I came across this question while thinking about the question whether the integrability is a local property. I first thought that the MSE question What's the definition of a "local property"? addresses this issue, but this question only answers what a local property for a topological space and not for a function is. Thus, my question is:

What is the definition for a property to be a local property of a function?

My attempt: Najib Idrissi gives in his answer to the question What's the definition of a "local property"? the answer

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 $P$.

From this one may deduce that

A property $P$ is local for a function $f:X\to Y$, iff from the fact, that $\{U_i\}$ is an open cover of $X$ and all $f|_{U_i}$ have the property $P$, follows that also $f$ has the property $P$.

Is this the right definition?

Update: First I had the following wrong argumentation in my question as pointed out by Daniel Fischer in the comments. However, the main question remains...

However, this does not seem to be the right definition. Let $X$ be the long line which we get by pasting $[0,1)$ uncountable many times together. We define $g:X\to\mathbb R$ with $g(x)=\sin(2\pi x)$ on each interval $[0,1)$. As J. Loreaux argued in his answer to the question Does this intuition for "calculus-ish" continuity generalize to topological continuity?, the function $g$ is not continuous.

This contradicts the above attempt for a definition of a function's local property since there is an (uncountable) open cover $\{U_i\}$ of the long line $X$ for which all $g|_{U_i}$ are continuous (for example when all $U_i$ are bounded) and continuity is one of the properties I would see as being a local property.

Also the Wikipedia article "Local property" does only answer what a local property for topological spaces is. It defines the term "being locally equivalent", but this is not what I am looking for.

So: What is the definition of a local property for a function?

  • 2
    $\begingroup$ There is no open cover such that $g\lvert_{U_i}$ is continuous for all $i$. On every neighbourhood of a limit ordinal, $g$ attains all values in $[-1,1]$ (infinitely often). $\endgroup$ Feb 15, 2016 at 21:57
  • $\begingroup$ I would say that it depends on the area. For example in dynamics, what you describe as a possible notion of local property is basically of no interest, and to be local usually means that it happens in some sufficiently small neighborhood (with given properties in that neighborhood). Sometimes this is called semilocal, and we reserve local for properties that need to be formulated in terms of germs (it also depends on schools). $\endgroup$
    – John B
    Feb 15, 2016 at 22:06
  • $\begingroup$ @DanielFischer: Oh, you are right... Would you say, that the above attempt of a definition is right? $\endgroup$ Feb 15, 2016 at 22:19
  • 1
    $\begingroup$ It's one way of looking at it. But one can also look in the other direction. When we say that continuity is a local property, then the open-cover-viewpoint tells you that global continuity is a local property. But we can also be interested in continuity at some point, say $x_0$, and that is local in the sense that it depends only on the behaviour of $f$ on arbitrarily small neighbourhoods of $x_0$. Taken to the end, it's a property of the germ of $f$ at $x_0$. $\endgroup$ Feb 15, 2016 at 22:25
  • $\begingroup$ An equivalent def'n that $P$ is local is :$ X$ has $ P$ iff for every $x\in X$ there is an open $U$ where $x\in U$, and $U$ has $P.$ For example a function is continuous iff it is locally continuous. $\endgroup$ Feb 15, 2016 at 23:28

1 Answer 1


To summarize the discussion in the comments: There are currently the following proposed definitions for saying that a property P is a local property of a function $f:X\to Y$:

  1. For any open cover $\{U_i\}$ $X$: All $f|_{U_i}$ have the property $P$, iff $f$ has the property $P$.
  2. Comment by user254665: $f$ has the property $P$, iff for every $x$ there is an open $U$ with $x\in U$ and $f|_U$ has the property $P$.
  3. Comment by Daniel Fischer: $P$ is a local property of $f$ in $x\in X$, iff $P$ is a property of the germ $[f]_x$ of $f$ in $x$. This means: If $f$ has the property $P$ in $x$ and $g$ coincide with $f$ in any neighborhood of $x$, then also $g$ has the property $P$ in $x$.

(1) and (2) are equivalent since (2) is a reformulation of (1).

(1) does not imply (3) and vice versa. Take for example the property "$f$ is continuous". It fulfills (1) but not (3). Take on the over hand the property "$f$ is continuous in $x$". It fulfills (3) but not (1) since $f|U_i$ is only continuous in $x$ for $x\in U_i$.

Note that I have added the implication "$f$ has property $P$ $\Rightarrow$ all $f|_{U_i}$ have property $P$" in (1).

  • $\begingroup$ A usage from Amann's Analysis I. Maybe it helps a bit. $\endgroup$
    – Eric
    Nov 7, 2017 at 7:53
  • $\begingroup$ @Eric This usage of "local" in Amann's Analysis coincides with the comment of user254665 $\endgroup$ Nov 9, 2017 at 14:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .