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Is this an infinite number of discontinuities? 
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#1
Aug2214, 09:41 AM

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So consider a function ##f(x)## which is continuous for all ##x## except on some finite interval, say ##[a, b]##. Imagine, for example, a function which goes to ##\infty## from the left at ##x=a##, is undefined from a to b, and then "comes from" infinity at ##x=b## and is defined and continuous everywhere else. Is this considered an infinite number of discontinuities, or a single discontinuity? My first thought is that it would be considered an infinite number of discontinuities, but I want to be sure



#2
Aug2214, 09:47 AM

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A discontinuity only really makes sense on the domain of the function. Since points in ##[a,b]## are not in the domain (the function is not defined there), those points cannot be discontinuities.
It follows that the function is continuous on its domain. 


#3
Aug2214, 09:48 AM

P: 508

Edit: Drats, sniped!



#4
Aug2214, 03:37 PM

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Is this an infinite number of discontinuities?
If what you want is a real function which is nowhere continuous only on a finite interval, consider the function [itex]f : \mathbb{R} \to \mathbb{R}[/itex] given by f(x) = 0 on [itex](\infty,0)[/itex], f(x) = x if x is rational, 0 if x is irrational on [0,1], and f(x) = 0 on [itex](1,\infty)[/itex]. This would amount to an infinite number of discontinuities.



#5
Aug2214, 05:44 PM

P: 153

Ok I see, thank you :)
See, my line of thought was that you could use the definition of continuity to argue that the function is "not continuous" at every point in the interval. I guess 'not continuous' and 'discontinuous' have two different meanings? You cannot speak of discontinuity at a point where the function isn't defined? Ok :) 


#6
Aug2314, 02:37 AM

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not continuous and discontinuous mean the exact same thing. But it doesn't make sense to speak of a function at a place where it isn't defined. That's like asking whether a real function is continuous or discontinuous at the imaginary number i.



#7
Aug2314, 04:36 AM

P: 1

As people said, if your function is not defined on the interval [a,b] it can't be discontinuous on the whole interval, it can however be discontinuous on points that are not in the function's domain if they are limit points (every open ball centered around these points contains one point of the domain distinct from itself).
A function that is discontinuous on [a,b], and in this case it should be defined at least on a subset that is dense on [a,b], like (a,b) or [itex](a,b) \cap \mathbb Q[/itex], will have infinite points of discontinuity, more specifically uncontably infinite. I assume the domain of your function is [itex]\mathbb R[/itex]. So I'll consider consider a function [itex]f:\mathbb R \rightarrow \mathbb R[/itex]. I believe a function like you intended to describe would be something like that: [itex]f(x)= \frac{1}{xa} \ \text{if} \ x < a[/itex], [itex]f(x)= 0 \ \text{if} \ x \in \mathbb Q \cap [a,b], \ f(x)=1 \ \text{if} \ x \in (\mathbb R  \mathbb Q) \cap [a,b][/itex] and [itex]f(x)= \frac{1}{bx} \ \text{if} \ x>b[/itex]. 


#8
Aug2314, 07:31 AM

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#9
Aug2314, 05:45 PM

P: 153




#10
Aug2314, 05:54 PM

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A function [itex]f : A \to B[/itex] (with A , B [itex]\subseteq \mathbb{R}[/itex]) is continuous at [itex]a \in A[/itex] if for all [itex]\epsilon > 0[/itex], there exists a [itex]\delta > 0[/itex] such that [itex]f(x)f(a) < \epsilon[/itex] for all x such that [itex]xa<\delta[/itex]. So, suppose [itex]a \in \mathbb{R}A[/itex]. What does "f is continuous at a" mean? Well, actually, continuity is not defined at a, because a function can only be continuous at a point in its domain by the very definition of continuity. So it's quite unreasonable to say that ( not (f is continuous at a) ), since (f is continuous at a) is nonsensical. 


#11
Aug2314, 07:44 PM

P: 153

I was taught that for a function to be continuous at a point, it's limit at that point must equal the actual function value at that point. Hence, I would conclude that the function ##f(x)=x^{2}## is not continuous at the point ##x=0## because clearly the 'function value' at that point cannot equal the limit at that point considering that the function is not defined there. I would also conclude that the function is not continuous at GREEN because neither the limit nor the function exist there. Do you understand how my misuse of terminology is not entirely "unreasonable" although it may be incorrect? Or is my English also in need of harsh criticism? Am I misunderstanding the meaning of "reason"?
It's also worth noting that while continuity in this sense is a mathematical term, it carries along with it an intuitive meaning. You can tell by visual inspection whether or not something is continuous because the mathematical definition of continuity is just an formal version of a conceptual idea that we all understood as a first step. Before continuity was given a formal definition, I'm sure there were plenty of mathematicians who considered asymptotes and other such points on a function as "points where the function is not continuous" even though the function itself is not defined there. 


#12
Aug2314, 08:50 PM

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P: 839

It's not what people expect when you use the term "discontinuity" without quantification. The presumption is that the function is indeed defined there, but the limits don't equal the value. The term "singularity" is more often used for this situation. However, it is a fairly common misuse of terminology that ##\frac{\sin x}{x}## has a removable discontinuity at x=0. "Removable singularity" is the correct term. None the less "continuous at a point" is different from "continuous function". ##f(x)=x^{2}## is not continuous at x=2 (under this interpretation), but it is a continuous function. 


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