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In another post someone mentioned that $x^x$ is continuous on the negative integers. All of the definitions for limit that I am aware say that the function has to be defined on an open interval around the point to even talk about a limit there. So I’m not sure how you can say that the function is continuous on the negative integers since there is no open interval around any negative integer where the function is defined.
I could be wrong about this, does someone have some input for me?
Indeed, the following function is continuous over $\mathbb C$: $$f(z)= \begin{cases} z^z,&z\ne0\\ 1,&z=0\\ \end{cases} $$
Let $z=re^{it}$.
It can be shown that $$f(z)=\text{exp}(e^{it}r\ln r+rt(i\cos t-\sin t))$$
The only ‘problematic point’ is $r=0$.
By showing $$|e^{it}r\ln r|=|r\ln r|\to 0 $$ as $r\to 0^+$, one can argue $f(z)$ is continuous everywhere on the complex plane. (Also note that the sum and composition of continuous functions are continuous.)
It does make sense to consider ‘continuity at a point’. Since the function is continuous everywhere, immediately it is continuous over $\mathbb Z^-$ as well.
