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!