# Are functions still continuous even with removable discontinuities?

If we define a continuous function as a function that is continuous at all points in its domain, then would a removable discontinuity not impact this?

For example consider the function $f(x) = x$. It is clearly continuous at every point in its domain, which is all real numbers.

But now let's have $f(x) = x \cdot \frac{x-2}{x-2}$ which means the function is undefined at $x=2$, but still equivalent to $f(x)=x$ everywhere else, so we exclude that point $x=2$ from the domain.

The domain is now something like $(-\infty, 2) \cup (2, \infty)$. Continuity at a point usually means the limits exist and they equal the function's value at that point. The number $2$ is not in the domain so we only have to consider numbers to the left and to the right of $2$, and no matter what number we pick, they will be defined and have limits equaling those values.

So would we still call such a function "continuous" despite having this $f(2)$ undefined?

• A function that is continuous with a removable discontinuity is continuous on its domain, but it is not continuous at the point of discontinuity, as it is not even defined there! However, it can be extended to a new continuous function by removing the singularity. Commented Feb 11, 2018 at 18:48
• I thought "continuous on its domain" was essentially the same as saying the function itself is a "continuous function" Commented Feb 11, 2018 at 18:49
• Yes. We only care about points that are in the domain of the function. Commented Feb 11, 2018 at 18:55
• In the Wikipedia page you link below, it is written that "The term removable discontinuity is sometimes an abuse of terminology for cases in which the limits in both directions exist and are equal, while the function is undefined at the point $x_0$. This use is abusive because continuity and discontinuity of a function are concepts defined only for points in the function's domain. Such a point not in the domain is properly named a removable singularity." A removable discontinuity isn't really a discontinuity. :\ Commented Feb 11, 2018 at 18:58
• The latter. The extension is a new function, with a different, larger domain. Commented Feb 11, 2018 at 19:06

Yes, of course. The only thing that matters for continuity are those points at which the function is defined. What occurs outside the domain is irrelevant.

• So discontinuous functions only really arise when we have jump discontinuities and essential/infinite discontinuities? Commented Feb 11, 2018 at 18:49
• No. THere's also that case in which the limit at a point doies not exist, Take, for instance $f(x)=\sin\left(\frac1x\right)$ if $x\neq0$, with $f(0)=0$. It is discontinuous at $0$ because $\lim_{x\to0}f(x)$ does not exist. Commented Feb 11, 2018 at 18:51
• (I had assumed that example was a form of an essential discontinuity but I could be mistaken) Commented Feb 11, 2018 at 18:51
• @user525966 I don't think I've ever heard of essential discontinuities. Commented Feb 11, 2018 at 18:53
• I had been going by this definition en.wikipedia.org/wiki/… Commented Feb 11, 2018 at 18:54

Yes. The function is continuous, because it is still continuous in every point of its (now smaller) domain $(-\infty,2)\cup(2, \infty)$.

I am aware that there are other "definitions" of continuity. The Wikipedia article is confusing in that sense, as it gives a number of definitions that are equivalent to each other, and some that are not:

The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers.

Ugh! Come on, it means that a function not defined on the whole $\mathbb R$ cannot be continuous!

I am sure there are other sources of information that muddle the waters more than they make things clear...

Obviously, if you look at the (continuous) function $f(x)=x, f:\mathbb R\setminus\{2\}\to\mathbb R$ - its graph is not connected, has two components, but that is not what we call discontinuity. A discontinuity would be a point $x_0$ such that $\lim_{x\to x_0}f(x)\ne f(x_0)$, and to even check this condition you must be able to calculate $f(x_0)$, i.e. the function $f$ must be defined in $x_0$ (in our case, $x_0\ne 2$).

Thus, my conclusion for the OP is this:

• Take the universally accepted definition of continuity: the function is continuous if it is continuous at every point of its domain. Or any definition equivalent to that one. (I personally like the topological one, where the function is continuous if the preimage of any open subset of the codomain is an open subset of the domain.)

• Accept that the graph of a continuous function need not be a connected set (in, say, $\mathbb R^2$) - though, obviously, it will be if the domain of the function itself is connected.

• Accept that in maths' education there are other non-equivalent definitions of continuity floating around; fight them with arguments (even if it seems like attacking windmills).

• I think the "intuitive" definition would require that the domain be convex, but not necessarily cover all reals. Commented Feb 12, 2018 at 0:51