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More specifically, I am wondering if $\sin(1/x) - x$ is discontinuous at $0$. I know that $f(x)= \sin(1/x)$ is discontinuous at $0$, but $f(x) = -x$ is continuous at all points. However if I add these two functions together, does that make sum of the function discontinuous?

user328651
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    If $f+g$ is continuous where $f$ is continuous, then $(f+g)-f=g$ is continuous. – Samuel Apr 05 '16 at 03:47
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    Actually, many people (including me) wouldn't agree that $\sin(1/x)$ is discontinuous at $0$, since the word "discontinuous" should only be used for points in the function's domain of definition. See discussion here: http://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow/1087642#1087642 – Hans Lundmark Apr 05 '16 at 08:51
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    Discontinuous in the sense that for any definition of $f(0)$, $f(x) = \sin (1/x)$ is discontinuous at $0$. This is not true for the function $\sin(x)/x$ for example. The latter is sometimes called a "removable" discontinuity. –  Apr 05 '16 at 14:24

3 Answers3

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First of all, we should note that $\sin(1/x)$, as it stands is continuous wherever it is defined. By that, I mean that $\sin(1/x)$ is not defined for $x=0$, so we have to assume that $x\neq 0$ when evaluating $\sin(1/x)$. Frequently, we avoid this problem by writing $$g(x)=\begin{cases} \sin(1/x) & x\neq 0\\ 0 & x=0.\end{cases}$$ This is probably the function you mean to use, or at least something like that. As you note, $g(x)$ is discontinuous at $0$. Suppose that $g(x)+x=f(x)$ was continuous. Then $g(x)=f(x)-x$. Since the difference of continuous functions is continuous, $g(x)$ is continuous. But this is a contradiction. Therefore, $g(x)+x$ cannot be continuous.

Plutoro
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    If one agrees with the statement that a continuous function is a function which is continuous at every point in its domain, then $\sin(1/x)+x$ is continuous! (But if you extend $\sin(1/x)$ to be defined on the whole real line, by assigning some arbitrary value at the origin, then it becomes discontinous.) – Hans Lundmark Apr 05 '16 at 08:55
  • @HansLundmark Very good point. Thanks for bringing that up. – Plutoro Apr 05 '16 at 17:10
  • Wouldn't the domain without 0 lead to an "almost everywhere" continuity, as there is no way to formulate an epsilon-delta continuity construction for a gapped domain? – rubenvb Apr 05 '16 at 21:21
  • @rubenv We can formulate $\epsilon-\delta$ continuity for any metric space, connected or not. In fact, by removing zero from the domain, the function is in fact everywhere continuous. – Plutoro Apr 05 '16 at 21:28
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If $f$ is continuous and $h$ is continuous then $g=h -f$ is continuous.

So if $f$ is continuous and $g$ is discontinuous $f+g = h$ can not be continuous else $h-f =g$ would also be continuous.

fleablood
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    So continuous +/- continuous = continuous, continuous +/- discontinuous = discontinuous, discontinuous +/- discontinuous = could be anything. – gnasher729 Apr 05 '16 at 08:04
  • Pretty much. And multiplication and division (except at zero). Basically i) continuous combined "nicely" with contionuous => continuous, ii) discontinuous means nothing other than not continuous and in general discontinuity and combinations of discontinuity imply nothing. However as cont % cont => cont then cont % disc = x => cont $%^{-1}$ x => disc => x is not cont => x is discontinuous by contrapositive. This is not by any characteristic of discontinuity but by the impossibility of the characteristic of continuity. – fleablood Apr 05 '16 at 15:52
  • Likewise discontinuous % x = continuous usually (not always) implies x ^ continuous = discontinuous => x is not continuous => x is discontinuous. (Assuming % is invertable in a way that preserves continuity.) – fleablood Apr 05 '16 at 15:56
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Recall the definition of continuous. We say a function is continuous at $x_0$, if $f(x_0)$ is well defined, and $\lim\limits_{x\rightarrow x_0}f(x)$ exists, where $\lim\limits_{x\rightarrow x_0}f(x)=f(x_0)$.

We should always define function formally. Let $f:\mathbb{R}-\{0\}\rightarrow \mathbb{R}$ such that $f(x)=\sin(1/x)$. When we defined $f$ in this way, we have $f$ is a continuous function. Let $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $g(x)=x$. We have this function is continuous. We cannot simply add $f$ and $g$, since their domain is different. We could restrict the domain of $g$ to the set $\mathbb{R}-\{0\}$. Where $g|_{\mathbb{R}-\{0\}}+f$ is continuous.

But in general, for any function $g,f:D\rightarrow R$, where $f$ is continuous and $g$ is discontinuous, we have $f+g$ is discontinuous.

Jialin Zhang
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    Your 'in general' statement suggests that there are cases in which a continuous function and a discontinuous function can produce a contininuous function, this is never the case. As Samuel had stated: If f+g is continuous where f is continuous, then (f+g)−f=g is continuous. – DrunkWolf Apr 05 '16 at 04:52
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    +1. This is easily the best answer so far, in my opinion. But for this definition to agree with the usual $\epsilon$ and $\delta$ stuff, one also needs to say explicitly that $f$ is continuous at $x_0$ if $x_0$ is an isolated point in the domain $D_f$ (in which case it doesn't make sense to talk about $\lim_{x\to x_0} f(x)$). – Hans Lundmark Apr 05 '16 at 08:56
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    @DrunkWolf If something holds in general it means that it holds for the general case, that is without further assumptions. This means that Eric's final statement states that there are no pair of continuous and discontinuous functions that sum to a continuous function. – Taemyr Apr 05 '16 at 12:35