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My question deals with a solution to an example problem from the text Introduction to Real Analysis by Bartle and Sherbert. From the text:

Example Problem: $ \phi(x):=\frac{1}{x}$ is not continuous at $x = 0$. Solution: Indeed, if $ \phi(x)=\frac{1}{x}$ for $x > 0$, then $\phi$ is not defined for $x=0$ , so it cannot be continuous there.

A second solution is given in which it's shown that $ \lim_{x \to 0} \frac{1}{x} $ does not exist in $\mathbb{R}$ and I completely understand the second solution. However, I don't fully understand the first solution unless it read like the modified solution below in which I switch the domain from $x > 0$ to $x = 0$:

Example Problem: $ \phi(x):=\frac{1}{x}$ is not continuous at $x = 0$. Solution: Indeed, if $ \phi(x)=\frac{1}{x}$ for $x \neq 0$, then $\phi$ is not defined for $x=0$ , so it cannot be continuous there.

The way the first solution reads right now, it seems like I can simply pick a convenient domain for the function. This doesn't make sense (unless I'm missing something) because then I could simply define a function with any $c \in \mathbb{R}$ missing so that the function will be not be defined (and thus not be continuous) at the point $c$.

Am I missing something? And if so, what?

hgil
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  • $\phi(x)=\frac{1}{x}$ is defined for $x\ne 0$; consequently it is defined also for $x>0$. – MattG88 Dec 01 '16 at 00:18
  • @MattG88 I understand that. My point of confusion is why the author only considers the domain $x>0$ as opposed to the larger domain $x \neq 0$. I think the issue actually starts in the problem statement by virtue of the author not stating the domain for the function as LeBtz mentions below). – hgil Dec 01 '16 at 02:58

1 Answers1

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If that is really what your book says you might want to read a different book.

For $\phi$ to be a function, it needs a domain. Without a domain it is not a function so one can not even talk about it being continuous. Let's say we do have a specified domain. If the domain doesn't contain $0$ then $\phi$ is neither continous in $0$ nor is it discountinuous there, one simply cant ask the question if it is continuous in a point which doesn't belong to its domain. That question about continuity in $0$ is just as meaningful as the question about continuity in "apple". We can't plug $0$ in our function just as we can't plug in "apple". There is no difference.

Lukas Betz
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  • Right and I know that the associated domain is necessary to define the function. The authors of this text have not specified the domain of this particular function ($\phi(x)=\frac{1}{x}$) in several instances until the solution statement. This has caused me a lot of confusion and I'm sure it has for other readers as well. Thanks for pointing out a distinction between being discontinuous at a point (for instance, between the function has no limit at the point) and not continuous at a point (for instance, because the point is not in the domain of the function). – hgil Dec 01 '16 at 02:51
  • Actually it looks like whether there is even a distinction with regard to the terms "discontinuous" and "not continuous" depends on the preference of the author (and their definitions). http://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – hgil Dec 01 '16 at 03:16