Let $(X,d)$ and $(Y,w)$ be two metric spaces, with A $\subseteq X $ and $f : A \to Y$. Also, let p be a limit point of A.

I have the following definition for functional limits.
$lim_{x\to p}f(x) = L$
$\equiv $ $\forall \varepsilon>0, \exists \delta>0, ( \forall x \in A (d(x,p)<\delta) \to w(f(x),L) <\varepsilon )$ .

Now my question.
I'm trying to see what will happen if we re-define the functional limits as to include p to be any point in the closure of A ( not necessarily a limit point, but possibly an isolated point ).
I'm thinking that whenever we allow p to be an isolated point of A, the symbolic definition will crash and cause wierd results. But i'm trying to see what will happen, more precisely.

What goes wrong in the definition whenever we allow p to be an isolated point ? I'm thinking that for those isolated points, the symbolic definition will become tautologically true, regardless of L, and hence f will be able to have ( by the old definition ) infinitely many limits as it approaches an isolated point in the domain of A. Is that accurate ? How is that true ?

P.S : I have a definition of an isolated point of A as a point such that there exists an open ball centered at it that contains no elements of A other than itself.

Thanks in advance


I don't recognize your definition of a limit. The definition I know requires $x\ne p$, as below:

$$ \forall \varepsilon>0, \exists \delta>0, ( \forall x \in A\ (0<d(x,p)<\delta) \implies w(f(x),L) <\varepsilon )$$

If $p$ is an isolated point of $A $ then the set $\{x \in A : 0<d(x,p)<\delta\}$ is empty for small $\delta$. As such, the requirement $w(f(x),L)<\varepsilon$ holds vacuously no matter what $L$ is. We lose uniqueness of limit; worse yet, since every real number of a limit of $f$ at $p$, the concept carries no information about $f$. I think this is enough to say that extending the definition was a bad idea.

The definition of continuity can be and is extended to isolated points; every function is continuous at every isolated point of its domain. This is consistent with the topological characterization of continuity (preimages of open sets are open), which applies to every kind of domain, including those with isolated points.

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  • $\begingroup$ In France the limit is usually defined with only $d(x,p)<\delta$, and not $0<d(x,p)<\delta$ (with the latter it is called "limite épointée"). $\endgroup$ – xxx Sep 23 '14 at 11:49
  • $\begingroup$ Exactly, i also had limit defined with d(x,p)<δ and not 0<d(x,p)<δ. Before i posted this question i was already thinking, if x would not be able to be equal to p, then i could see how the definition would crash, but since i had limit defined allowing x=p, i don't see how isolated points crash the definition. Should that mean that there's no problem allowing isolated points in my definition ? $\endgroup$ – nerdy Sep 23 '14 at 15:42
  • $\begingroup$ @nerdy If you are using the French definition, everything's fine. The French limit of $f$ at an isolated point is equal to the value of $f$ at that point. $\endgroup$ – user147263 Sep 23 '14 at 16:05
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    $\begingroup$ @nerdy Since $x=p$ qualifies for $d(x,p)<\delta$, the implication requires $w(f(p),L)<\epsilon$. For all $\epsilon$. Hence $f(p)=L$. $\endgroup$ – user147263 Sep 29 '14 at 22:44
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    $\begingroup$ @nerdy Notice that $x\ne p$ in $\forall x \in A \cap( p-\delta,p)$. Therefore, if $p$ is an isolated point of $A$, the implication becomes true regardless of $L$. $\endgroup$ – user147263 Sep 29 '14 at 22:59

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