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I have a question to ask about a function.

Suppose a function $$f(x) = \frac{x^2 - x}{ x - 1},$$ we can simplify this function to be $f(x) = x$. Yet, we say that this function is discontinuous at $x = 1$ but after the simplification, we say that the function $f(x)$ is continuous.

Which one is correct? The fact that $f$ is discontinuous or continuous?

Ted
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Chris Myung Hyun Kim
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    What is correct depends on what the person grading will mark as correct. It is safest to say that the function is not continuous at $x=1$ because it is not defined there. But some people (and programs) will not even notice a removable discontinuity. – André Nicolas Apr 21 '14 at 08:00

3 Answers3

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The original function has domain $x: x \neq 1$ while the second function has domain all real numbers. So they are two different functions, and they are not equal. $f$ is discontinuous at $x = 1$. Here $f(x) = \dfrac{x^2 - x}{x - 1}$

DeepSea
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The original function $$f\left(x\right)=\frac{x^{2}-x}{x-1}$$ has $\mathbb{R}\backslash\left\{ 1\right\} $ as (maximal) domain and is continuous. It is not defined on $\left\{ 1\right\} $ and consequently statements like '$f$ is (dis)continuous at $1$' don't make sense. It can only be (dis)continuous at points that belong to its domain.

drhab
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    Exactly! Saying $f$ is discontinuous at $1$ is senseless. It makes as much sense as saying it is discontinuous at $\begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix}$. – Git Gud Apr 21 '14 at 08:22
  • @GitGud Actually, I wonder if this is true. I've never seen discontinuity defined, other than as the negation of continuity. Since it is not continuous at $1$, it must be discontinuous. Similarly the function $f$ is also discontinuous at $\begin{bmatrix} 1 & 2\ 0 & 1\end{bmatrix}$ and $\operatorname{Spec}(\Bbb Z)$ and everything else too. – Mario Carneiro Apr 21 '14 at 16:10
  • @MarioCarneiro Given $a\in \mathbb R$, assuming $a\in \text{dom}(f)$, $a$ is continuous if, and only if, $\text{usual statement}$. The $a$ is given before you define continuity. If you replace $a$ by $1$ you get: assuming $1\in \text{dom(f)} \ldots$. Whatever comes next won't mean anything. Granted the conditional statement is true, but that says nothing of the consequent which is what concerns us. – Git Gud Apr 21 '14 at 16:24
  • @GitGud Ugh, conditional definitions are annoying and nonsensical. Propositional formulas are true or false; they are never "undefined", whatever that means. A predicate is an abbreviation for a propositional formula, and as such always has meaning. I can write down the expression "$f$ is continuous at $1$" down as a formula, and it will be either true or false. Since $1\notin {\rm dom}(f)$, it will be false. Thus the function is not continuous at $1$. – Mario Carneiro Apr 21 '14 at 16:48
  • @MarioCarneiro "I can write down the expression "$f$ is continuous at $1$" down as a formula, and it will be either true or false." It will not be false, part of the formula is $f(1)$, this doesn't make sense. I agree with the first part of your comment, but I claim that what has meaning is the formula for "$1\in \text{dom}(f)\implies f\text{ is continuous}$". This will either be true or false. Saying $f$ is discontinuous at $1$ is saying the consequent of this assertion is false. Saying it is continuous is saying the consequent is true. – Git Gud Apr 21 '14 at 16:56
  • @GitGud There is no such thing as a "part of the formula that doesn't make sense". A propositional formula is a collection of $\in,=,\wedge,\dots$ with variables; it is always true or false. There is no "doesn't make sense". Since "$f$ is continuous at $1$" is (an abbreviation for) a formula, it too is either true or false, even though $f(1)$ is embedded somewhere in the formula. The actual result of that part of the expression won't matter, because of the guard $1\in{\rm dom}(f)$, which will make the formula false regardless of what $f(1)$ actually ends up evaluating to. – Mario Carneiro Apr 21 '14 at 17:01
  • @MarioCarneiro "because of the guard 1∈dom(f), which will make the formula false regardless of what f(1) actually ends up evaluating to" - It seems to me that you're basically saying that $0\neq 1\implies \frac 1 0\neq 0$ (feel free to replace the consequent with any meaningless statement you wish) is a meaningful statement. If this is not what you're saying, then I argue as follows. – Git Gud Apr 21 '14 at 17:12
  • Of course in a formal system a formula will be either true or false. You say $f$ is continuous if the formula ends up being true and $f$ is discontinuous if it ends up being false. But this is metalanguage. If you define it like this, I must agree with you. But epistemologically I don't think this reflects what's being done (in a platonistic) kind of way. I disagree with this way of abbreviating continuity in the metalanguage. – Git Gud Apr 21 '14 at 17:14
  • @GitGud Of course $0=1\implies \frac10\ne0$ is a meaningful true statement. $0\ne1\implies\frac10\ne0$ is a meaningful statement, but it is not clear whether it will be true or false; this depends on the exact definition of division, zero, etc, and if objects like $0$ are taken as axiomatic, it may even be unprovable in either direction. As for why I am appealing to a formal system - formal systems are the solid logical foundation upon which the more wishy-washy textual mathematics is built, at least in principle, so I use them to inform me on the "high philosophy" aspects of how math works. – Mario Carneiro Apr 21 '14 at 17:18
  • @MarioCarneiro I assume $\sf ZFC$. I won't ask you formalize the statement because I'm pretty sure that whatever you get will not, in my opinion, be descriptive of what $0\neq 1\implies\frac10\ne0$ is trying to convey. – Git Gud Apr 21 '14 at 17:21
  • @GitGud Relevant: an actual formal system proof of $1/0$, by me, incidentally – Mario Carneiro Apr 21 '14 at 17:23
  • @MarioCarneiro Can you indeed refer to places where discontinuous is defined as not continuous? Then why using this term 'discontinuous'? We could just do with 'not continuous', right? The term discontinuous is used if you want to say something about the function. Not something about its domain. To obstruct the escaperoute '...or is not defined at $1$' you use the term discontinuous instead of not continuous. – drhab Apr 22 '14 at 08:43
  • @drhab You are right as far as my having no source. In fact, I've never seen the term discontinuous ever defined. Usually, when an adjective is defined, the negated versions though english prefixes are implicitly assumed to have the opposite meaning, but this is rarely spelled out. In order to have the 'correct' meaning in this context, you would have to define discontinuous as its own predicate; something like '$f$ is discontinuous at $x$ iff $x\in{\rm dom}(f)$ and $f$ is not continuous at $x$'. Then if $x\notin{\rm dom}(f)$ then $f$ is not continuous or discontinuous at $x$. – Mario Carneiro Apr 22 '14 at 14:53
  • I really like this answer to the fact that it is very explicit in the answer :) Thanks. – Chris Myung Hyun Kim Apr 23 '14 at 02:01
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The "two" functions you are talking about are exactly equal (as far as analysis is concerned) as long as you want their domains equal (when you specify a function in the form $f(x)=...$ one really should qualify x for completeness). Take the limits from the left and right at 1 and you'll get 1 so this is a perfectly legitimate function over all of $\mathbb R$ if you define it to be such.

Sean D
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