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From Wikipedia:

Suppose $X,Y$ are topological spaces with $Y$ a Hausdorff space. Let $p$ be a limit point of $Ω⊆X$, and $L ∈Y$. For a function $f : Ω → Y$, it is said that the limit of $f$ as $x$ approaches p is L (i.e., $f(x)→L$ as $x→p$) and write $$ \lim_{x \to p}f(x) = L $$ if for every open neighborhood $V$ of $L$, there exists an open neighborhood $U$ of $p$ such that $f(U∩Ω- \{p\}) ⊆ V$.

I wonder if people also often generalize the definition of a limit of a function $f$ to the case when $p$ is an isolated point of $\Omega$?

Can the above definition except that $p$ is a limit point of $\Omega$ can be applied to the case when $p$ is an isolated point of $\Omega$?

Specifically, is the "openness" of $V$ wrt the topology of $X$ or wrt the subspace topology on $\Omega$?

Thanks and regards!

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If $p$ is an isolated point, then we can always find $U$ such that $U\cap\Omega-p = \varnothing$, in which case the condition $f(U\cap\Omega-p)\subseteq V$ is trivially satisfied. This means that $\lim\limits_{x\to p}f(x) = L$ for all $L$, which makes the notion of "limit" as an isolated point rather useless. (Intuitively: the limit asks what the function $f$ is doing "near", but not at the point $p$; if $p$ is isolated, then there is no "near" there). –  Arturo Magidin Feb 14 '12 at 21:47
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As to your last question: doesn't matter: because $A\subseteq \Omega$ is open with respect to the topology on $\Omega$ if and only if there is a set $V$ that is open in $X$ such that $V\cap \Omega = A$. Since wee are only considering what happens inside $\Omega$, whether you look at $A$ or at $V$, you get the same intersection. –  Arturo Magidin Feb 14 '12 at 21:50
    
The process of removing the point $p$ smells a bit fishy to me, if only because of this kind of ambiguities. The definition in the French Wikipedia and in my topology courses do not use this: we want $f(\Omega \cap U) \subset V$. Is this a French/English difference, or the English Wikipedia using a weird definition? –  D. Thomine Feb 14 '12 at 22:03
    
@D.Thomine: nice question. I also want to know why –  Tim Feb 14 '12 at 22:07
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@D.Thomine: The problem with that definition is that functions that are defined but not continuous at $p$ cannot have a limit at $p$. The entire notion of "removal discontinuity" goes away. –  Arturo Magidin Feb 14 '12 at 22:08

1 Answer 1

This question has been answered in comments:

If $p$ is an isolated point, then we can always find $U$ such that $U∩Ω−p=∅$, in which case the condition $f(U∩Ω−p)⊆V$ is trivially satisfied. This means that $\lim_{x→p}f(x)=L$ for all $L$, which makes the notion of "limit" as an isolated point rather useless. (Intuitively: the limit asks what the function $f$ is doing "near", but not at the point $p$; if $p$ is isolated, then there is no "near" there).

As to your last question: doesn't matter: because $A⊆Ω$ is open with respect to the topology on $Ω$ if and only if there is a set $V$ that is open in $X$ such that $V∩Ω=A$. Since we are only considering what happens inside $Ω$, whether you look at $A$ or at $V$, you get the same intersection. – Arturo Magidin Feb 14 '12 at 21:50

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