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Is a property called local if and only if for every point there exists a neighbourhood for which the property is true?

For example: Let $X,Y$ be topological spaces. Then $f: X \to Y$ is continuous if and only if for every $x \in X$ there exists a neighbourhood $U$ such that $f \mid_U$ is continuous?

If so what would be another example of a local property?

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    $\begingroup$ I suggest to dig especially into the differences between connected, locally connected, semilocally connected. Which if these is local? $\endgroup$ – Hagen von Eitzen Dec 18 '14 at 11:33
  • $\begingroup$ @HagenvonEitzen THank you so much for your comment! I will get back to you shortly, with the results of my investigation. $\endgroup$ – a student Dec 19 '14 at 0:31
  • $\begingroup$ Your question is also adressed in en.wikipedia.org/wiki/Local_property $\endgroup$ – Stephan Kulla Feb 15 '16 at 15:15
  • $\begingroup$ @tampis Thank you for your comment. $\endgroup$ – a student Feb 16 '16 at 1:11
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Not only should the property be true for a neighborhood of each point.
It must also be the case that having a neighborhood with the given property around each point implies that the entire space satisfies the property (satisfying a property locally is not the same as the property being local).

For example, being open is a local property (by which I mean that whether a subspace is open can be checked around each point of said subspace).

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    $\begingroup$ Could you give an example where "the property be true for a neighborhood of each point" is not enough? $\endgroup$ – JiK Dec 18 '14 at 10:49
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    $\begingroup$ @JiK As mentioned in another answer, a manifold is something that is homeomorphic to $\mathbb{R}^n$ in a neighborhood of each point. But for example the unit circle is not itself homeomorphic to $\mathbb{R}$. $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 10:56
  • $\begingroup$ @TobiasKildetoft: I just searched and saw your answer. I feel it may related to the question that currently bother me. Can you help me at there? :) $\endgroup$ – Eric Nov 7 '17 at 5:13
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To add to the existing answers, be careful about two different terminologies:

  • a property $P$ is local when, for all spaces $X$, if $\{U_i\}$ is an open cover of $X$ and all the $U_i$ have $P$, then $X$ has $P$. (cf. also Tobias Kildetoft for an equivalent characterization)

  • if $P$ is some property of topological spaces, a space $X$ is said to be locally $P$ if every point of $X$ has a neighborhood basis of subspaces satisfying $P$.

So for example the property $P$ of being connected is not local, but you can talk about a space being locally connected, which is completely unrelated to the connectedness of the space.

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  • $\begingroup$ Good answer! Sadly, the second description of how “locally” is used and understood isn’t always applicable: Many topologists think of a locally compact space as one with each point having a compact neighbourhood, not an entire local base of compact neighbourhoods. $\endgroup$ – k.stm Dec 18 '14 at 8:41
  • $\begingroup$ @k.stm The thing is that for the property "compact", the two definitions are equivalent in good cases. For a general property, you really want a whole basis of neighborhoods. If you want to be consistent, definition (3) is the correct one, as you can see by the fact that "Compactness implies conditions (1) and (2), but not (3)." $\endgroup$ – Najib Idrissi Dec 18 '14 at 8:43
  • $\begingroup$ I think they are not equivalent in general. But I agree with you in that I prefer definition (3) over the others. $\endgroup$ – k.stm Dec 18 '14 at 8:50
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    $\begingroup$ @k.stm They are equivalent for Hausdorff spaces. But really, everyone should just drop topological spaces and use simplicial sets... $\endgroup$ – Najib Idrissi Dec 18 '14 at 8:51
  • $\begingroup$ Are you sure about the part "for all spaces $X$" in your first definition? Take the following example from this answer: Let $X$ be the long line by pasting $[0,1)$ uncountable many times together. Define $f:X\to\mathbb R$ so that $f(x)=\sin(2\pi x)$ on each interval $[0,1)$. As described in math.stackexchange.com/a/166340/32951 this function is not continuous although one finds an open cover $\{U_i\}$ of $X$ where all $f|_{U_i}$ are continuous. $\endgroup$ – Stephan Kulla Feb 15 '16 at 18:23
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Your intuition is correct. Another example is the definition of a manifold; something which is locally homeomorphic to $\mathbb{R}^{n}$.

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    $\begingroup$ There is a difference between a property being local and a space satisfying a property locally. $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 8:10
  • $\begingroup$ Property $p$ is local on $X$. The space $X$ satisfies property $p$ locally. təˈmātō,-ˈmätō $\endgroup$ – parsiad Dec 18 '14 at 8:12
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    $\begingroup$ No, a property being local is not related to which space we are looking at. Being local is about whether the property holding locally implies it holding globally. $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 8:13

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