Question: I have a request for an example. I have the following definition of an ordinal: 

A set $\alpha$ is said to be an $\textit{ordinal}$ if

1.) $\alpha$ is well-ordered by $\in$, and

2.) $\beta\in\alpha\rightarrow\beta\subseteq\alpha$

My question is, is there such thing as a set where 1 holds and 2 doesn't? As far as I can tell it seems that satisfying 1 would make 2 a given. Am I correct or no?

Answer: 
It comes down to what we mean by a *function*. In specifying the function, we need to specify not just the "rule", but also the *domain*, i.e., what are you allowed to put into the function?

In particular, to answer your question directly: if we want $f(x)=x+3$ to be defined at $x=2$, then we aren't allowed to say things like $$``f(x)=f(x)\cdot\dfrac{x-2}{x-2},\,''$$ because the expression on the right is not defined at $x=2$.

I think it might be the norm to brush over this kind of thing in introductory calculus courses, because in such a course we tend to assume that the function is defined "everywhere it makes sense" (a frame of mind which is convenient, but carries many problems with it besides just this).

To be a bit more explanatory, if you just say the function $f$ is specified by $$f(x) = \frac{x^{2}+x-6}{x-2},$$ then you haven't specified the function enough to be satisfactory for a really rigorous treatment. In particular, is your function defined at $x=2$, or is it not? Given the way that you have written it down, I would assume that it is not, since, as you point out, we cannot divide by $0$.

The way we deal with this in a rigorous manner is to require the definition of the function to specify the domain. For example, you could say that $f$ has domain $\mathbb{R}$ and is given by $$f(x) = \begin{cases}\dfrac{x^{2}+x-6}{x-2}, & x\neq2, \\ 5, & x=2.\end{cases}$$ Note that the only reason this makes sense is that your function is specified for all $x$ in the domain (in particular, it is clear that there is, for this value of $f(2)$, no discontinuity at all). If we had said "$f$ has domain $\mathbb{R}$ and is given by $$f(x) = \dfrac{x^{2}+x-6}{x-2}$$ for all $x$" then we would run into problems: we claim that $f$ has domain $\mathbb{R}$, but $2\in\mathbb{R}$ and $f(2)$ is not defined.

By contrast, you could specify your function as the function with, for example, domain $[0,1]$ given by $$f(x)=\dfrac{x^{2}+x-6}{x-2},$$ when we see that there is, in a sense, no discontinuity at $x=2$ since $f(2)$ has not been defined. In this case, it *makes sense* to say $$f(x)=f(x)\cdot\dfrac{x-2}{x-2},$$ since this expression now makes sense for all of the values *in the domain of the function*. It also now makes sense to make the cancellation and say $f(x)=x+3$, because our knowledge of the domain of the function ensures that we are not dividing by $0$.
I think there's some ambiguity here, and I think it is somewhat of a beneficial ambiguity. Here's a question - does $f(x)$ represent exactly the set of computations on the right side, or does the right side merely represent a suggestion of values? I think that applied mathematicians often switch between these two ideas quite often and rapidly, and for good reason. Sometimes what the right-hand side says can be expanded upon. Sometimes these expansions make sense, sometimes they don't, and sometimes it depends on context.

For example, the factorial function can be made continuous. Sometimes it makes sense to consider the continuous version of factorial in order to make your answers make sense or apply in more situations. In other cases, it might not.

Or even take ordinary functions. We sometimes pretend that discrete functions are continuous so we can apply the manipulations/simplifications of calculus. These are not strictly correct, but here, we are using *both* the right hand side and the results as being more signifiers than exact.

So, in your case, the original function *does* have a discontinuity at $x = 2$ if you take the right-hand side to be determined by explicit calculation. However, if you just take it as a more general signifier of values, then you could say that it doesn't really have a discontinuity at all.

I think that pure mathematicians would be uncomfortable with this, but I think that it is a pretty common activity (even if never explicitly explained) in applied mathematics.