# Proving that if $f:X \rightarrow Y$ is a continuous function on $X$ such that $\lim \limits_{p \to p_0} f(p) = P$ for $p_0 \in X$, then $f(p_0)=P$.

Let $(X', d)$ and $(Y, \partial )$ be metric spaces so that $X \subseteq X'$.

I'm trying to prove that if $f:X \rightarrow Y$ is a continuous fucntion on $X$ such that $\lim \limits_{p \to p_0} f(p) = P$ for $p_0 \in X$, then $f(p_0)=P$.

My attempt:

Suppose not, so that $f(p_0) = Z \ne P$. Then $\partial (Z, P) = r >0$.

$f$ being continuous on $X$ means $\forall \epsilon > 0, \exists \delta_1 > 0$ such that $d(x, x_0)< \delta_1 \Rightarrow \partial (f(x), f(x_0))< \epsilon$,

and $\lim \limits_{p \to p_0} f(p) = P$ means $\forall \epsilon > 0, \exists \delta_2 > 0$ such that $d(p, p_0)< \delta_2 \Rightarrow \partial (f(p), P)< \epsilon$.

In both cases let $\epsilon = \frac{r}{2}$ and let $\delta = \min (\delta_1, \delta_2)$. Choose $q \ne p_0$ so that $d(q, p_0)<\delta$. Then $\partial (f(q), Z) < \frac{r}{2}$ and $\partial(f(q), P) < \frac{r}{2}$. So $$r=\partial(Z, P) \le \partial(f(q), Z) + \partial(f(q), P) < \frac{r}{2} + \frac{r}{2}$$ a contradiction.

The proof is dependent on the idea that there exists a $q$ such that $d(q, p_0)<\delta$. What if this isn't the case? If the theorem fails when such $q$ doesn't exist, could someone provide an example? And if the theorem always holds, then what would be a complete proof (one that doesn't depend on the existence of $q$)?

I would appreciate any help/thoughts!

• The limit of a function at a point only makes sense when the point is not isolated in the domain of the function. That is, you can assume that such a $q$ exists. – amsmath Sep 7 '18 at 4:06
• @amsmath. Really? The formal definition of a limit doesn't seem to necessarily imply that. – Leo Sep 7 '18 at 4:13
• It does. Read here (en.wikipedia.org/wiki/…) under "A more general definition..." The defintion of "limit" you use above is not entirely correct. – amsmath Sep 7 '18 at 4:15
• @amsmath. That definition concerns itself only with $ℝ$ as a metric space, where you can always find $q$. – Leo Sep 7 '18 at 4:20
• No, it doesn't. Read my comment more carefully. Alternatively, read this: en.wikipedia.org/wiki/… – amsmath Sep 7 '18 at 4:22