Is Neighborhood an open set?












3














I am reading a book where it is written that ,




Let $(X,d)$ be any metric space $a in X$ then for any $r gt 0$ the set $S_r(a)$ ={$x in X$ : $d(x,a) lt r$} is called an open ball of radius $r$ centered at $a.$
&
Let $(X,d)$ be any metric space and $x in X$. A subset $N{(a)}$ of $X$ is called a neighborhood of a point $a$ , if there exist an open ball $S_r(a)$ centered at $a$ and contained in $N{(a)}$ i.e $S_r(a)$ $subseteq$ $N{(a)}$.




But in Rudin ,it is given that in a metric space $X$
a neighborhood of $a$ is a set $N_r(a)$ containing of all q such that $d(a,q) lt r$, for some $r gt 0$ ,the number $r$ is called the radius of $a$ .
According to the definition of Rudin every neighborhood is an open set.
But according to the text which I am reading, does it tell that every neighborhood is an open set?






general-topology







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  • 1




    There tend to do be two definitions of "neighborhood of $a$." I believe the older older one has this connotation - just anyset containing an open ball around $a.$ The newer one, as per Rudin, has neighborhoods being open and containing $a.$ The older form has more use when discussing "continuity at a point," rather than continuity on a whole space. But it also turns out that continuitity at a point can be made into continuity on the whole space under a different topology, so it is less relevant. Hence, I believe Rudin's approach is now preferred.
    – Thomas Andrews
    2 hours ago
















3














I am reading a book where it is written that ,




Let $(X,d)$ be any metric space $a in X$ then for any $r gt 0$ the set $S_r(a)$ ={$x in X$ : $d(x,a) lt r$} is called an open ball of radius $r$ centered at $a.$
&
Let $(X,d)$ be any metric space and $x in X$. A subset $N{(a)}$ of $X$ is called a neighborhood of a point $a$ , if there exist an open ball $S_r(a)$ centered at $a$ and contained in $N{(a)}$ i.e $S_r(a)$ $subseteq$ $N{(a)}$.




But in Rudin ,it is given that in a metric space $X$
a neighborhood of $a$ is a set $N_r(a)$ containing of all q such that $d(a,q) lt r$, for some $r gt 0$ ,the number $r$ is called the radius of $a$ .
According to the definition of Rudin every neighborhood is an open set.
But according to the text which I am reading, does it tell that every neighborhood is an open set?






general-topology







share|cite|improve this question




















  • 1




    There tend to do be two definitions of "neighborhood of $a$." I believe the older older one has this connotation - just anyset containing an open ball around $a.$ The newer one, as per Rudin, has neighborhoods being open and containing $a.$ The older form has more use when discussing "continuity at a point," rather than continuity on a whole space. But it also turns out that continuitity at a point can be made into continuity on the whole space under a different topology, so it is less relevant. Hence, I believe Rudin's approach is now preferred.
    – Thomas Andrews
    2 hours ago














3












3








3


1





I am reading a book where it is written that ,




Let $(X,d)$ be any metric space $a in X$ then for any $r gt 0$ the set $S_r(a)$ ={$x in X$ : $d(x,a) lt r$} is called an open ball of radius $r$ centered at $a.$
&
Let $(X,d)$ be any metric space and $x in X$. A subset $N{(a)}$ of $X$ is called a neighborhood of a point $a$ , if there exist an open ball $S_r(a)$ centered at $a$ and contained in $N{(a)}$ i.e $S_r(a)$ $subseteq$ $N{(a)}$.




But in Rudin ,it is given that in a metric space $X$
a neighborhood of $a$ is a set $N_r(a)$ containing of all q such that $d(a,q) lt r$, for some $r gt 0$ ,the number $r$ is called the radius of $a$ .
According to the definition of Rudin every neighborhood is an open set.
But according to the text which I am reading, does it tell that every neighborhood is an open set?






general-topology







share|cite|improve this question















I am reading a book where it is written that ,




Let $(X,d)$ be any metric space $a in X$ then for any $r gt 0$ the set $S_r(a)$ ={$x in X$ : $d(x,a) lt r$} is called an open ball of radius $r$ centered at $a.$
&
Let $(X,d)$ be any metric space and $x in X$. A subset $N{(a)}$ of $X$ is called a neighborhood of a point $a$ , if there exist an open ball $S_r(a)$ centered at $a$ and contained in $N{(a)}$ i.e $S_r(a)$ $subseteq$ $N{(a)}$.




But in Rudin ,it is given that in a metric space $X$
a neighborhood of $a$ is a set $N_r(a)$ containing of all q such that $d(a,q) lt r$, for some $r gt 0$ ,the number $r$ is called the radius of $a$ .
According to the definition of Rudin every neighborhood is an open set.
But according to the text which I am reading, does it tell that every neighborhood is an open set?






general-topology




general-topology






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share|cite|improve this question













share|cite|improve this question




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edited 3 hours ago









Thomas Shelby

2,017219




2,017219










asked 3 hours ago









Supriyo BanerjeeSupriyo Banerjee

836




836








  • 1




    There tend to do be two definitions of "neighborhood of $a$." I believe the older older one has this connotation - just anyset containing an open ball around $a.$ The newer one, as per Rudin, has neighborhoods being open and containing $a.$ The older form has more use when discussing "continuity at a point," rather than continuity on a whole space. But it also turns out that continuitity at a point can be made into continuity on the whole space under a different topology, so it is less relevant. Hence, I believe Rudin's approach is now preferred.
    – Thomas Andrews
    2 hours ago














  • 1




    There tend to do be two definitions of "neighborhood of $a$." I believe the older older one has this connotation - just anyset containing an open ball around $a.$ The newer one, as per Rudin, has neighborhoods being open and containing $a.$ The older form has more use when discussing "continuity at a point," rather than continuity on a whole space. But it also turns out that continuitity at a point can be made into continuity on the whole space under a different topology, so it is less relevant. Hence, I believe Rudin's approach is now preferred.
    – Thomas Andrews
    2 hours ago








1




1




There tend to do be two definitions of "neighborhood of $a$." I believe the older older one has this connotation - just anyset containing an open ball around $a.$ The newer one, as per Rudin, has neighborhoods being open and containing $a.$ The older form has more use when discussing "continuity at a point," rather than continuity on a whole space. But it also turns out that continuitity at a point can be made into continuity on the whole space under a different topology, so it is less relevant. Hence, I believe Rudin's approach is now preferred.
– Thomas Andrews
2 hours ago




There tend to do be two definitions of "neighborhood of $a$." I believe the older older one has this connotation - just anyset containing an open ball around $a.$ The newer one, as per Rudin, has neighborhoods being open and containing $a.$ The older form has more use when discussing "continuity at a point," rather than continuity on a whole space. But it also turns out that continuitity at a point can be made into continuity on the whole space under a different topology, so it is less relevant. Hence, I believe Rudin's approach is now preferred.
– Thomas Andrews
2 hours ago










2 Answers
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There are different notions of neighbourhoods floating around. Usually one calls Rudin's approach open neighbourhoods to avoid confusion. Whereas the one you just cited is just the ordinary neighbourhood definition (if people use open neighbourhoods instead of neighbourhoods always, they usually say that in the introduction). Generally speaking a neighbourhood of $x$ is just a set $X$ such that it contains an open set $U$ with $x in U$.



In particular every neighbourhood contains an open neighbourhood, and so passing from general neighbourhoods to open ones contained in them is not too hard. Passing to closed neighbourhoods in a given neighbourhood however is more difficult and is one of the reasons one likes to have a regular Hausdorff (also known as $T_3$) spaces.






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  • sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
    – Enkidu
    3 hours ago





















2














Clearly not. Take $X = mathbb R$ with the usual metric, then $(-1, 1]$ contains a ball $(-1/2, 1/2)$ centered at $0$, so $(-1, 1]$ is a neighborhood of $0$, but itself is not open.






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    2 Answers
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    5














    There are different notions of neighbourhoods floating around. Usually one calls Rudin's approach open neighbourhoods to avoid confusion. Whereas the one you just cited is just the ordinary neighbourhood definition (if people use open neighbourhoods instead of neighbourhoods always, they usually say that in the introduction). Generally speaking a neighbourhood of $x$ is just a set $X$ such that it contains an open set $U$ with $x in U$.



    In particular every neighbourhood contains an open neighbourhood, and so passing from general neighbourhoods to open ones contained in them is not too hard. Passing to closed neighbourhoods in a given neighbourhood however is more difficult and is one of the reasons one likes to have a regular Hausdorff (also known as $T_3$) spaces.






    share|cite|improve this answer























    • sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
      – Enkidu
      3 hours ago


















    5














    There are different notions of neighbourhoods floating around. Usually one calls Rudin's approach open neighbourhoods to avoid confusion. Whereas the one you just cited is just the ordinary neighbourhood definition (if people use open neighbourhoods instead of neighbourhoods always, they usually say that in the introduction). Generally speaking a neighbourhood of $x$ is just a set $X$ such that it contains an open set $U$ with $x in U$.



    In particular every neighbourhood contains an open neighbourhood, and so passing from general neighbourhoods to open ones contained in them is not too hard. Passing to closed neighbourhoods in a given neighbourhood however is more difficult and is one of the reasons one likes to have a regular Hausdorff (also known as $T_3$) spaces.






    share|cite|improve this answer























    • sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
      – Enkidu
      3 hours ago
















    5












    5








    5






    There are different notions of neighbourhoods floating around. Usually one calls Rudin's approach open neighbourhoods to avoid confusion. Whereas the one you just cited is just the ordinary neighbourhood definition (if people use open neighbourhoods instead of neighbourhoods always, they usually say that in the introduction). Generally speaking a neighbourhood of $x$ is just a set $X$ such that it contains an open set $U$ with $x in U$.



    In particular every neighbourhood contains an open neighbourhood, and so passing from general neighbourhoods to open ones contained in them is not too hard. Passing to closed neighbourhoods in a given neighbourhood however is more difficult and is one of the reasons one likes to have a regular Hausdorff (also known as $T_3$) spaces.






    share|cite|improve this answer














    There are different notions of neighbourhoods floating around. Usually one calls Rudin's approach open neighbourhoods to avoid confusion. Whereas the one you just cited is just the ordinary neighbourhood definition (if people use open neighbourhoods instead of neighbourhoods always, they usually say that in the introduction). Generally speaking a neighbourhood of $x$ is just a set $X$ such that it contains an open set $U$ with $x in U$.



    In particular every neighbourhood contains an open neighbourhood, and so passing from general neighbourhoods to open ones contained in them is not too hard. Passing to closed neighbourhoods in a given neighbourhood however is more difficult and is one of the reasons one likes to have a regular Hausdorff (also known as $T_3$) spaces.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 3 hours ago

























    answered 3 hours ago









    EnkiduEnkidu

    1,12319




    1,12319












    • sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
      – Enkidu
      3 hours ago




















    • sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
      – Enkidu
      3 hours ago


















    sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
    – Enkidu
    3 hours ago






    sorry meant $T_3$ or regular Hausdorff, also I already changed it. Thanks for the remark.
    – Enkidu
    3 hours ago













    2














    Clearly not. Take $X = mathbb R$ with the usual metric, then $(-1, 1]$ contains a ball $(-1/2, 1/2)$ centered at $0$, so $(-1, 1]$ is a neighborhood of $0$, but itself is not open.






    share|cite|improve this answer


























      2














      Clearly not. Take $X = mathbb R$ with the usual metric, then $(-1, 1]$ contains a ball $(-1/2, 1/2)$ centered at $0$, so $(-1, 1]$ is a neighborhood of $0$, but itself is not open.






      share|cite|improve this answer
























        2












        2








        2






        Clearly not. Take $X = mathbb R$ with the usual metric, then $(-1, 1]$ contains a ball $(-1/2, 1/2)$ centered at $0$, so $(-1, 1]$ is a neighborhood of $0$, but itself is not open.






        share|cite|improve this answer












        Clearly not. Take $X = mathbb R$ with the usual metric, then $(-1, 1]$ contains a ball $(-1/2, 1/2)$ centered at $0$, so $(-1, 1]$ is a neighborhood of $0$, but itself is not open.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        xbhxbh

        5,8281522




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