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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            Bundesstraße 106

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            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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            Verónica Boquete

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            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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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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