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I opened this question because I am still very confused by the answers and the comments from the following post: differentiability check.

In the problem, the OP asks the following simple question:

"What is the number of points at which $f(x) =\frac{1}{(x-2)}$ not differentiable?"

The OP himself and a number of people said that the correct answer is none, pointing out that differentiability of $f$ at $x=2$ is not well-defined, since $2$ doesn't belong to the domain of $f$.

I think the question here is ambiguous and it depends on the precise way "$f$ is differentiable at $x$" is defined. For example, if the definition is "$x$ is differentiable if and only if ($x$ is in the domain of $f$ and $f'(x)$ exists)," then $f$ is not differentiable at $2$. But then $f$ is also not differentiable at a 2x2 matrix, or any complex number (assuming that we defined the domain of $f$ to be $\mathbb{R}\setminus 2$).

However, if the question was

"What is the number of points in $\mathbb{R}$ at which $f(x) = \frac{1}{(x-2)}$ not differentiable?,"

then I believe that the answer is unequivocally 1. @GitGub said that merely adding the words "in $\mathbb{R}$" isn't sufficient to make a difference, but I don't really understand his argument. You can see his arguments here.

Here's my reasoning:

Let $E = \{ x\in dom(f): \text{$f'(x)$ exists.} \}$.

$2\in \mathbb{R}$, but $2\not\in E$. Therefore, $2\in \mathbb{R} \setminus E =\{2\}$.

The question is asking for the number of elements in $\mathbb{R} \setminus E$, which is 1.

I feel like I am missing some very basic understanding of logic. Can someone help me clarify this?

Edit: It doesn't feel like anyone is really addressing my reasoning. From the way I see it, the definition of "the set where $f$ is differentiable" is not the issue. Note the way I defined :

$E = \{ x\in dom(f): \text{$f'(x)$ exists.} \}$.

I have defined the set $E$, the set where $f$ is differentiable to be the points in the domain of $f$, as both @Vladimir, @Github, and @Hurkly is saying that some/many people would.

What I understood as the original issue was that just asking what $E^c$ (complement) is not well-defined, because there is no universe for the set $E$. When the universe is not specified, the most sensible interpretation would be to look at $dom(f) - E$.

The issue I am having is, if the question explicitly mentions $ \mathbb{R}$, "How many points in $\mathbb{R}$ is $f$ not differentiable," i.e. when a universe is given, why isn't this question logically equivalent to "How many points in $\mathbb{R} \setminus E$?"

Vladimir asked: "What if $g(x) =x$ was defined with $dom(g) = R\setminus\{2\}$."?

It doesn't change anything. $E = \{ dom(g): g'(x)\text{ exists}\}$ would be "$R\setminus\{2\}$", and $R\setminus E = \{2\}$.

EDIT2: I feel like reference to calculus is detracting the entire discussion.

Here is what I believe to be the essence of my question.

I have sets $A$ and $B$, with $A\subset B$, a proper subset.

Let $P$ be a predicate such that $P_A(x)$ is the sentence: "For $x\in A$, $x$ has property $P'$."

Let $E = \{ x\in A: P_A(x)\}$.

Then $E$ is a subset of $A$ and is also a subset of $B$.

Take a point $y \in B$.

I feel that @Hurkyl and @Github would say, for $y\not\in A$, $P_A(y)$ is not a well-defined statement.

But then, why isn't asking "Is $P_A(y)$ true" equivalent to asking "Is $y \in E$"?

Community
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Braindead
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2 Answers2

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What you are discovering is one of the little lies that crop up for the sake of simplification. We often talk about functions -- mathematical objects that associate to each point of its source a unique point in its target -- and define things like limits for functions.

But in practice, we very frequently are not dealing with functions: we are dealing with partial functions -- mathematical objects that associate to some points of its source a unique point, and to other points of its source no points at all.

(partial functions which are defined at every point of its source -- i.e. ones that are functions in the ordinary sense -- are called "total functions")

Your function $1/(x-2)$ is one of these: it is merely a partial function on $\mathbb{R}$, because it is not defined at $x=2$.

Now, there are a number of ways to deal with this issue. We could modify the definition of limit to work with partial functions. We could leave definitions unchanged but restrict our attention to the domain of the partial function (i.e. the set of points of the source where it is defined). We could even consider alternative objects, like functions modulo negligible functions.

Unfortunately, as far as I know, these things are rarely done explicitly in introductory courses -- and I'm not even sure what is done implicitly is even done in a consistent fashion.

Really, for a question like the one your topic is about, the answer requires guessing what the asker really means to ask, and there isn't a unique answer. I would guess it's most common to mean to ask the number of points for which $f$ is not defined plus the number of points for which $f$ is defined but not differentiable.

As an aside, in my opinion,the line of reasoning that involves "$f(x)$ is not defined at $2$ so it doesn't make sense to say if it is differentiable or if it is not differentiable there" really should be pushed further to say "it doesn't make sense to ask how many points of $\mathbb{R}$ that $f(x)$ is not differentiable at".

  • I appreciate your lengthy response, but I feel like it doesn't really address my reasoning. Whether we are working with partial functions, full functions, or how the differentiability is defined, I feel like it makes perfect sense to ask the question: "Given a set $E\subset\mathbb{R}$, what are the points in $\mathbb{R}\setminus E$?" – Braindead Jun 22 '14 at 06:40
  • Are you saying that "How many points of $\mathbb{R}$ that $f(x)$ is not differentiable at" is different from asking "How many points are in $\mathbb{R}\setminus E$?" I really don't see the difference between the two. – Braindead Jun 22 '14 at 06:44
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    @Braindead: the issue is that you have interpreted the question about differentiability in that way, but there are other interpretations. Some would replace $\mathbb{R}$ with the domain of the function. – Carl Mummert Jun 22 '14 at 11:12
  • @CarlMummert I get that some would interpret the original question with domain of the function. I just don't understand why when $\mathbb{R}$ is explicitly mentioned (i.e. "How many points in $\mathbb{R}$ is $f(x)$ is not differentiable at"), the question would be interpreted in any other way than "How many point are in $\mathbb{R} \setminus E$." – Braindead Jun 22 '14 at 14:08
  • @Braindead: By the usual definition of differentiable, "$f(x)$ is not differentiable at $2$" means that there exists $\epsilon > 0$ such that for all $\delta > 0$ there exists $x$ such that both $0 < |x| < \delta$ or $\left| (f(2+x) - f(2))/x \right| > \epsilon$. But $f(2)$ is not defined, so how can $f(x)$ possibly be not differentiable at $2$? If you apply the convention that using terms that only apply to functions when working with a partial function, you mean to restrict to its domain: thus when you ask where $f$ is not differentiable on $\mathbb{R}$, you use the domain rather than R. –  Jun 22 '14 at 14:28
  • @Hurkyl As I said before, the issue I am having with is logic. Let $q(2,x)$ be the difference quotient. You are saying that the statement, " $\lim\limits_{x\to 0} q(2,x)$ does not exist" is not well-defined, because $q(2,x)$ doesn't exist, correct? – Braindead Jun 22 '14 at 14:44
  • @Hurkyl Suppose that $E = { x \in A : P_A(x) }$, where $P_A$ is a predicate that says, "For elements $x\in A$, $x$ has property $P$." I take $A\subset B$. If I were to ask "How many points of $x\in B$ does not satisfy $P_A(x)$?" wouldn't this be the definition of the set $B\setminus{E}$? – Braindead Jun 22 '14 at 14:52
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    @Braindead: Taking everything literally, no: if $x$ is not in the domain of $P_A$ (or is a variable not restricted to the domain of $P_A$), then the expression $P_A(x)$ is nonsense, as is any expression that contains $P_A(x)$. One can adopt a convention that one might mean certain things other than what is literally written; e.g. that ${ x \in B \mid \not P_A(x) }$ is shorthand for ${ x \in A \mid x \in B \wedge \not P_A(x) }$, but it's just that: a convention, and it's not the only convention one might use. –  Jun 22 '14 at 15:18
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    It's a fine convention to use, if you know you're using it. And it's not without its problems, which are directly related to the reasons why it is nonsense if interpreted literally rather than conventionally. An example is that $Q_A(x) := \neg P_A(x)$ is also a predicate with domain $A$, and with this convention, $B \setminus { x \in B \mid P_A(x) } \neq { x \in B \mid Q_A(x) }$. And this particular mismatch introduces serious ambiguity when trying to translate English to math when negations are used, since English doesn't really have a way to easily distinguish between the two. –  Jun 22 '14 at 15:20
  • Thank you, I think your last point addresses my confusion. If you move the comment to the body of your answer (or create another answer), I will accept it. – Braindead Jun 22 '14 at 16:17
  • @Braindead: here is a parallel situation. Many people would say that the set of $x \in \mathbb{R}$ for which $1/x > 5$ is not the complement of the set of $x \in \mathbb{R}$ for which $1/x \not > 5$. This is because $0$ is not in either set. To these people one cannot even write $1/0$, much less compare it with $5$. Similarly, some number of people view "differentiable" and "not differentiable" as opposites of each other only on the domain of the function. – Carl Mummert Jun 22 '14 at 18:22
  • @CarlMummert But that's the thing: I WOULD say 0 belongs to the set ${x\in\mathbb{R}: 1/x \not > 5}$, and the crux of the whole thing is that I am not understanding why it wouldn't. – Braindead Jun 22 '14 at 18:46
  • @CarlMummert So I think the issue is my way of processing logic: $1/x \not > 5$ is true when $1/x > 5$ is not true (note that I am not equating "not true" with "false."). For $0$, the statement $1/x > 5$ is not true (as in it cannot called a "true statement" for whatever reason). Therefore, for $x=0$, $1/x \not > 5$ is true. – Braindead Jun 22 '14 at 18:56
  • @Braindead: the other way someone could see it is that "$1/0 > 5$" is not false so much as meaningless. A common way that mathematicians think is that the negation of a meaningless statement is also meaningless. So they would agree that "not" interchanges "true" and "false", but since $1/0 > 5$ is not false, these people don't view its negation as true. You will find, if you correspond with other mathematicians, that apart from a few conventions such as vacuous quantifiers there is a strong cultural avoidance of meaningless statements. – Carl Mummert Jun 22 '14 at 18:59
  • @Hurkyl: I am certain that the implicit conversion is not done consistently. I know I have seen people (who were not raised in the U.S. system) say things like "$\sin(x)/x$ is continuous at $0$" - because they always implicitly extend the function by continuity even to points where the formula doesn't apply. I doubt that a contemporary U.S. text would make that claim. – Carl Mummert Jun 22 '14 at 19:02
  • @CarlMummert I think the conclusion to all this is that given $E = { x: P(x)}$ with universe $U$, $E^c = U \setminus E \ne { x : \text{not } P(x)}$. – Braindead Jun 22 '14 at 19:17
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The ambiguity in the question "What is the number of points at which $f(x)=1/(x−2)$ is not differentiable?" resides not only in how you define "the domain of $f$" and "$f$ is differentiable at $x$" but also in the word "points". Once you define all three, everything becomes clear. If you define point to be an element of $R$, dom$f=R\setminus\{2\}$, and differentiable at $x\equiv (x\in\operatorname{dom}(f)$ and $f'(x)$ exists), then the answer is $1$. If you define points as elements of the set including real numbers, square matrices, and planets of the solar system, then the answer will be different;)

In short: Much ado about nothing.

As an afterthought: let $g(x)=x$ with dom$(g)=R\setminus\{2\}$. What is the number of points $x\in R$ at which $g$ is not differentiable? Clearly, the answer (with the above definition of differentiability) is $1$. Much of the confusion in the previous discussion in this and the other topics stems from the fact that one (erroneously) asks about points $x$ an which $f(x)$ is differentiable. So apparently if $f(x)$ is undefined at that particular $x$, then the question makes no sense. But the right question is about points $x$ at which $f$ (the function), not $f(x)$ (a number, or an undefined entity), is differentiable.

Vladimir
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  • I feel like a broken record. In the case where you define $g: R\setminus{2}$ with $g(x) = x$, if the question was "How many points of R is $g$ differentiable," for me this is clearly asking "How many points are in $R\setminus E$, where $E = { dom(g) : \text{$g'(x)$ exists.}}$. Since $E \subset dom(g)$, $2\not\in E$, so answer would be 1, regardless of the definition of differentiability. No? – Braindead Jun 22 '14 at 14:14
  • @Braindead: But I agree, the answer is 1. What is wrong? Not regardless of the definition of differentiability, however. I might wish to give the following definition: $g$ is differentiable at $x$ iff $x\in$dom$g$, $x\ne3$, and $g$ is differentiable at $x$ in the common sense of the word. Then the answer would be $2$, the points $2$ and $3$ being the points where $g$ is not differentiable; $2$ does not lie in the domain of $g$, and $3$ --- no function is differentiable at $x=3$ according to the definition I adopt. So, that depends... – Vladimir Jun 22 '14 at 14:17
  • Yes, it depends on the definition of "differentiability" in the sense that if you completely change the meaning, then yes the problem does depend on it. What I mean is that irregardless of whether the definition of differentiability at $x$ only makes sense in the domain of $f$ or it makes sense on points outside of the domain of $f$ (thinking of $f$ as a partial function), it doesn't affect the outcome of my question, because I specified the universe to be $R$. – Braindead Jun 22 '14 at 14:35
  • Yes, that was exactly my point. – Vladimir Jun 22 '14 at 14:38