Difference between finite and discrete set of discontinuities I am reading this paper. In page 395 and 396, theorems $1$ and $2$ use the terms 'finite' and 'discrete' to refer to sets, in this case sets of discontinuities. What I don't understand is: what is the difference between these two? I think I would understand a lot better if someone would provide an example to illustrate the difference. 
 A: Discrete just have to be separated from one another, finite means the total number must be $<\infty$. So $f(x)=x$ has no discontinuities at all, and $f(x)={1\over x}$ has one discontinuity at $0$ and again, this number is finite.
On the other hand the floor function $f(x)=[x]$ (definition below if you haven't seen it) has infinitely many of them--one at each integer--but they're all separated by a distance of $1$, so they are discrete, but not finite. Another example is the tangent function, which is discontinuous everywhere $\cos x=0$ i.e. at ${\pi\over 2}+\pi k,\, k\in\Bbb Z$, again the number is finite, but discrete since the set of discontinuities doesn't "bunch up too much," they stay "spread out" in some sense.
And it's not just evenly spaced ones that are discrete, you can have some odd balls in there as well, like ${1\over x}+\tan x$ which has an extra discontinuity at $0$, but they're still not clumping up.
Since it seems you are reading material at a relatively high level, I'll assume you know what an accumulation point is.

Formally a set, $S$ in a metric space is discrete if every $x\in S$ has some $\epsilon=\epsilon(x)$ so that $B_x(\epsilon)\cap S=\{x\}$.

Note in the examples I gave, we can take $\epsilon = {1\over 2}$ for all of them, independently of $x$.



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*In case you have not seen the greatest integer function for some reason, recall it is defined as follows:


Find $n\in\Bbb Z$ (which clearly exists) such that $n\le x < n+1$, then $f(x)=n$ by definition.
