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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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    @D.Thomine By the way, perhaps a translation might be "punctured limit", in analogy to the fact that if $U$ is a neighborhood of $p$, then $U-\{p\}$ is called a "punctured neighborhood of $p$".2012-02-15

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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