2
$\begingroup$

I read this content at the bottom of this page enter image description here just wonder why axiom of continuity could guarantee no gaps exist on the real axis? any proofs ?

$\endgroup$
5
  • $\begingroup$ What is your definition of a "gap"? For different definitions there will be different proofs. $\endgroup$ May 25, 2016 at 13:19
  • $\begingroup$ @Abstraction well, it means the real axis is continuous and there is no gaps exist on it. $\endgroup$
    – iMath
    May 25, 2016 at 13:23
  • $\begingroup$ I dont understand the statement because any set by itself is continuous. "Gaps" only make sense for functions under some kind of algebra. $\endgroup$
    – Masacroso
    May 25, 2016 at 13:25
  • 1
    $\begingroup$ In "Understanding Analysis" by Stephen Abbott you can read about several equivalent statements of the "completeness" of the real numbers, of which, the Nested Intervals Property is one. If you construct the real numbers say by Dedekind cuts in the rational numbers, then you can prove that the real numbers have the least upper bound property, which is another equivalent version of completeness. Many authors simply assume completeness as an axiom of the real numbers. $\endgroup$
    – Simon
    May 25, 2016 at 14:42
  • $\begingroup$ Its great that someone is interested to know about continuity and gaps. Rarely do textbooks emphasize on this aspect and jump on using the axiom of continuity as if it were available for free. +1 $\endgroup$
    – Paramanand Singh
    May 25, 2016 at 15:03

3 Answers 3

4
$\begingroup$

The axiom of continuity goes by the name of axiom of completeness also. This is the only property of real numbers which distinguishes it from rational numbers. There are many forms of it which I have discussed in this answer. Since one of the answers has already talked about the gap arising out of an empty intersection of nested intervals, I will talk about two other approaches to gaps in this answer.


In order to understand the concept of "gap" we need to use the geometric language. While we don't need to worry so much about foundations of geometry and definitions of ideas like points and lines, it is important to understand that we do have an intuitive idea of a straight line.

And let's say we would like to represent each point of a line via a specific number and we will call such a line as number line. Thus we can take any point say $O$ on a line and represent it by number $0$. By convention such a number line is shown horizontally and then a point on the right of $O$ is chosen (say $A$) and represented by number $1$. If $P$ is a point to the right of $O$ such that length of $OP$ is $n$ times that of $OA$ (where $n$ is a natural number) then we represent $P$ by number $n$. If the point $P$ was to the left of $O$ then it is represented by integer $-n$.

In so doing we see that some points of the line have been represented by integers, but there are many points on the line still left which need to be represented. Then we get the idea of using rational numbers to represent points of the number line. Using Euclidean constructions it is possible to mark a point $Q$ to the right of $O$ such that length of $OA$ is $n$ times that of $OQ$ (where $n$ is some specific natural number). We then represent point $Q$ by rational number $1/n$. If $P$ is a point to the right of $O$ such that length of $OP$ is $m$ times that of $OQ$ (where $m$ is again a specific natural number) then $P$ is represented by rational number $m/n$. If in the last construction $P$ were on the left of $O$, it would have been represented by $-m/n$. In this manner we have represented some more points of the line using rational numbers.

Our goal is to ensure that we represent all the points of the number line via some sort of numbers. The mechanism described in the last paragraph manages to represent many points on the line via rational numbers, but there are Euclidean constructions available to create a point $P$ such that line segment $OP$ is incommensurable with line segment $OA$ (two line segments $LM$ and $RS$ are said to be commensurable if there is a line segment $XY$ available such that length of $LM$ as well as that of $RS$ is a multiple of length of $XY$). One such construction is as follows: construct $OB$ perpendicular to $OA$ and of same length as $OA$, choose $P$ on the number line such that length of $OP$ is same as that of $AB$. This shows that there is a point $P$ on number line which is not represented by a rational number.

Now we can choose any point $Q$ on number line such that length of $OQ$ is a multiple of that of $OP$ and then this new point $Q$ will also not be represented by a rational number. Thus it is possible to find as many points as we please on the number line which are not represented by any rational number.

If we keep only those points of the number line which are represented by rational numbers and remove other points, then visually the number line will look almost the same but there is a deep dissatisfaction that the line has now lost infinitely many of its points and has gaps. Thus in order to model the number line appropriately we need to extend our number system by considering numbers of different kind which will be able to represent the points of the line which are not represented by rational numbers. This is one aspect of "gap" which is easy to understand but its appreciation depends on geometric intuition.


So let's try a non-geometric approach. We know that there is no rational number whose square is $2$. Thus if $x$ is a positive rational number then either $x^{2} < 2$ or $x^{2} > 2$. Put all positive rational numbers $x$ in a set $A$ if $x^{2} < 2$. All the other positive rational numbers are put in set $B$. Clearly both $A, B$ are non-empty and disjoint. Together they contain all positive rational numbers. And most importantly every member of $A$ is less than any member of $B$. Clearly $1 \in A, 2 \in B$ and as we start moving from $1 \in A$ and taking greater and greater rational number we will eventually reach $2 \in B$ and it makes perfect sense to assume that while transitioning from members of $A$ to members of $B$ there will be exactly two exclusive possibilities:

  • There is a last member say $a \in A$ such all positive rational numbers less than $a$ belong to $A$ and all rational numbers greater than $a$ belong to $B$.
  • There is a first member say $b \in B$ such all positive rational numbers less than $b$ belong to $A$ and all rational numbers greater than $b$ belong to $B$.

Thus there has to be some rational number which acts as the boundary between sets $A$ and $B$. To our utter surprise the example we have chosen above does not have any such boundary number (it is not obvious but not too difficult to prove this). On the other hand if we define $A, B$ as $$A = \{x \mid x \in \mathbb{Q}^{+}, x^{2} \leq 1\}\, B = \{x \mid x \in \mathbb{Q}^{+}, x^{2} > 1\}$$ then there is a rational number which acts as boundary between $A$ and $B$ (obviously the boundary number is $1$ and it belongs to $A$).

BTW we dealt with positive numbers just for ease of understanding and we can very well include both positive and negative rational numbers in our discussion. Thus if all the rational numbers are divided into two disjoint and non-empty sets $A$ and $B$ such that each member of $A$ is less than any member of $B$ then based on specific choices of $A, B$:

  • It is possible that there is a rational number say $r$ which acts as a boundary between $A$ and $B$ i.e. all rational numbers less than $r$ lie in $A$ and all rational numbers greater than $r$ lie in $B$. The rational number $r$ itself may lie in $A$ or in $B$.
  • It is possible that there is no rational number which acts a boundary between $A$ and $B$.

Thus the rational number system possesses a "gap" and we say that the system lacks continuity (continuity in going from $A$ to $B$) or completeness and hence we need to enlarge our number system to fill such gaps (i.e create a number system where partitioning all numbers into two sets $A$ and $B$ as mentioned above always leads to a boundary number).

The real numbers are designed/constructed to fill this gap and it guarantees the existence of a boundary number between $A$ and $B$ whenever the real numbers are partitioned into two sets $A$ and $B$ in the manner mentioned above.


Most textbooks postulate the axioms which assert that there are no gaps in the real number system but they neither illustrate the gap nor show how the gap is filled. Perhaps the textbook from which the image in question has been taken also follows the same approach and that is the reason for the appearance of this question. It is important to understand the issue of "gap" in the rational number system and understand at least one approach as to how the gap is filled via the system of real numbers.

$\endgroup$
2
  • $\begingroup$ thanks very much , you always provide a clear explanation, hope we could have further communication , mail to me [email protected] $\endgroup$
    – iMath
    May 28, 2016 at 7:51
  • $\begingroup$ @iMath: Ok will get in touch on mail. $\endgroup$
    – Paramanand Singh
    May 28, 2016 at 7:57
3
$\begingroup$

In the context of this passage: "Postulate of Nested Intervals" and "Axiom of Continuity" are synonyms.

The phrase "it guarantees that no gaps exist on the real axis" is is not a formal mathematical statement, unlike the statement of the postulate itself. Instead it is written in an attempt to make this postulate/axiom appeal to your intuition.

The intuition is that if the intersection of $I_1,I_2,I_3,...$ were empty, then the "missing point which ought to be there in the intersection" would represent a gap in the real line. Like many intuitions, especially intuitions about things which are not supposed to exist, this is somewhat murky. But perhaps it can be understood with an example: the real line would have a gap at $0$ if the intersections of the intervals $$I_1=[-1,+1],\quad I_2=[-1/2,1/2], \quad I_3=[-1/3,1/3], ...$$ were empty.

$\endgroup$
1
  • 2
    $\begingroup$ Contrast this with the rationals $\mathbb Q$, which does have a gap, namely at the intersection of $[3,4], [3.1,3.2], [3.14,3.15], [3.141,3.142], ...$. $\endgroup$ May 25, 2016 at 13:32
1
$\begingroup$

wikipedia gives a very good example on explaining why real number system lack of irrational numbers is not complete(has gaps!) https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers#Nested_intervals_theorem

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .