I am a seventeen year old high school student and I was studying some Real Analysis on my own. In the process, I encountered the Dedekind Cut being used to construct the Reals. I just can't get the hang of the definition that $\mathbb R =\{ \alpha \mid \alpha \text{ is a cut} \}$.

Why does this work? Why is this a good definition for the reals? What got me really thinking was the fact that cuts are subsets of $\mathbb Q$. Why are they used to construct the reals?

Also, to define the reals, we are to consider cuts at exactly the "real" points on the number line, right? If we don't know what they are (the Reals) how can we find a cut corresponding to a real (at the "rational" points there seems to be no problem) and how can I Prove that they are unique.

Maybe my naive thoughts are hindering my progress but I just can't seem to understand the cuts being used here. So could you please elaborate your answers a bit more than necessary so that I can get the concept right.

Any help is much appreciated!

Thanks in advance.

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    $\begingroup$ I don't have time to write a full answer right now, but this may help. When you construct the real numbers you need to construct them out of things you have at hand. If you are constructing the reals, presumably you don't yet have real numbers to work with (why construct them if you did?), so you must use $\mathbb{Q}$. Dedekind cuts are simply a way to represent real numbers in terms of rational numbers. $\endgroup$
    – tomcuchta
    Dec 30, 2011 at 20:59
  • $\begingroup$ Believe it or not it did help a bit. Thanks anyway! $\endgroup$
    – user11470
    Dec 30, 2011 at 21:01
  • $\begingroup$ Set theoretic definition of numbers and In set theory, how are real numbers represented as sets? should be helpful reads. $\endgroup$
    – Asaf Karagila
    Dec 30, 2011 at 21:09
  • $\begingroup$ There is a book written by Terrence Tao, "Analysis 1" published by Hindustan Book Agency. You would certainly find it very interesting. He is a great mathematician who not only see things clearly but write them out clearly as well. $\endgroup$
    – Dilawar
    Jan 3, 2012 at 8:14
  • $\begingroup$ related: gutenberg.org/files/21016/21016-pdf.pdf (translation of Dedekind's paper) $\endgroup$
    – Grigory M
    Jan 4, 2012 at 11:41

8 Answers 8


Devlin K.: The Joy of Sets (Springer, Undergraduate Texts in Mathematics)

In naive set theory we assume the existence of some given domain of 'objects', of which we may build sets. Just what these objects are is of no interest to us. Our only concern is the behavior of the 'set' concept. This is, of course, a very common situation of mathematics. For example, in algebra, when we discuss a group, we are (usually) not interested in what the elements of the group are, but rather in the way the group operation acts upon those elements.

The above quote is mentioned in connection with "definition" of sets, but it shows that this situation is quite common in mathematics.

It is not important how the real numbers are represented, the important thing are their properties.

In the case of Dedekind cuts the starting point is that we suppose we already have defined the rational numbers $\mathbb Q$, and we what somehow get a new set $\mathbb R$, which will have nicer properties. This means that we want somehow define a set $\mathbb R$ together with operations $+$ and $\cdot$ and relation $\le$, such that

  • they have "all familiar properties"; i.e. $(\mathbb R,+,\cdot)$ is an ordered field;
  • they "contain" rational numbers; which formally means that there is an injective map $e:\mathbb Q\to\mathbb R$, which preserves addition, multiplication and inequality;
  • they "improve" the set of rational numbers in the sense that it contains all "missing" numbers; every non-empty subset of $\mathbb R$ which is bounded from above has a supremum, see wikipedia: Least-upper-bound-property.

Note that rational numbers do not have least-upper-bound-property, the set $\{x\in\mathbb Q; x^2<2\}$ does not have a supremum in $\mathbb Q$.

We can give many different definitions which will fulfill the above properties; theoretically they are all equally good; for practical purposes some of them might be easier to work with.

The construction of reals using Cauchy sequences has a similar spirit, in this case the property which we want to add is completeness as a metric space. (Rational numbers do not have this property.)

Let me mention two books, which deal with this topic:

  • Artmann B: The concept of number (Ellis Horwood, 1988). This books mentions several constructions of reals (Dedekind cuts, Cauchy sequences, decimal representation, continued fractions). Advantages and disadvantages of various approaches are mentioned in this book. (Although all construction lead to "the same" - isomorphic - set of reals, some properties of $\mathbb R$ are easy to prove and some might be more difficult, depending on the chosen approach.)

  • Ethan D. Bloch: The Real Numbers and Real Analysis, Springer, 2001. This book is intended as a textbook for a course in real analysis, but it discusses the two most usual definitions of real numbers in detail in the first two chapters.

  • $\begingroup$ Martin: Do you mind if I change the books.google.sk links in your answer to books.google.com? :) $\endgroup$
    – Srivatsan
    Dec 30, 2011 at 23:07
  • $\begingroup$ Srivatsan: Feel free to make any edits you consider an improvement. (In this case, I've already changed the links.) $\endgroup$ Dec 31, 2011 at 0:04
  • $\begingroup$ Martin: Thanks for your edit. $\endgroup$
    – Srivatsan
    Dec 31, 2011 at 0:29
  • $\begingroup$ I'd like to note that the proof which Ethan D. Bloch provides for the fact that if $A < B$, then $A + C < B + C$ for Dedekind cuts is in fact incorrect. Rudin's Principles of Mathematical Analysis, however, provides a correct construction. $\endgroup$
    – jsmith
    Oct 9, 2015 at 3:12
  • $\begingroup$ @jsmith Just to check which proof you mean, you are talking about part 12 of Theorem 1.7.6, right? $\endgroup$ Oct 9, 2015 at 11:01

Let's use the historically incorrect but nowadays popular definition of a cut as a subset $C$ of $\mathbb{Q}$ such that the following conditions hold:

  1. $\emptyset\neq C\neq \mathbb{Q}$. So a cut is a proper subset.
  2. If $q\in C$ and $p\in\mathbb{Q}$ and $p<q$ then $p\in C$.
  3. $C$ has no maximum, there is no element in $C$ larger than all other elements in $C$.

Now we let all cuts be real numbers. It is a messy task to define (!) all algebraic operations for cuts, but the order is straightforward: $C_1\leq C_2$ if and only if $C_1\subseteq C_2$. The idea is that we can identify every real number with the set of all rational numbers below them. We have used cuts to define real numbers, so how can we formulate the intuition that all real numbers can be represented by cuts? What we can do, is showing how we can construct cuts from decimal numbers. So let $0.d_1 d_2 d_3 d_4\ldots$ be a decimal number. It is certainly at least as large as $0.d_1$ and $0.d_1 d_2$ and so on. And every number larger than each of these should be at least as large as the corresponding number. So we can let the real number (and hence cut) we are looking for be the smallest number at least as large as each of $0.d_1 d_2$ and so on. Now $0.d_1$ is a rational number so we can identify it with the cut $C_1=\{q\in\mathbb{Q}:q<0.d_1\}$ and similarly, we can construct all cuts $C_2$, $C_3$ and so on. Now we want to find a real number, a cut $C$, corresponding to the decimal number. Since it is at least as large as $0.d_1$ we have $C_1\subseteq C$. Since it is at least as large as $0.d_1d_2$ we have $C_2\subseteq C$. Continuing this way, we say that we must have $\bigcup_{n=1}^\infty C_n\subseteq C$. We also want the cut to be not larger than necessary. So we can actually take $C=\bigcup_{n=1}^\infty$ (of course, one must verify that $\bigcup_{n=1}^\infty C_n$ is a cut). In this sense, the set of all cuts contains "every real number".

Now, why is the representation unique? Let $C_1$ and $C_2$ be two different cuts and let $C_1\subset C_2$. Then there is a rational number $q$ in $C_2$ that is not in $C_1$. Also, since the cut $C_2$ has no maximum, there exists a rational number $q'$ in $C_2$ that is larger than $q$ and a even larger number $q''$. Let $C_q$ and $C_{q'}$ be the corresponding cuts. Then $C_1\subseteq C_q\subset C_{q'}\subset C_{q''}\subseteq C_2$, so there lies essentially a rational number $q'$ between $C_1$ and $C_2$, which shows that they are different real numbers.

So, intuitively, we can represent every real number by a cut. And in mathematics, we are bold enough to actually use the cuts as our definition. There are other ways to define the real numbers, but they are in some sense the same, and this "the same" needs lengthy explanation in itself.

  • $\begingroup$ Historically incorrect? $\endgroup$
    – Asaf Karagila
    Dec 30, 2011 at 21:47
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    $\begingroup$ Dedekind split $\mathbb{Q}$ into two sets $(L,R)$ to definea cut. But things are actually simpler if one works only with $L$, which is what I have done and what one finds in most books today. The part $R$ is superfluous. $\endgroup$ Dec 30, 2011 at 21:51
  • 1
    $\begingroup$ @Michael: As long as one doesn’t mind occasionally having to deal with $\mathbb{Q}\setminus A$, the one-sided approach works fine for everything and overall seems to impose a smaller notational burden. $\endgroup$ Jan 2, 2012 at 20:45
  • 3
    $\begingroup$ @Adam: I consider Dedekind cuts a far more intuitive construction, since I think of the reals first in terms of their order and only secondarily in terms of their arithmetic, and overall easier to work with to get the whole package. $\endgroup$ Jan 2, 2012 at 20:47
  • 1
    $\begingroup$ @FawzyHegab I guess it depends on what you mean by a real number. What one can show is that all complete ordered fields are isomorphic and as far as one only uses the field and order structure, there is only one notion of real numbers. $\endgroup$ Sep 20, 2013 at 0:48

One clue that might lead you to Dedekind cuts is as follows.

Suppose you are oblivious to the real numbers, and only know about the rationals. One day in physics class you're studying the path of a projectile under constant acceleration. In an example you come up with the equation $y = 2 - x^2$ where and $y$ is height and $x$ is time. Solving for height equal to zero you get $x^2 = 2$ or more specifically $2{q^2} = p^2$ if you express $x$ as $\frac{p}{q}$ (where $p$ and $q$ are positive integers).

But then comes the problem. That last equation has no solution. The proof isn't automatic, but essentially $2{p^2}$ has an odd number of $2$'s in it's factorization, and $q$ must have an even number. This is because of what happens if you square the factorization of an integer... the powers afterwards must all be even, right? So it's established that this equation (and many like it) have no solution.

Now if you dig a pit and repeat the experiment so the projectile can pass below height zero you find a very odd situation. A physical object is above a certain plane and then after a period of time is below a certain plane. In the course of it's travel, did it just skip through that plane? Physical intuition suggests that it could not... even in a strictly geometric sense it's hard to imagine a point just phasing through a plane like that. So perhaps the rationals need to be extended to these new 'real' quantities just like the integers needed to be extended to the rationals.

With this in mind, you have a dilemma: how does one approach these new 'real' quantities if they haven't yet been defined. One idea is to use elimination. Whatever this new quantity is (soon $\sqrt{2}$) in comparison to the rationals it is greater than any rational $r$ with $r^2 < 2$ and less than any rational $s$ with $s^2 > 2$. In the experiment, these are the (rational) times before and after the point passed through the plane. This gives you two intervals of rational numbers... an interval being defined by the property that if it contains two numbers, it contains any number in between those two.

Now, to repeat, we can't solve for the time of intersection using the rationals but we can pick two rational times just before and just after our 'missing' intersection. With some work, we can even pick these two as close apart as we want. This is a good clue that as far as approximation goes, the two intervals are doing a good job. We might also notice that we can just keep track of one of the intervals, say the first one, and we'll remember the second for free: it's just the set of rationals greater than all the rationals in the first set.

After more examples like this, we would see more and more examples of intervals of rational numbers that are unbounded below and bounded above. Some are new, some we have already seen before all this speculation, like $(-\infty,r)$. This is convenient, as it provides a way to represent the rationals using these new intervals.

The last step is to somehow define the various mathematical operations for these placeholder intervals so that we can work with the new quantities by using their known properties, and not by how we represent them using Dedekind cuts. This is fairly obvious for Dedekind cuts that represent positive quantities if you remember to first remove the negative, then do the operation on the sets involved, and then add the negatives back in.

[Note: My explanation is all very loose and rough. Also, it turns out that adding roots of polynomials whenever they cross the $x$-axis is not enough. The real numbers add these, and more, so that there is also always a solution to $f(x) = 0$ whenever you have $f(-1) > 0$ and $f(1) < 0$ for a decreasing function $f$ defined on $[-1,1]$, not just polynomials, etc...]


When we are constructing the reals, we are really, in essence trying to reverse engineer something we already know the properties of from first principals (at least this is how it happened historically). I think mathematicians had a good idea of the properties of real numbers before they had an actual construction for them. Thus "constructing the reals" really means creating a consistent mathematical object that has all the properties that the real numbers should have.

There is more than one construction of $\mathbb{R}$ from $\mathbb{Q}$ (equivalence classes of Cauchy sequences of rationals for instance), but the various methods are equivalent in the sense that the final product has all the same important properties that we care about. In this way we really do not care about uniqueness. You'll come across this same idea again and again in the context of isomorphisms and homeomorphisms.

Think about what truly distinquishes $\mathbb{R}$ from $\mathbb{Q}$...completeness. That is, every subset of $\mathbb{R}$ that is bounded above has a least upper bound. $\mathbb{Q}$ certainly doesn't have this as Dan showed. Dendekind cuts fill in the gaps so to speak by using the flexibility allowed in constructing sets of rational numbers so that the real number they represent is the one that would cause the cut to have a least upper bound if added to the cut.

This always made the most sense to me the first time around when it was put in the context of operations. Let $\alpha, \beta$ be cuts representing the real numbers $1, \pi$ respectively. What does it mean to add $\alpha$ and $\beta$, equivalently $1+\pi$? $$\alpha + \beta =1+\pi = \{x \in \mathbb{Q}|x=r+s \; \text{ such that } \; r \in \alpha, \;\; s \in \beta\}$$ $$ = \{x \in \mathbb{Q}| x <1+\pi\}$$

In essence, after constructing the real numbers and knowing there properties we can manipulate them without thinking about them actually being sets or equivalence classes.


Since the OP is beginning to study real analysis I would strongly advise to read first chapter of "A Course of Pure Mathematics" by G. H. Hardy where these notions are explored in very great detail with almost no symbolism and just plain prose.

The basic confusion seems to be that OP is thinking that real numbers are already there (as points on some number line) and we are somehow trying to make a cut corresponding to a real number. This line of thought has to be avoided seriously. What we have is a system $\mathbb{Q}$ of rational numbers which satisfy common properties of a field. Then we define certain subsets of $\mathbb{Q}$ with certain well defined properties and call it a cut. This cut is a real number. It does not correspond to a real number but is the real number itself.

Another two points which must be noted are:

1) For some Dedekind cuts we can associate a rational number to it (in one-one correspondence) and hence we can identify the rationals with such cuts

2) There are remaining Dedekind cuts for which there can't be such correspondence with rationals and these we call as irrational numbers. These cuts do not correspond to irrationals because irrationals don't exist separately and are available only via this definition of cuts which don't correspond to rationals.

I am also happy that OP is following a trend of reading construction of reals before delving into deeper theorems of analysis. The true understanding of the non-algebraic nature of reals can't be had without an understanding of its construction. I personally don't like the modern trend of treating real numbers axiomatically because their completeness properties are so non-obvious (and form the basis of almost any serious theorem of analysis) to be taken as axioms (especially for beginner high school student).


Let me add a systematic way how you may arrive at the Dedekind cuts:

We start with arbitrary sets of rational numbers. Now we can at first distinguish two types of sets: Those which are bounded from above (that is, for the set $S$ there exists a rational number $q$ so that $x\le q$ for all $x\in S$) and those which aren't. (In the same way, sets may be bounded from below, but that doesn't add something qualitatively new, therefore we concentrate no on being bounded from above).

Looking closer, there are three types of sets bounded from above:

  • The first type has a maximum, that is, there exists a rational number $m\in S$ which is an upper bound. An example of such a set is $S_1=\{x\in\mathbb Q: x\le 0\}$ with the maximum $0$.
  • The second type has no maximum, but at least a supremum: There exists an upper bound $s$ so that no $x<s$ is an upper bound for $S$. Obviously, if a maximum exists, that maximum is the supremum. An example is the set $S_2=\{x\in \mathbb Q: x<0\}$ with the supremum $0$ which is no maximum because $0\notin S_2$.
  • The third type has not even a supremum. An example is the set $S_3=\{x\in\mathbb Q: x<\sqrt{2}\}$.

Now we don't really like statements of the form "only some of X have Y, other's don't". Now it is quite easy to see that we can't do anything on the fact that not all above-bounded sets have a maximum because the set of rationals is dense (if in a set with maximum the maximum is an accumulation point of the set, just removing the maximum from the set doesn't change the supremum, therefore we are left with a set without maximum). However maybe we can do something about the fact that some above-bounded sets don't even have a supremum.

So our goal is to find some sort of number so that every set has a supremum.

As a complication, several sets can have the same supremum, for example the set $\{x\in\mathbb Q: x<0\}$ has the same supremum (namely $0$) as the set $\{x\in\mathbb Q: -1\le x\le 0\}$ and the set $\{-1/n: n\in\mathbb N\}$. Now quite obviously, if $p\in S$, then for the supremum it doesn't matter what goes on below $p$, therefore all sets which differ only below $p$ all have the same supremum. One such set is the one which contains all numbers below $p$. In other words, for each set $S$ with supremum we can find another set with the same supremum, namely the set $D_S:=\{x\in Q: x<y \text{ for some } y\in S\}$. It is only reasonable to demand that the same is true for our extended numbers containing the supremum of any subset.

Now under the non-empty sets bounded from above, those sets $D_S$ are characterised by the property that if $p\in D_S$ then for any $q<p$ also $q\in D_S$. Moreover, any set $D$ with that property obviously equals $D_D$, thus we can limit our consideration to that type of set.

However there can still be two sets of this type with the same supremum. For example, $D_1=\{x\in\mathbb Q: x<0\}$ and $D_2=\{x\in\mathbb Q: x\le 0\}$ both have the supremum $0$. The difference between those is that the set $D_2$ even has a maximum. Now it is easy to check that for each $D$ which has a maximum, that maximum is an accumulation point of $D$, and therefore we can remove the maximum without changing the supremum. In other words, we can just demand that our sets do not have a maximum.

So we are now left with all subsets $D$ of $\mathbb Q$ that have the following four properties:

  • $D$ is not empty
  • $D$ is bounded from above
  • If $p\in D$ and $q<p$ then $q\in D$
  • $D$ does not have a maximal element.

Such sets $D$ are called Dedekind cuts. One can show that for any rational number $q$ there exists exactly one Dedekind cut which has that number as supremum; in addition, we have explicitly constructed a map from an arbitrary above-bounded set $S$ of rational numbers top a Dedekind cut (add all numbers which are below any element of $S$; if the resulting set has a maximum, remove that from the set), and this operation should not change the supremum. Therefore it is reasonable that for our new "supremum numbers" (which we chose to call the "real numbers"), there should be exactly one such number for each Dedekind cut. Therefore we can identify the Dedekind cut with the corresponding real number, that is, we can use the Dedekind cut as a representation of that number (not unlike the fact that the digit string 125 is a representation of the number one hundred and twenty five — and "one hundred and twenty five" is yet another representation of that same number).

Now we "only" have to determine how to compare, add and multiply real numbersin the Dedekind cut representation, and we are finished.

Comparison is easy: If $a\le b$ then the set of rational numbers $<a$ should clearly be a subset of the set of rational numbers $<b$. We want that to be true for real numbers as well, therefore $D_a \le D_b$ iff $D\subseteq D_b$.

Addition also is easy: If $x<a$ and $y<b$, then $x+y<a+b$, and you can go arbitrary close ($(a-\epsilon) + (b-\epsilon) = a+b-2\epsilon$ which gets arbitrary close to $a+b$ if $\epsilon$ gets arbitrary small); again we want to extend that to real numbers, so you add two Dedekind cuts by simply taking the set of all sums of their elements.

Multiplication is more involved, and I omit it here.

Now, while our construction guarantees that we have a supremum for any set of rational numbers, a priori we don't know whether this is also true for any set of real numbers. However one can easily check that it is indeed the case: The supremum of a above-bounded set of real numbers is just the real number described by the union of their Dedekind cuts.


I haven't looked at the construction in some time, but I seem to recall that a "cut" corresponds to a division of the rational numbers into two distinct parts. Where the "cut" occurs may or may not be a rational number, it is simply a division of the rational numbers into sets $A,B$ such that if $x \in A$ and $y \in B$, then $x < y$.

For example, take the following "cut":

  • $\{x \in \mathbb{Q}|x^{2} < 2$ or $x < 0\}$

  • $\{x \in \mathbb{Q}|x^{2} > 2\}$

This set clearly partitions the rationals, but the "cut" itself, the actual point of partition, does not correspond to any rational number. By taking the set of all possible partitions, we are able to not only include the rationals (for we certainly "cut" the rational numbers at rational numbers), but we can also include irrational numbers such as $\sqrt{2}$, as in the example above. Thus we arrive at the set of real numbers, the set of all partitions, the set of all "cuts", in the rational number line.

  • 1
    $\begingroup$ What you wrote is not a cut. You want $\{x\in\mathbb Q|x^2<2\text{ or }x<0\}$. $\endgroup$ Dec 30, 2011 at 21:08
  • 2
    $\begingroup$ A cut is not just any old partition of the rationals: A cut $(A, B)$ satisfies the important condition that every member of $A$ is strictly smaller than every member of $B$. $\endgroup$
    – Srivatsan
    Dec 30, 2011 at 21:08
  • $\begingroup$ Whoops...That's what happens when you rush things =) $\endgroup$ Dec 30, 2011 at 21:10
  • 2
    $\begingroup$ In your definition of cut, you also want to insist that $A$ has no largest element. Otherwise, at the rational numbers themselves, with your definiton, you'd get two cuts: one where the critical point was in the lower half, and one where it was in the upper half. This quibble about the nature of cuts is exactly what leads to the surreal numbers, where those two cuts do indeed correspond to different surreal numbers. $\endgroup$
    – JDH
    Dec 30, 2011 at 21:48
  • 1
    $\begingroup$ $\{x\in\mathbb Q:x^2>2\}$ is incorrect: You mean $\{x\in\mathbb Q:x^2>2\text{ and }x>0\}$. $\endgroup$ Dec 30, 2011 at 23:21

I would like to add another answer, that's perhaps not obvious as being helpful - it doesn't matter!

I should elaborate further ...

It's important that we can model specific structures (i.e. the real numbers in ZFC) so we have both a foundation and a semi-good case to say that the axioms we want the reals to satisfy (complete ordered field) aren't going to get us into trouble (e.g. a contradiction).

Now, the theorems you prove in real or complex analysis depend upon our axioms for a complete ordered field (where the complex numbers are just the cartesian-product of the real numbers). That is we better not worry ourselves at all whether pi is a cut that has -20,00 in; all that matters is we know pi exists working from our axioms for a complete ordered field.

It is extremely important you realise completeness is non-trivial as well. Let's emphasis why - if you only used ordered field axioms to prove - for instance - that 2 has a square root, you could prove it for the rational numbers also (which is false). It also implies the reals are uncountable and makes an easy proof of the fact any two complete ordered fields are isomorphic. However, the most useful thing is that every sequence that should converge, does converge.

  • $\begingroup$ * Or to be really precise - which I should - every Cauchy sequence of real numbers converges to some real number. $\endgroup$
    – Adam
    Dec 30, 2011 at 22:13