Vtyjkyuk

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 $epsilon = fracr2$ and let $delta = min (delta_1, delta_2)$. Choose $q ne p_0$ so that $d(q, p_0)<delta$. Then $partial (f(q), Z) < fracr2$ and $partial(f(q), P) < fracr2$. So $$r=partial(Z, P) le partial(f(q), Z) + partial(f(q), P) < fracr2 + fracr2$$ 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!