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: 
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> What's stopping you from multiplying [a continuous function] by $1$ in the form $\frac{x−2}{x−2}$? Can multiplying by $1$ really introduce a discontinuity to the function?
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Ah, but $\frac{x−2}{x−2}$ is *not equal to* $1$. Or, to be more, precise, it is only equal to $1$ for values of $x$ that are different from $2$. If $x=2$, then $\frac{x−2}{x−2}$ is undefined.

And that's more or less the answer to your question: If you take a given function $f(x)$ that is continuous on all of $\mathbb{R}$, and you multiply it by a function $g(x)$ that is equal to $1$ everywhere except at a single isolated point, at which point $g(x)$ is undefined, then the product $f(x)g(x)$ will be exactly the same as $f(x)$ everywhere except at that point, where it, too, will be undefined.
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.