1
$\begingroup$

Given metric spaces $(X,d_X), (Y, d_Y),$ where $S\subseteq X,f:S\to Y,$ and $x_0$ a limit point of $S$, we say $y_0$ is a limit of $f$ at $x_0$ when $$\forall \epsilon>0, \exists \delta>0 \text{ s.t. } f(B_\delta(x_0)\cap S\setminus \{x_0\})\subseteq B_\epsilon(y_0).$$

To be sure, since $$f(\emptyset)=\emptyset \subseteq B_\epsilon(y_0)\quad \forall y_0\in Y, $$

can we say any point $y_0\in Y$ is a limit of $f$ at $x_0$ when $x_0$ is not a limit point of $S$? Or does the definition of a limit of a function not apply to non limit points in the first place?

$\endgroup$

1 Answer 1

1
$\begingroup$

If $x_0\in X$ is not a limit point of $S$ , then $\exists\delta>0$ such that $B_{\delta}(x_0) \cap S\setminus \{x_0\}=\emptyset$

Then $f(\emptyset) \subset B_{\epsilon}(y_0) $ for all $\epsilon>0$

Hence if $x_0\in X$ is not a limit point of $S$ then every point $y_0\in Y$ is a limit point of $f$ at $x_0$.

So the definition is vacuously satisfied and "non-limit points" are useless in this context.

$\endgroup$
2
  • $\begingroup$ Right I got that; I guess my question was whether the "limit of a function" formally applies to non limit points; the notes I have assume $x_0$ is a limit point in the definition. $\endgroup$ Commented Aug 15, 2022 at 8:04
  • $\begingroup$ According to my answer if $x_0$ is a non limit point then every $y\in Y$ is a limit point of $f$ at $x_0$. So the definition is vacuously satisfied and "non-limit points" are useless here. $\endgroup$ Commented Aug 15, 2022 at 8:25

You must log in to answer this question.

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