Def.: let be $(A,\tau)$,$(C,\zeta)$ two topological spaces, $f \in C^E$, with $E \subseteq A$, and $x_0$ an accumulation point of $E$, a point $l \in C$ is limit of $f$ as $x$ approaches $x_0$ if $$\forall Y \in \displaystyle\mathcal{U}_{(C,\zeta)}(l) (\exists X \in \displaystyle\mathcal{U}_{(A,\tau)}(x_0)(f((X-\{x_0\} )\cap E) \subseteq Y))$$

$\displaystyle\mathcal{U}_{(A,\tau)}(x_0)$ is neighbourhood system for $x_0$

$\displaystyle\mathcal{U}_{(C,\zeta)}(l)$ is neighbourhood system for $l$

Is correct? Thanks in advance!

up vote 1 down vote accepted

Yes, that is a correct definition of the limit of $f$ at $x_0$.

Another way to put the definition is to say that $\tilde{f}\colon E\cup \{x_0\} \to C$ defined by

$$\tilde{f}(x) = \begin{cases} f(x) &, x \in E\setminus\{x_0\} \\ l &, x = x_0\end{cases}$$

is continuous.

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    It agrees with…, at least.... – Henno Brandsma Jun 16 '14 at 12:04
  • @Daniel Fischer, Is it possible to define the limit of $f:E\to D$ with $D \subseteq C$? – mle Aug 1 '14 at 18:59
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    @mle Sure. $D$ is a topological space too when endowed with the subspace topology induced by $C$ (or any other topology, but then $D\subseteq C$ is irrelevant). You just replace $\mathcal{U}_{(C,\zeta)}(l)$ with $\mathcal{U}_{(D, \zeta')}(l)$ in the definition. – Daniel Fischer Aug 1 '14 at 19:07
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    @mle For the standard definition, you need $l\in D$. If you want to allow a limit of $f\colon E\to D$ to lie in $C\setminus D$, you need some modification (like you need for limits of $\pm\infty$ for real-valued functions). You can then stick to $$\forall Y \in \displaystyle\mathcal{U}_{(C,\zeta)}(l) (\exists X \in \displaystyle\mathcal{U}_{(A,\tau)}(x_0)(f((X-\{x_0\} )\cap E) \subseteq Y))$$ which effectively means that for the definition of limits, we forget that the range of $f$ is contained in the subset $D$ of $C$. But that ignores that only elements in the closure of $D$ can be limits, – Daniel Fischer Aug 1 '14 at 19:31
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    so one could make a compromise and consider $f\colon E \to \overline{D}$. Or you can replace the $$f((X-\{x_0\})\cap E)\subseteq Y$$ with $$f((X-\{x_0\})\cap E) \subseteq (Y\cap D).$$ (That doesn't change the meaning, but it emphasises different points of view.) – Daniel Fischer Aug 1 '14 at 19:31

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