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The def on wiki: "If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V \subseteq X$ that includes an open set $U$ containing $p$."

And it says: "Note that the neighbourhood $V$ need not be an open set itself."

Later, it says: "A set that is a neighbourhood of each of its points is open."

The claims above seem odd; one says a neighborhood need not be open; the other, be open. I know the difference is one says a point; the other, a set. Is there any other strict explanation?

Thanks

Asaf Karagila
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sleeve chen
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    The second sentence does not say that a neighbourhood of any point is open. It says that if it is a neighbourhood of each of its points, then it is open. Not every neighbourhood of some point has this property. – Tunococ Sep 07 '14 at 09:43
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    consider, in $\mathbb{R}^2$, the closed disk of radius 1 about the origin. This is a neighborhood of the point $(0,0)$, since it contains the open disk of radius 1 about the point $(0,0)$. However, this is not a neighborhood for the point $(1,0)$ since there is no open set about the point $(1,0)$! Therefore, the closed disk is not open. Good. haha – Eoin Sep 07 '14 at 09:44
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    @chen No. The second says that a neighbourhood is only open if it is a neighbourhood of each of its points. For example, let $X=\mathbb{R}$ and $V=[0,1]$. Then $V$ is a neighbourhood of each of the points in $(0,1)$, but it is not a neighbourhood of 0 (or 1). – almagest Sep 07 '14 at 09:45
  • @almagest. I am confused about your answer for the following: [0,1],which is the neighborhood of each of points in (0,1) is closed, not open. However, before it, you say a neighbourhood is only open if it is a neighbourhood of each of its points. – sleeve chen Sep 07 '14 at 10:19
  • Which half of it: that $V$ is a nei of 1/2, or that it is not a nei of 0? – almagest Sep 07 '14 at 10:20

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The statement you have doesn't contradict the fact that a neighborhood need not be open, it just happens to be open, given the added condition that it is a neighborhood to each of its points:

Let $V$ be a neighborhood of some point $p$. Suppose $V$ is a neighborhood for each of its points, that means if $x \in V$, then there exists $U$ open with $x \in U \subseteq V$. Conclude that $V$ is open by definition of open set.

Robert Cardona
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