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?

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    $\begingroup$ 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. $\endgroup$ Jan 9, 2019 at 17:03
  • $\begingroup$ I might be missing something but Rudin's definition, as stated in the question, does not require $N_r(a)$ to be open, but merely to contain $B(a, r)$. $\endgroup$ Jan 9, 2019 at 20:47

2 Answers 2


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.

  • $\begingroup$ I'm curious - if you define neighborhoods to necessarily be open, then what do you do once you get into the topic of filters and want to define the neighborhood filter? With the more permissive definition, you just say the set of all neighborhoods of $a$ forms a filter, end of story. $\endgroup$ Jan 9, 2019 at 23:38

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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