0
$\begingroup$

The usual definition of a neighborhood N of a point $x$ in a topological space $(X,\tau)$ is a set that includes an open set $U\in \tau$ that contains $x$ \begin{equation} x \in U \subset N \end{equation} A neighborhood itself may or may not be open. My question is why not define neighborhoods as open neighborhoods (open sets that contain the specified point). What would we lose in the simple definition (any statements/proofs relying on the more general definition)?

$\endgroup$
4
  • 3
    $\begingroup$ Actually, both definitions are used. From my experience, the definition that a neighborhood is an open set containing $x$ is actually more common, and is used in more books. This is the definition I have always used. $\endgroup$
    – Mark
    Nov 11, 2023 at 0:57
  • 2
    $\begingroup$ Offhand, we lose not much, though sometimes one does want to speak of, e.g., a compact neighborhood. <> Inversely, is anything much gained by adding a hypothesis that seldom (if ever) simplifies local arguments? $\endgroup$ Nov 11, 2023 at 0:58
  • 1
    $\begingroup$ In functional analysis it is convenient to use definition that does not require openness. In other contexts authors usually define neighborhoods to be open. $\endgroup$
    – Matsmir
    Nov 11, 2023 at 0:59
  • $\begingroup$ See my answer to math.stackexchange.com/questions/4354620/…. $\endgroup$
    – Ningxin
    Nov 11, 2023 at 7:31

0

You must log in to answer this question.

Browse other questions tagged .