*Any* function from $E \to E^{\prime}$ is continuous at $p_0$ if $p_0$ is not a cluster point. Can someone help me conceptualize this? Full statement: Let $E$,$E^{\prime}$ be metric spaces, let $p_0 \in E$, let $\{p_0\}^c$ be the completement of $\{p_0\}$ in E, and let $f:\{p_0\} \to E^{\prime}$
My textbook explains that any function is continuous at $p_0$ if $p_0$ is not a cluster point. I see how the math works out since we know that $d(p,p_0) < \delta$ since $p=p_0$.
Therefore $d^{\prime}(f(p),f(p_0)) = 0$
But conceptually, doesn't $p_0$ not being a cluster point imply that there is a break in the function? I'm envisioning some function in $\Bbb{R}^2$ where there is a hole at the limit but the function is defined for some different value of y (on a standard x,y plane). 
Sorry if my confusion isn't entirely clear.


But conceptually, doesn't $p_0$ not being a cluster point imply that there is a break in the function? I'm envisioning some function in $\Bbb{R}^2$ where there is a hole at the limit.

There's no break or hole because there's no limit at which to be near $p_0$.  Instead, $p_0$ is separated from the rest of $E$.  
An example of $E$ might be $(0,1) \cup \{2\}$, as a subspace of $\mathbb{R}$.  Here $p_0 = 2$ is a point of $E$ which is not a cluster point of $E$.  A function $f \colon E \to \mathbb{R}$ can take any value it likes at $2$, without interfering with the values on $(0,1) = \{2\}^c$.    
