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The definition of neighbourhood:

If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$, which includes an open set $U$ containing $p$.

Does this say in other words that, a neighbourhood is any open set that contains $p$? Why it has to be for a set $S$, $S\subset U\subset V$ and not $S\subset V$?

LearningMath
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  • Nhoods are basically a nuisance and of secondary importance. Open sets usually suffice for nhoods and are easier to use. – William Elliot Feb 23 '18 at 02:57
  • @bof I now get it, I'm sorry, I'm not studying math, I'm learning it on my own so I have many holes in my knowledge and understanding and I try to fill them as much as I can. – LearningMath Feb 23 '18 at 16:42
  • https://math.stackexchange.com/questions/1449382/why-consider-neighbourhood-containing-an-open-set – Clemens Bartholdy Sep 02 '24 at 05:34

2 Answers2

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Usually "neighborhood" (please excuse my American spelling) is allowed to be either open or not open. So it's not enough to say $p \in V$, because then $V$ could be the set consisting only of the point $p$, which wouldn't capture the sense of the word "neighborhood" as a region of points close to $p$. Therefore we require $p \in U \subset V$ where $U$ needs to be open, and $V$ can be open or closed.

Often it won't make much of a difference, and you can just consider the open set $U$ directly.

See the question here, and particularly Pedro's answer, for some further insight.

BallBoy
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    It doesn't even necessarily have to be either open or closed, for example $[-1, 1)$ is a neighborhood of 0. (One easy result which might be useful for thinking about it is: $N$ is a neighborhood of $x$ if and only if $x$ is in the interior of $N$.) – Daniel Schepler Feb 23 '18 at 01:34
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What it means is, if $U$ is an open set and $p \in U$ then if a subset $V$ of $X$ contains $U$, $V$ is a neighbourhood of $p$. Choose $p=0$, according to the definition, a subset $V\subset R$ like $V=[-0.5,0.5]$, includes the open set $(-0.5, 0.5)$ which contains $p$ ($p$ belongs to the open set $(-0.5,0.5)$). So $V$ is a neighbourhood of $p$.

tobiasbriones
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