Understanding Removable discontinuity of the given function: ##e^x##

[chwala]

TL;DR Summary

See attached.

I would like to understand the highlighted part. In my understanding, this function does not seem to have a hole! Having said this, i can state that $x_0=1$ and we have our defined $f(x_0)=2$. It follows that,
$f(1^{+}) = e$
$f(1^{-}) = e$

thus $f(x_0^{+})=f(x_0^{-})≠f(x_0)$ thus the function has a removable discontinuity as per definition.

 

[PeroK]

Is there a question?

Clearly, if we take any function that is continuous on all of $\mathbb R$ and replace the function value at one point, then that must be a removable discontinuity!

 

[chwala]

Is there a question?

Clearly, if we take any function that is continuous on all of $\mathbb R$ and replace the function value at one point, then that must be a removable discontinuity!

Noted

Ok @PeroK ...if that is the case, then that's clear...i had assumed that the given function had to be undefined at some point implying a hole at some place ...then only can we check for removable discontinuity.

 

[PeroK]

Ok @PeroK ...if that is the case, then that's clear...i had assumed that the given function had to be undefined at some point implying a hole at some place ...then only can we check for removable discontinuity.

Either undefined or not the value needed for continuity.

Note that technically a function has to be defined at a point in order to be discontinuous. So, technically, the function $1/x$ is not discontinuous at $x =0$. That said, it still tends to be called a removable discontinuity.

 

[fresh_42]

Either undefined or not the value needed for continuity.

Note that technically a function has to be defined at a point in order to be discontinuous. So, technically, the function $1/x$ is not discontinuous at $x =0$. That said, it still tends to be called a removable discontinuity.

Both are wrong in my opinion. $x \longmapsto 1/x$ has a discontinuity at $x=0$ despite the fact that it is not defined. It is not continuous, therefore dis-continuous.

$x\longmapsto 1/x$ has not a removable discontinuity at $x=0.$ There is no value of $x$ that could be used instead in order to make the function continuous.

 

[chwala]

Either undefined or not the value needed for continuity.

Note that technically a function has to be defined at a point in order to be discontinuous. So, technically, the function $1/x$ is not discontinuous at $x =0$. That said, it still tends to be called a removable discontinuity.

Getting confused here...if we have a function, say $f(x)=\dfrac{1}{x}$, then for sure at $x=0$ the function is undefined and will therefore be discontinous at this point.

Something am missing?

 

[chwala]

There is this link which has an example that i am following; talking of how to remove discontinuities;

https://www.cuemath.com/calculus/removable-discontinuity/

 

[chwala]

Both are wrong in my opinion. $x \longmapsto 1/x$ has a discontinuity at $x=0$ despite the fact that it is not defined. It is not continuous, therefore dis-continuous.

$x\longmapsto 1/x$ has not a removable discontinuity at $x=0.$ There is no value of $x$ that could be used instead in order to make the function continuous.

Going with the link on post $7$ look at the first example...We can for sure, by re defining the function $f(x)=\dfrac{1}{x}$ at the point $x=0$ make it continuous ...or is that not the case?

 

[fresh_42]

Getting confused here...if we have a function, say $f(x)=\dfrac{1}{x}$, then for sure at $x=0$ the function is undefined and will therefore be discontinous at this point.

Something am missing?

No.

 

[fresh_42]

Going with the link on post $7$ look at the first example...We can for sure, replace $x$ with a value and thus make it continous...or is that not the case?

Yes, by $f_{mod}(1)=e^1.$ That's why it is removable. If we define $f_{mod}(x)=e^x$ for all $x\in \mathbb{R},$ then the modified function $f_{mod}$ is continuous and $f\_{mod}\equiv f$ everywhere except at $x=1.$ That is: we have removed the discontinuity of $f$.

 

[PeroK]

Both are wrong in my opinion. $x \longmapsto 1/x$ has a discontinuity at $x=0$ despite the fact that it is not defined. It is not continuous, therefore dis-continuous.

$x\longmapsto 1/x$ has not a removable discontinuity at $x=0.$ There is no value of $x$ that could be used instead in order to make the function continuous.

I've found sources like MIT maths that agree with you, but that appears to lead to a contradiction when we consider functions on disconnected sets:

https://math.mit.edu/~jorloff/suppnotes/suppnotes01-01a/01c.pdf

"We say a function is continuous if its domain is an interval, and it is continuous at every
point of that interval."

They do concede, however, that $1/x$ is continuous on every point of its domain.

I don't know that a function can be continuous at every point of its domain, yet not a continuous function.

That means that there is no such thing as a continuous function on a disconnected set!? But, there is no general demand that a metric space be connected in order to define continuous functions on it. So, no demand that a subset of $\mathbb R$ be an interval in order to have continuous functions defined on it.

In any case, I've never heard that a function must be defined on an interval to be continuous. Certainly that definition does not extend very well to metric spaces where there is not generally the notion of an interval.

PS I would say that $1/x$ is a continuous function on a disconnected set. That is a position that is generalizable to metric spaces.

 

[PeroK]

PPS The consensus on stack exchange is with me. A function can only be continuous or discontinuous on its domain. That must be right and the MIT guy must be wrong. It makes no sense generally to talk about a function having properties outside of its domain.

https://math.stackexchange.com/ques...-point-is-it-also-discontinuous-at-that-point

 

[PeroK]

My suggested approach is that discontinuities for points outside the domain refer to what happens if we try to extend the function to all of $\mathbb R$. If we try to extend $1/x$ to all of $\mathbb R$, then we find an infinite discontinuity at $x = 0$, regardless of the value chosen for the function at that point.

 

[fresh_42]

This is more of a philosophical discussion, or a linguistical, maybe even historical if you will. $x\longmapsto 1/x$ is continuous by its topological definition. However, as soon as we say "continuous / discontinuous at $x=0$" we phrase something that brings the contradiction with it. It has, in my opinion, no longer a topological meaning. The function does not exist at $x=0$ so the question does not make sense.

What is left, is the graphical intuition that sees a hole. Saying "discontinuous at $x=0$" refers to this observation rather than to the preimages of open sets. If we define: "A function is discontinuous if there exists an open set whose preimage is not open." then you are right; if we define: "A function is discontinuous if it is not continuous" then I am right since not defined implies not existent, implies not continuous.

I admit that this could be discussed endlessly.

In any case, it is not removable since $x=0$ is an essential singularity.

 

[PeroK]

"A function is discontinuous if it is not continuous" then I am right since not defined implies not existent, implies not continuous.

In this case, you take $\mathbb R$ as a universal set and say that only functions defined on all of $\mathbb R$ are continuous? And discontinuous at al points where a function is not defined. That is tenable, but definitely not standard. The MIT paper demands only that the domain is an interval. Although, even that seems suspect to me.

 

[fresh_42]

In this case, you take $\mathbb R$ as a universal set and say that only functions defined on all of $\mathbb R$ are continuous? And discontinuous at al points where a function is not defined. That is tenable, but definitely not standard. The MIT paper demands only that the domain is an interval. Although, even that seems suspect to me.

No, that was not what I was saying. I mean non-existence automatically implies the absence of any other property like continuity. It is a matter of language. The elements of the empty set have theoretically a purple eye color, but that is not what is commonly thought about the empty set.

Edit: But in a way, yes. We introduce the real number line at school as something like a universal set in real calculus long before we speak about topology where continuity belongs to. Note that I do not use universal in its mathematical sense here!

 

[chwala]

Am following the discussion keenly...I really appreciate the insight on this...

 

[pbuk]

It makes no sense generally to talk about a function having properties outside of its domain.

I disagree, it makes a great deal of sense for example to distinguish between the behavior of $\frac 1 x$, which has an infinite discontinuity; $\frac{x^2}x$, which has a removable discontinuity; and the [edit]Dirac delta functionHeaviside step function, which has a jump discontinuity. In each case the discontinuity is at $x = 0$ despite the fact that this value is not in any of these functions' domains.

PPS The consensus on stack exchange is with me. A function can only be continuous or discontinuous on its domain. That must be right and the MIT guy must be wrong.

Or alternatively they are both right depending on the context. I think Mathworld puts it quite well:

Note that the given definition of removable discontinuity fails to apply to functions

for which

and for which

fails to exist; in particular, the above definition allows one only to talk about a function being discontinuous at points for which it is defined. This definition isn't uniform, however, and as a result, some authors claim that, e.g.,

has a removable discontinuity at the point

. This notion is related to the so-called sinc function.

 

[PeroK]

I disagree,

But, then, you have to accept the consequences. For example, I might prove that a set $X$ is disconnected by finding a "continuous", onto function $f: X \to \{0, 1\}$. Then, you could not accept that proof because $f$ is not continuous at points where it is not defined. Or, at least, it would be debatable whether $f$ is continuous because it's continuous on its domain; or, not continuous, because it's discontinuous at points where it is not defined.

My point is that to argue that a function has discontinuities outside its domain (however appealing that might be in elementary mathematics) eventually comes back to bite you when you try to extend the notion of continuity.

With your approach, I'm not really sure you know when a function is continuous and when it's not, because any continuous function becomes discontinuous when you consider elements outside its domain.

 

[PeroK]

For example, I might prove that a set $X$ is disconnected by finding a "continuous", onto function $f: X \to \{0, 1\}$.

E.g. with $X = (-\infty, 0) \cup (0, \infty)$ and$$f(x)=\begin{cases} 0 & x < 0 \\ 1 & x > 0 \end{cases}$$I claim that's a continuous function and it proves that $X$ is disconnected. Whereas, if you claim it is not a continuous function, I say that is plain wrong!

 

[pbuk]

But, then, you have to accept the consequences.

I am prepared to accept the consequences of using different definitions in different contexts.

If you treat labels as more important than their definitions then you reduce mathematics to stamp collecting.

 

[PeroK]

I am prepared to accept the consequences of using different definitions in different contexts.

If you treat labels as more important than their definitions then you reduce mathematics to stamp collecting.

I don't see how stating that the function $\frac 1 x$ is a continuous function relates to stamp collecting? It's a true statement based on the rigorous definition of continuity.

 

[pbuk]

I don't see how stating that the function $\frac 1 x$ is a continuous function relates to stamp collecting?

Because you want to tell me I can't say the Heaviside step function has a jump discontinuity at $x = 0$ because it is already stuck into the page of continuous functions in your album.

 

[WWGD]

Both are wrong in my opinion. $x \longmapsto 1/x$ has a discontinuity at $x=0$ despite the fact that it is not defined. It is not continuous, therefore dis-continuous.

$x\longmapsto 1/x$ has not a removable discontinuity at $x=0.$ There is no value of $x$ that could be used instead in order to make the function continuous.

I guess its partly an issue of convention. As I understand it, a function must be defined everywhere in its domain

 

[PeroK]

Because you want to tell me I can't say the Heaviside step function has a jump discontinuity at $x = 0$ because it is already stuck into the page of continuous functions in your album.

The Heaviside step function is defined on all of $\mathbb R$ and is discontinuous at $x = 0$. By definition of continuity. $\frac 1 x$ has a disconnected set as its domain. And is continuous on that domain. This is basic stuff.

This reminds me of the $0.999 \dots = 1$ debate. You've immersed yourself in naive notions of continuity (like the guy from MIT, who has even posted his pdf online). If you focus only on what the rigorous maths is telling you, then $1/x$ is a continuous function. There is no other conclusion.

If you treat $X = (-\infty, 0) \cup (0, \infty)$ as a metric space and consider the set of all continuous functions on $X$, then the function $1/x$ is in that set. As is the Heaviside function if you exclude $0$ from its domain. You cannot get round that by invoking schoolboy notions of continuity.

 

[WWGD]

As a general comment, in my experience, Continuity requires or assumes a specific domain. The statement ' f is continuous' by itself , alone, is ambiguous.

 

[PeroK]

As a general comment, in my experience, Continuity requires or assumes a specific domain. The statement ' f is continuous' by itself , alone, is ambiguous.

And, critically, the domain of $1/x$ cannot include $0$.

 

[fresh_42]

I guess its partly an issue of convention. As I understand it, a function must be defined everywhere in its domain

I really do not want to re-enter this discussion. Continuity is not a matter of convention. It is a clear definition of morphisms in the category of topological spaces.

However, kids are taught differently. There is a real number line, the entire line, and eventually, long after, it becomes a topological space, as $\mathbb{R}-\{0\},$ too. The common language may differ here from the exact definition, simply because nobody made the effort to teach it right! Personally, I prefer to consider $x\longmapsto 1/x$ as continuous, since that is what it is.

My reasoning was about the term discontinuous. Sure, it means not continuous, but it is not automatically clear in which total logical set. I can say that $x\longmapsto 1/x$ is discontinuous at $0$ because the function does not exist there. It means I demand existence as part of the definition, i.e. no existence ergo no property. This is logically wrong since statements over the empty set are automatically true, but it is how common language is used, and what people mean if they say that $1/x$ is discontinuous at $x=0.$ They describe the graph, not the topological property. School talk: continuous is what can be drawn without interruption. Interruption means discontinuous.

That's it from my perspective: common language at the school level (discontinuous), or scientific language (continuous). It is this discrepancy that fuels our discussion here, and if you meant this by convention, then yes, although I wouldn't use that word here. Convention sounds too much as a matter of taste.

I would prefer to teach it right rather than twice. Add it to the list. It is a long list I find plain wrong in mathematical education. Almost every single concept has to be re-learned: Euclidean geometry, zero is not part of multiplicative groups, depending on age: yes, there are negative numbers, yes, there are complex numbers, rational numbers are a representation system of equivalence classes, not numbers, integration is not the reverse of differentiation, differentials are derivations, etc.

 

[WWGD]

I really do not want to re-enter this discussion. Continuity is not a matter of convention. It is a clear definition of morphisms in the category of topological spaces.

However, kids are taught differently. There is a real number line, the entire line, and eventually, long after, it becomes a topological space, as $\mathbb{R}-\{0\},$ too. The common language may differ here from the exact definition, simply because nobody made the effort to teach it right! Personally, I prefer to consider $x\longmapsto 1/x$ as continuous, since that is what it is.

My reasoning was about the term discontinuous. Sure, it means not continuous, but it is not automatically clear in which total logical set. I can say that $x\longmapsto 1/x$ is discontinuous at $0$ because the function does not exist there. It means I demand existence as part of the definition, i.e. no existence ergo no property. This is logically wrong since statements over the empty set are automatically true, but it is how common language is used, and what people mean if they say that $1/x$ is discontinuous at $x=0.$ They describe the graph, not the topological property. School talk: continuous is what can be drawn without interruption. Interruption means discontinuous.

That's it from my perspective: common language at the school level (discontinuous), or scientific language (continuous). It is this discrepancy that fuels our discussion here, and if you meant this by convention, then yes, although I wouldn't use that word here. Convention sounds too much as a matter of taste.

I would prefer to teach it right rather than twice. Add it to the list. It is a long list I find plain wrong in mathematical education. Almost every single concept has to be re-learned: Euclidean geometry, zero is not part of multiplicative groups, depending on age: yes, there are negative numbers, yes, there are complex numbers, rational numbers are a representation system of equivalence classes, not numbers, integration is not the reverse of differentiation, differentials are derivations, etc.

edit:
I don't mean to start something here. I meant the approach while using the definition.
I guess it depends. A correct definition of continuity is that the inverse image of an open(closed) set is open(closed). Or , for 1st Countable spaces, sequential continuity: $({x_n} \rightarrow x) \rightarrow (f(x_n) \rightarrow f(x))$. Just what consequences of using the latter do you oppose, or just what about it is wrong?

Would you ignore then the (co)domain and consider the inverse image of , say $(-1,1)?$, or, conclude $x \rightarrow 1/x$ is dicontinuous ,since $x_n \rightarrow 0 1/x \rightarrow 1/0$ doesn't hold?

If you convince me my usage is faulty in some sense, I'll change it; otherwise, unless I'm writing something in Category Theory, I don't see the need for it. Believe me, there are PHD Mathematicians that are not aware of precise Category-Theoretical Definitions.

 

[fresh_42]

A correct definition of continuity is that the inverse image of an open(closed) set is open(closed).

Yes, and that is all there is to say.

Except nitpicking ...

Continuity requires the mention of which two spaces and which two topologies. It is similar to linear independence. People "always" forget to mention the field, or at least the characteristics would be crucial! Another one for the list.

Edit: $\left(\dfrac{1}{x}\right)^{-1}\left((-\varepsilon ,0)\cup (0,\varepsilon )\right)=(-\infty ,-1/\varepsilon )\cup (1/\varepsilon ,\infty ).$ Open.

 

[pbuk]

If you focus only on what the rigorous maths is telling you, then $1/x$ is a continuous function. There is no other conclusion.

I am not disputing that. I am not even disputing that it is possible to define a discontinuity at $x$ in such a way that $x$ must be in the domain of $f$.

All I am saying is that if you remove that requirement from the definition you have something that can be even more useful and does not lead to any mathematical inconsistency. To illustrate its utility, if I say "$x^2 / x$ has a removable discontinuity at $x = 0$" then surely everyone understands what I mean.

I don't think it is better to have to say "$f: \mathbb R | 0 \to \mathbb R; x \mapsto \frac{x^2}x$ does not have a removable discontinuity at $x = 0$ because 0 is not in the domain of $f$, however $$ g: \mathbb R \to \mathbb R; \\
x \mapsto \begin{cases}
\frac{x^2}x & x \ne 0 \\
0 & x = 0
\end{cases}
$$(noting that $g(x) = f(x)$ for all $x$ in the domain of $f$) does have a removable discontinuity at $x = 0$".

 

[PeroK]

All I am saying is that if you remove that requirement from the definition you have something that can be even more useful and does not lead to any mathematical inconsistency. To illustrate its utility, if I say "$x^2 / x$ has a removable discontinuity at $x = 0$" then surely everyone understands what I mean.

Yes, but technically that function is a continuous function (on its domain). If you look at the definition of "removable discontinuity" and "continuity" you see that a continuous function, technically, may have a removable discontinuity. The contradiction is only linguistic.

I don't think it is better to have to say "$f: \mathbb R | 0 \to \mathbb R; x \mapsto \frac{x^2}x$ does not have a removable discontinuity at $x = 0$ because 0 is not in the domain of $f$, however $$ g: \mathbb R \to \mathbb R; \\
x \mapsto \begin{cases}
\frac{x^2}x & x \ne 0 \\
0 & x = 0
\end{cases}
$$(noting that $g(x) = f(x)$ for all $x$ in the domain of $f$) does have a removable discontinuity at $x = 0$".

That function does have a removable discontinuity at $x = 0$. But, that does not imply it is not a continuous function. That's the point I made in post #4:

Note that technically a function has to be defined at a point in order to be discontinuous. So, technically, the function $1/x$ is not discontinuous at $x =0$. That said, it still tends to be called a discontinuity.

 

[pbuk]

I agree with everything you say in #32, particularly that we may speak of a removable discontinuity of a continuous function and

The contradiction is only linguistic.

 

[Svein]

In complex analysis the definition of a removable singularity goes like this: Suppose that f(z) is analytic in the region Ω', obtained by omitting a point a from a region Ω. A necessary and sufficient conditiion that there exist an analytic function in Ω which coincides with f(z) in Ω' is that $\lim_{ z\to a}(z-a)f(z)=0$. The extended function is uniquely defined.
Nor do we have a problem with $\frac{1}{z}$. Cauchy's formula says that $f(z)=\frac{1}{2\pi i}\int_{C}\frac{f(\zeta) d\zeta}{\zeta - z}$ for all f(z) that are analytic inside of the closed curve C.

 

[Office_Shredder]

@Svein I'm not sure how that addresses the question of whether $1/z$ is a continuous function though, which is the analogous question to $1/x$ in $\mathbb{R}$