Generalizations and extensions are commonplace in mathematics, but I am not clear on what the difference is between the two. Are they really only synonyms? If not, I would appreciate having an example of each, with it being made clear why the generalization should not be called an extension, and why the extension should not be called a generalization.

  • $\begingroup$ Can you give examples of what you mean? I can explain the difference between generalization and abstraction. For example to go from $\mathbb R$ to $\mathbb R^n$ is generalization; to go from $\mathbb R$ to topological spaces is abstraction. So what exactly is extension in this context? $\endgroup$
    – user4894
    Apr 20, 2017 at 23:30
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    $\begingroup$ I may extend a result from a smaller to a larger set for example, or I can have the result generalised so that it is a corollary of a more general statement. But in either case there is nothing to stop me from calling the generalisation an extension of my 'smaller' result. $\endgroup$
    – AnyAD
    Apr 20, 2017 at 23:41
  • $\begingroup$ @user4894: Some would say that for n > 3, the situation is already pretty abstract. My take is that it is good to be aware, as you point out, of the two poles of generalization and abstraction, but that there is an unbroken transition between them, much like the unbroken transition between vapor and liquid above the critical point in the P-T phase diagram of water. $\endgroup$
    – user27325
    Apr 21, 2017 at 15:26

4 Answers 4


Let's start with two examples:

  • Suppose you have a function $f\colon(0,1)\to\mathbb R$. You can turn this into a function $\tilde f\colon\mathbb R\to\mathbb R$ by defining, for example, $\tilde f(x)=f(x)$ when $x\in(0,1)$ and $\tilde f(x)=0$ otherwise. The function $\tilde f$ is an extension of $f$, not a generalization.

  • Maybe you have proved earlier that if $x\leq y\leq x$ for some real $x$ and $y$, then $x=y$. But in fact the same holds true if $x$ and $y$ are matrices. There is a natural definition of "$\leq$" for matrices. In the case of $1\times1$ matrices you get the original version back. The result for matrices is a generalization of the result on the real line. I would not object to calling this an extension, but I would prefer to use the word "generalization" instead.

I would say that in general objects are extended and results are generalized.

You might want to extend your field from the reals to the complex (or make some other field extension), or you might want to extend the domain of definition of some function. Generalization is a more general result that contains the original result as a special case. If the original result is not included, I would not call the new result a generalization but and analogous result inspired by the original. Similarly, if a new function is somehow inspired by an old one but does not "contain" it, I would prefer to call it an induced function rather than an extension.

  • $\begingroup$ Very nice. I would accept your answer, but I've been chewed out in the past for accepting an answer before others have had a chance to weigh in (like, at least waiting 24 hours). Although I am a satisficer, not a maximizer, I think I'm going to cave in to public pressure on this one. I am, however, up-voting your answer immediately. $\endgroup$
    – user27325
    Apr 21, 2017 at 1:15
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    $\begingroup$ @EsperantoSpeaker1 Take your time. Accepting immediately would only discourage others from giving their answers. I really appreciate the feedback, but I'm not in a hurry to get 15 more internet points. $\endgroup$ Apr 21, 2017 at 1:24
  • $\begingroup$ @JoonasIlmavirta Regarding your first example $f\colon(0,1)\to\mathbb R$: Then it's often, and often tacitly, assumed that it acts on the corresponding power sets $\,\mathcal P\big((0,1)\big)\to\mathcal P(\mathbb R)\,$ in the obvious way $\,S\mapsto\{\,f(x)\mid x\in S\,\}$. Would you call this an extension, or a generalisation? As for every $x\in(0,1)$ we have $\{x\}\in \mathcal P\big((0,1)\big)$ this smells more a (natural) extension, doesn't it? $\endgroup$
    – Hanno
    Apr 21, 2017 at 6:21
  • $\begingroup$ @Hanno I would call that an extension much rather than a generalization. But my preference would be to call it the induced map between the power sets. $\endgroup$ Apr 21, 2017 at 6:46
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    $\begingroup$ @EsperantoSpeaker1 Words like "extension" and "generalization" are soft, and they can often be replaced by other words. Analytic continuation is a good example of extension. I thought perihelion and aphelion were reserved for objects orbiting a sun (Greek: gee$\approx$Earth, helion$\approx$Sun). $\endgroup$ Apr 21, 2017 at 16:51

I agree with Joonas that "extension" can be used to refer to objects, such as functions extended to a larger domain while preserving some properties, or in field extensions (a larger field containing the original one). However, we do often use the same word to refer to generalizations of a theorem as well, namely we could say something like "We can easily extend this result to an arbitrary metric space.". Hence you see things like "extended mean-value theorem" and "extended Bayes theorem" and "extended Green's theorem". Also, the Wikipedia article about the binomial theorem even uses both words together: "The generalized binomial theorem can be extended to the case where x and y are complex numbers.".

The bottom line is that the words "generalization" and "extension" turn out to just have the natural English meaning even when used in mathematics, most of the time. It may nevertheless be worth pointing out that there is one technical meaning of "extension" in the field of logic, which has little to do with the common English meaning.


Let us think of a mathematical object as a set on which some operations are defined. An extension is a bigger object of which the original object is part, while a generalization is a less tightly defined object, of which the original object is an example. In an extension, the operations defined on the original set are extended to apply in a similar manner to the extended set. In both cases, some of the characteristic properties of the original operations hold, but generally not all of them.

The canonical example is the counting numbers: $1$, $2$, $3$, ... . This set ($\Bbb Z_+$) of numbers is extended first by $0$ to the natural numbers ($\Bbb N$) and then by the negative numbers to form the integers ($\Bbb Z$). The latter set may naturally be generalized to varying degrees; in particular, it is an example of a commutative ring with identity, which is not a completely defined object. Another example of such a ring is the ring of polynomials in one variable over the integers, $\Bbb Z[x]$, which is itself an extension of the integers. But not all cases of a generalization need to be an extension: for instance, the "clock-arithmetic" ring $\Bbb Z_{12}$ is not an extension of $\Bbb Z$.


This thread has been dead for a long time, but I will post my answer nevertheless.

To define generalisation, assume that we have a definition $D_0$ that is given by a set of properties $\{d_1,\ldots\ ,d_n\}$ that an object must satisfy to be framed by the definition. This definition leads to a set of elements $C_0$ that satisfies the properties of $D_0$. Now remove some $d_i$ from the list of properties. This leads to a new definition $D_1$ and a new set of elements $C_1$. Necessarily, $C_0 \subseteq C_1$. In this case, we say that $D_1$ is a generalisation of $D_0$. That is; whenever you remove constraints, you generalise, and in the generaliation, you include more elements.

An extension can be a generalisation. Starting with the natural numbes $\mathbb{N}=\{1,2,\ldots\}$ and extending to the integers $\mathbb{Z}=\{\ldots,-1,0,1,\ldots\}$ causes more numbers to be included in the set; hence a generalisation. This is because $\mathbb{N}=\{n \in \mathbb{Z} | n \ge 1\}$, and we obtain $\mathbb{Z}$ by removing the requirement $n \ge 1$.

But extension can be used about non-generalising actions too. If $C^1([0,1])$ are the once continuously differentiable function on $[0,1]$, we can make an extension to $C^1(\mathbb{R})$ by requiring $f'(0)=f'(1)=0$, and defining $$ \hat f (x) = \begin{cases} f(x) & \text{for $x \in [0,1]$} \\ f(0) & \text{for $x \le 0$} \\ f(1) & \text{for $x \ge 1$.} \end{cases} $$ But this extension cannot be considered to be a generalisation.

$\Sigma$: My impression is that extension should be understood in the general English meaning of increasing the reach of something; being a domain, the number of members or anything alike. Whereas generalisation should be understood as defined above.


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