I did search for whether this question was already answered but couldn't find any.

Does a function have to be "continuous" at a point to be "defined" at the point?

For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.

Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.

I ask because at http://en.wikipedia.org/wiki/Continuous_function#Examples there is the text:

the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$.

and the caption of the associated graph reads:

The function is not defined for $x = -2$.

Do I interpret this to mean that a function can not be defined at a point of discontinuity or merely that the function is intentionally not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?

  • 4
    $\begingroup$ It's the other way around: a function must be defined at a point $x$ in order to look whether it's continuous at $x$. "Point of discontinuity" has very little to do with continuity of the function. $\endgroup$
    – egreg
    Jun 16, 2013 at 14:53
  • $\begingroup$ For an example, the floor function is defined in $x=1$, say, but not continuous at that particular point. $\endgroup$ Jun 16, 2013 at 19:20
  • 1
    $\begingroup$ A function is continuous/discontinuous only at points in its domain. So a function is always defined at every point of discontinuity (if there are any). $\endgroup$
    – Ink
    Jun 16, 2013 at 19:42

6 Answers 6


The most common definitions of continuity agree on the fact that a function can be continuous only on points of its domain.

Asking whether $f(x)=1/x$ is continuous is like asking what's the preferred food of unicorns.

You're being misled by the phrase "point of discontinuity". Well, the truth is that a continuous function can many points of discontinuity. It's just an unfortunate terminology that I find being an endless source of misunderstandings. The terminology is due to an old fashioned way of thinking to continuity: it marks a “break” in the graph. However, the concept that a function is continuous if “it can be drawn without lifting the pencil” is a wrong way to think to continuity. The function $$ f(x)= \begin{cases} 0 & \text{if $x=0$,}\\ x\sin(1/x) & \text{if $x\ne0$} \end{cases} $$ is everywhere continuous, but nobody can really think to draw its graph without lifting the pencil. Can you?

The fact that $1/x$ (defined on the real line except $0$) has a point of discontinuity doesn't mean that the function is not continuous somewhere. Indeed it is continuous at each point of its domain.

Prompted by a comment, I'll add that a function can be defined at a point an not be continuous at it. The easiest example is the Dirichlet function $$ D(x)= \begin{cases} 0 & \text{if $x$ is irrational,}\\ 1 & \text{if $x$ is rational} \end{cases} $$ which is continuous nowhere.

So a function can certainly be noncontinuous (I purposely avoid discontinuous) at a point where it is defined.

Returning to the function $f(x)=1/x$, one can specify any subset of the real numbers as its domain, so long as it doesn't contain $0$. When no domain is explicitly specified, it's customary to use the largest subset of the reals where the expression makes sense, in this case it is $\mathbb{R}\setminus\{0\}$.

It's surely possible to define a function $g$ that extends $f$ in $0$; the function $g$ cannot, however, be continuous, because the limit of $g$ at $0$ can't be the value $g(0)$.

  • $\begingroup$ OP's question goes the other way. You are discussing whether a function needs to be defined to be continuous. $\endgroup$ Jun 16, 2013 at 16:00
  • $\begingroup$ @RossMillikan The initial statement should suffice to remove the OP's doubts. The rest explains why a doubt can arise. $\endgroup$
    – egreg
    Jun 16, 2013 at 16:10
  • $\begingroup$ Thanks egreg for your reply and Ross Millikan for your comment. A few more clarifications: 1) I never asked whether $f(x) = 1/x$ is continuous! 2) Why isn't it possible to draw the graph of $x \sin\left({1 \over x}\right)$ (wolframalpha.com/input/?i=plot+xsin%281%2Fx%29&lk=4) without lifting the pen? It may be difficult for you or me but a sure-handed artist could do it I think. 3) You say "$1/x$ defined on the real line *except $0$" -- that is precisely the point I want to know about -- why do you add that clause "except $0$"? $\endgroup$
    – jamadagni
    Jun 16, 2013 at 18:20
  • $\begingroup$ @jamadagni I added something about $1/x$. $\endgroup$
    – egreg
    Jun 16, 2013 at 21:04
  • $\begingroup$ However, doesn't a "point of discontinuity" mean one where the formally-defined continuity property does not hold? "Must lift the pencil" is an intuitive idea with limitations. But if that's the case, then how can a continuous function have many "points of discontinuity"? Your given sine example would have none, "pencils" notwithstanding. $\endgroup$ Jun 16, 2013 at 21:33

No, it has not.

You can define a function at any point in any way you wish.

For example, you can define function $\mathrm{sign}(x)$ as $$ \mathrm{sign}(x) = \left\{\begin{array}{rl}-1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0\end{array}\right. $$ and then, after defintion, study it for continuity. (It will be continious at all points except $x = 0$, but it is surely defined there).

But a function has to be defined at a point in order to study continuity at this point, that's why they write "The question of continuity at $x=−2$ does not arise, since $x=−2$ is not in the domain of $f$."


Talking about your question is meaningless, unless you say something about the domain of your function. The expression1/x is not defined for x=0 of course.

A function does not have to be continous in some point, to be defined there, e.g. take the characteristic function of the rational numbers in the set of the real numbers.

Furthermore a function has to be actually defined at some point to discuss whether you function is continous or not in that point.

  • $\begingroup$ Thanks for your reply. I do not however understand why "talking about my question is meaningless" -- I never asked about the continuity of a particular function which you would refer to as "my function". I was asking in principle whether a function has to be continuous to be defined. You and others have replied to that. $\endgroup$
    – jamadagni
    Jun 16, 2013 at 18:10

Continuity is the definitive concept formalizing the following idea: If $x$ is near $x_0$ then $f(x)$ is near $f(x_0)$. If this idea has to make sense for some particular instance then the minimum requirement for $f$ is that $f(x_0)$ be defined. When $f(x_0)$ is defined, but there are no points $x$ "near" $x_0$ where $f$ is defined as well, then $x_0$ is an isolated point of the domain of $f$, and $f$ is declared continuous at $x_0$ by definition.

The interesting case is when there are points $x\in{\rm dom}(f)$ arbitrarily close to $x_0$. What should be the exact meaning of "If $x$ is near $x_0$ then $f(x)$ is near $f(x_0)$"? We want to have some control over the output error $|f(x)-f(x_0)|$ when $x$ is "near" $x_0$. An estimate of the form $$|f(x)-f(x_0)|\leq |x-x_0|$$ would be fine, and we would also accept $$|f(x)-f(x_0)|\leq C\ |x-x_0|\tag{1}$$ for some constant $C$, say $C=20$. But unfortunately our idea of "If $x$ is near $x_0$ then $f(x)$ is near $f(x_0)$" encompasses examples not covered by $(1)$, e.g. the function $f(x)=\sqrt{x}$ $\>(x\geq0)$ at $x_0:=0$. That's why in the end mathematicians have settled to the "cumbersome" definition that $f$ is declared continuous at $x_0$ if, given any output tolerance $\epsilon>0$, we can find an input allowance $\delta>0$ (depending on $\epsilon$) such that $|x-x_0|<\delta$ guarantees $|f(x)-f(x_0)|<\epsilon$.


Surely, this is not a precise mathematical question - or, at least, as a precise mathematical question it is rather trivial: there are lots of discontinuous functions that are perfectly well-defined everywhere, if you want them to.

However, there is a deeper perspective coming from algebraic geometry and functional analysis, in which one works not with a single function, but rather with a ring $A$ and thinks of it as a ring of functions. Then one just defines the "points where these functions are defined" to correspond to maximal (or prime, if you really want them) ideals of that ring. This is what people call spectrum of a ring and Gelfand spectrum of a Banach algebra (let's stick to Banach algebras for the sake of simplicity, instead of polynormed algebras or even worse). The Gelfand spectrum of a Banach algebra carries a natural compact Hausdorff topology, generated by the functions in the algebra, and this topology is just designed to make these functions continuous. However, this construction depends only on $A$ being a Banach algebra, so if originally it was also an algebra of functions defined on some space, you may find that the spectrum is much richer than that space. In this case the spectrum provides the natural compactification, and in some cases it may be so large and complicated that it actually looks a little frightening at first sight (in particular, the spectra of $\ell^\infty$ or $L^\infty$ certainly do), but it is a really, really useful thing.

An alternative, but closely related, point of view is to fix a ring $A$ and also a field (or ring) $\mathbb{K}$, for your purpose usually $\mathbb{R}$ or $\mathbb{C}$, and define a $\mathbb{K}$-point of $A$ to be a ring homomorphism $A \to \mathbb{K}$ (see functor of points). If $\mathbb{K}$ carries a topology, then so does this space of $\mathbb{K}$-points, and again, this topology is designed to make the functions continuous. In ths case of complex Banach algebras the space of $\mathbb{C}$-points is the same as the Gelfand spectrum.


First, the formal question "must a function be defined at a point to be continuous there?" has answer "yes", from the definition of "continuity", since the latter definition refers to the value of the function at the point.

But this formal sense of the question does not go far enough, in my opinion. That is, I'd argue that at a (single) point where a function is not continuous, its point-wise value is "ambiguous"... I'll argue in favor of this informal qualification of the formal answer:

Even in calculus (and certainly in complex analysis) we do speak of "removable singularities/discontinuities", where either a function continuous at a point was whimsically redefined just at that point to make it discontinuous, or, more importantly, where the function had no formal definition at some particular point, but could be defined "by continuity" there. That is, given $f$ defined in a neighborhood of a point $x_o$, if there is a unique value $y_o$ such that the extension of $f$ defined by $f(x_o)=y_o$ is continuous at $x_o$, then this extension is "extension by continuity". Of course, there may fail to be such an extension.

But, in many practical circumstances, there is such an extension. There is at most one such, if there is any.

Thus, depending how we think about it, the value of such $f$ at $x_o$ is entirely determined (by continuity) by near-by values... even if $f$ was not originally "defined" there.

From another side: to redefine a function at a single point does not change its integrals against other functions. Thus, somehow, values at single points are partly irrelevant to the interaction with other objects.

Another: for pointwise evaluation purposes, we might like the map $f\rightarrow f(x_o)$ to be continuous, under some metric on functions themselves. The simplest metric to put on functions on an interval $[a,b]$ is $d(f,g)=\sup_{x\in[a,b]}|f(x)-g(x)|$, and this makes pointwise evaluation continuous.

With other (otherwise reasonable) metrics and such on "functions", pointwise evaluation is often not a continuous map from functions to values.

In that, vein, when we take limits of sequence of functions, we might hope that $\lim_n f_n(x_o)=(\lim_n f_n)(x_o)$. One must stipulate in what sense the limit of functions is taken. If we use the sup-norm metric, this property does hold, and, in fact, the "limit function" is itself continuous.

So, in effect, _in_practice_, a purported pointwise value at a point where a function is not continuous either can be corrected so that the function is continuous, or else has no useful specific value.

  • $\begingroup$ Thank you for your reply. However I confess I lack the mathematical training to fully understand your reply. (My Ph D was in humanities -- I only work with some maths out of my own interest.) Thanks for writing this in detail anyway. $\endgroup$
    – jamadagni
    Jun 16, 2013 at 18:09

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