for a set not for a sequence ??
    – GEdgar
    Oct 24 '12 at 21:24
















So: "accumulation point" is for a set not for a sequence ??
– GEdgar
Oct 24 '12 at 21:24




So: "accumulation point" is for a set not for a sequence ??
– GEdgar
Oct 24 '12 at 21:24










4 Answers
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4














The sequence $(-1)^nleft(1+frac{1}{n}right)$ has exactly two A.Ps $1$ and $-1$.



More generally if $(a_n)_{nin mathbb{N}}$ is a sequence s.t. $a_n to a neq0 $ and the set ${a_n:n in mathbb{N}}$ is infinite then the sequence $((-1)^na_n)_{nin mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.






share|cite|improve this answer























  • how can this be proved?
    – Maximiliano
    Oct 24 '12 at 21:14












  • Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
    – P..
    Nov 26 at 0:53





















6














Exactly two accumulation points can be done using the sequence
$$frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots,$$
and in many other ways. In this case the accumulation points are $0$ and $1$.



Here I intend the set $S$ of the definition to be ${frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.






share|cite|improve this answer























  • I presume your set is the unit interval?
    – Lord_Farin
    Oct 24 '12 at 20:57










  • No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
    – André Nicolas
    Oct 24 '12 at 21:05










  • Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
    – Lord_Farin
    Oct 24 '12 at 21:05



















1














This seems a little vague.
What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?



Intuitively, accumulations points are the points of the set S which are not isolated.



e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) cup$ {2}



from there its pretty easy to construct such a sequence.






share|cite|improve this answer





















  • Yes an example of a sequence
    – Maximiliano
    Oct 24 '12 at 20:58



















1














Point $p$ is accumulation point of a set $S$, if every punctured neighborhood of the point $p$ contains points of $S.$ (In other words, if there exists sequence ${x_n} subset S$ that converges to $p.$)
Set $$S={left lbrace sin{dfrac{1}{n}}, cos{dfrac{1}{n}} right rbrace}_{n in mathbb{N}}=left lbrace sin{1}, cos{1},sin{dfrac{1}{2}}, cos{dfrac{1}{2}}, ldots, sin{dfrac{1}{n}}, cos{dfrac{1}{n}} , ldots right rbrace$$ has exactly two accumulation points.






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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    The sequence $(-1)^nleft(1+frac{1}{n}right)$ has exactly two A.Ps $1$ and $-1$.



    More generally if $(a_n)_{nin mathbb{N}}$ is a sequence s.t. $a_n to a neq0 $ and the set ${a_n:n in mathbb{N}}$ is infinite then the sequence $((-1)^na_n)_{nin mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.






    share|cite|improve this answer























    • how can this be proved?
      – Maximiliano
      Oct 24 '12 at 21:14












    • Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
      – P..
      Nov 26 at 0:53


















    4














    The sequence $(-1)^nleft(1+frac{1}{n}right)$ has exactly two A.Ps $1$ and $-1$.



    More generally if $(a_n)_{nin mathbb{N}}$ is a sequence s.t. $a_n to a neq0 $ and the set ${a_n:n in mathbb{N}}$ is infinite then the sequence $((-1)^na_n)_{nin mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.






    share|cite|improve this answer























    • how can this be proved?
      – Maximiliano
      Oct 24 '12 at 21:14












    • Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
      – P..
      Nov 26 at 0:53
















    4












    4








    4






    The sequence $(-1)^nleft(1+frac{1}{n}right)$ has exactly two A.Ps $1$ and $-1$.



    More generally if $(a_n)_{nin mathbb{N}}$ is a sequence s.t. $a_n to a neq0 $ and the set ${a_n:n in mathbb{N}}$ is infinite then the sequence $((-1)^na_n)_{nin mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.






    share|cite|improve this answer














    The sequence $(-1)^nleft(1+frac{1}{n}right)$ has exactly two A.Ps $1$ and $-1$.



    More generally if $(a_n)_{nin mathbb{N}}$ is a sequence s.t. $a_n to a neq0 $ and the set ${a_n:n in mathbb{N}}$ is infinite then the sequence $((-1)^na_n)_{nin mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 at 0:49

























    answered Oct 24 '12 at 21:10









    P..

    13.4k22348




    13.4k22348












    • how can this be proved?
      – Maximiliano
      Oct 24 '12 at 21:14












    • Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
      – P..
      Nov 26 at 0:53




















    • how can this be proved?
      – Maximiliano
      Oct 24 '12 at 21:14












    • Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
      – P..
      Nov 26 at 0:53


















    how can this be proved?
    – Maximiliano
    Oct 24 '12 at 21:14






    how can this be proved?
    – Maximiliano
    Oct 24 '12 at 21:14














    Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
    – P..
    Nov 26 at 0:53






    Prove that $a_{2n}rightarrow a , a_{2n+1}rightarrow −a ,a_{2n}neq a ,a_{2n+1}neq −a$, for (almost) all $ninmathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{ninmathbb N}$.
    – P..
    Nov 26 at 0:53













    6














    Exactly two accumulation points can be done using the sequence
    $$frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots,$$
    and in many other ways. In this case the accumulation points are $0$ and $1$.



    Here I intend the set $S$ of the definition to be ${frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.






    share|cite|improve this answer























    • I presume your set is the unit interval?
      – Lord_Farin
      Oct 24 '12 at 20:57










    • No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
      – André Nicolas
      Oct 24 '12 at 21:05










    • Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
      – Lord_Farin
      Oct 24 '12 at 21:05
















    6














    Exactly two accumulation points can be done using the sequence
    $$frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots,$$
    and in many other ways. In this case the accumulation points are $0$ and $1$.



    Here I intend the set $S$ of the definition to be ${frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.






    share|cite|improve this answer























    • I presume your set is the unit interval?
      – Lord_Farin
      Oct 24 '12 at 20:57










    • No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
      – André Nicolas
      Oct 24 '12 at 21:05










    • Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
      – Lord_Farin
      Oct 24 '12 at 21:05














    6












    6








    6






    Exactly two accumulation points can be done using the sequence
    $$frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots,$$
    and in many other ways. In this case the accumulation points are $0$ and $1$.



    Here I intend the set $S$ of the definition to be ${frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.






    share|cite|improve this answer














    Exactly two accumulation points can be done using the sequence
    $$frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots,$$
    and in many other ways. In this case the accumulation points are $0$ and $1$.



    Here I intend the set $S$ of the definition to be ${frac{1}{3}, frac{2}{3},frac{1}{4},frac{3}{4}, frac{1}{5},frac{4}{5},frac{1}{6}, frac{5}{6},frac{1}{7}, frac{6}{7},dots}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 24 '12 at 21:09

























    answered Oct 24 '12 at 20:54









    André Nicolas

    451k36421805




    451k36421805












    • I presume your set is the unit interval?
      – Lord_Farin
      Oct 24 '12 at 20:57










    • No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
      – André Nicolas
      Oct 24 '12 at 21:05










    • Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
      – Lord_Farin
      Oct 24 '12 at 21:05


















    • I presume your set is the unit interval?
      – Lord_Farin
      Oct 24 '12 at 20:57










    • No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
      – André Nicolas
      Oct 24 '12 at 21:05










    • Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
      – Lord_Farin
      Oct 24 '12 at 21:05
















    I presume your set is the unit interval?
    – Lord_Farin
    Oct 24 '12 at 20:57




    I presume your set is the unit interval?
    – Lord_Farin
    Oct 24 '12 at 20:57












    No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
    – André Nicolas
    Oct 24 '12 at 21:05




    No, my set $S$ is the set ${frac{1}{3},frac{2}{3},frac{1}{4},dots}$.
    – André Nicolas
    Oct 24 '12 at 21:05












    Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
    – Lord_Farin
    Oct 24 '12 at 21:05




    Of course; I just wanted to explicate stuff in case OP desired that $p in S$.
    – Lord_Farin
    Oct 24 '12 at 21:05











    1














    This seems a little vague.
    What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?



    Intuitively, accumulations points are the points of the set S which are not isolated.



    e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) cup$ {2}



    from there its pretty easy to construct such a sequence.






    share|cite|improve this answer





















    • Yes an example of a sequence
      – Maximiliano
      Oct 24 '12 at 20:58
















    1














    This seems a little vague.
    What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?



    Intuitively, accumulations points are the points of the set S which are not isolated.



    e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) cup$ {2}



    from there its pretty easy to construct such a sequence.






    share|cite|improve this answer





















    • Yes an example of a sequence
      – Maximiliano
      Oct 24 '12 at 20:58














    1












    1








    1






    This seems a little vague.
    What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?



    Intuitively, accumulations points are the points of the set S which are not isolated.



    e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) cup$ {2}



    from there its pretty easy to construct such a sequence.






    share|cite|improve this answer












    This seems a little vague.
    What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?



    Intuitively, accumulations points are the points of the set S which are not isolated.



    e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) cup$ {2}



    from there its pretty easy to construct such a sequence.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 24 '12 at 20:57









    Henrique Tyrrell

    784623




    784623












    • Yes an example of a sequence
      – Maximiliano
      Oct 24 '12 at 20:58


















    • Yes an example of a sequence
      – Maximiliano
      Oct 24 '12 at 20:58
















    Yes an example of a sequence
    – Maximiliano
    Oct 24 '12 at 20:58




    Yes an example of a sequence
    – Maximiliano
    Oct 24 '12 at 20:58











    1














    Point $p$ is accumulation point of a set $S$, if every punctured neighborhood of the point $p$ contains points of $S.$ (In other words, if there exists sequence ${x_n} subset S$ that converges to $p.$)
    Set $$S={left lbrace sin{dfrac{1}{n}}, cos{dfrac{1}{n}} right rbrace}_{n in mathbb{N}}=left lbrace sin{1}, cos{1},sin{dfrac{1}{2}}, cos{dfrac{1}{2}}, ldots, sin{dfrac{1}{n}}, cos{dfrac{1}{n}} , ldots right rbrace$$ has exactly two accumulation points.






    share|cite|improve this answer




























      1














      Point $p$ is accumulation point of a set $S$, if every punctured neighborhood of the point $p$ contains points of $S.$ (In other words, if there exists sequence ${x_n} subset S$ that converges to $p.$)
      Set $$S={left lbrace sin{dfrac{1}{n}}, cos{dfrac{1}{n}} right rbrace}_{n in mathbb{N}}=left lbrace sin{1}, cos{1},sin{dfrac{1}{2}}, cos{dfrac{1}{2}}, ldots, sin{dfrac{1}{n}}, cos{dfrac{1}{n}} , ldots right rbrace$$ has exactly two accumulation points.






      share|cite|improve this answer


























        1












        1








        1






        Point $p$ is accumulation point of a set $S$, if every punctured neighborhood of the point $p$ contains points of $S.$ (In other words, if there exists sequence ${x_n} subset S$ that converges to $p.$)
        Set $$S={left lbrace sin{dfrac{1}{n}}, cos{dfrac{1}{n}} right rbrace}_{n in mathbb{N}}=left lbrace sin{1}, cos{1},sin{dfrac{1}{2}}, cos{dfrac{1}{2}}, ldots, sin{dfrac{1}{n}}, cos{dfrac{1}{n}} , ldots right rbrace$$ has exactly two accumulation points.






        share|cite|improve this answer














        Point $p$ is accumulation point of a set $S$, if every punctured neighborhood of the point $p$ contains points of $S.$ (In other words, if there exists sequence ${x_n} subset S$ that converges to $p.$)
        Set $$S={left lbrace sin{dfrac{1}{n}}, cos{dfrac{1}{n}} right rbrace}_{n in mathbb{N}}=left lbrace sin{1}, cos{1},sin{dfrac{1}{2}}, cos{dfrac{1}{2}}, ldots, sin{dfrac{1}{n}}, cos{dfrac{1}{n}} , ldots right rbrace$$ has exactly two accumulation points.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Oct 24 '12 at 21:33

























        answered Oct 24 '12 at 21:05









        M. Strochyk

        7,67711119




        7,67711119






























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            Le Mesnil-Réaume

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            Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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            web3.py web3.isConnected() returns false always

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            1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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