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.
Simply crossing out common factors from the original function does not change the fact the x cannot exist at x=2...

By crossing out the common factor of (x-2) from both the numerator and the denominator, you need to state that x cannot equal 2 (otherwise division of the common factor cannot take place; cannot divide by zero). This accounts for the removable discontinuity at x=2

Also, when you multiply by $(x-2)/(x-2)$, you are multiplying by 1, except for when x equals 2... then you are really multiplying by something that is undefined.