[Math] Is Neighborhood an open set

general-topology

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 $a.$
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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?

Best 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.