Can the continuity of functions be defined in the field of rational numbers?

In summary, the function of continuity is defined as follows: A function is continuous at some point if for all positive values of ##\epsilon##, there is an integer ##N## with the property that whenever a natural number ##n## is greater than or equal to ##N##, ##f(x_n)## is within distance ##\epsilon## of ##f(x_0)##, where ##x_n## is a sequence of points converging to ##x_0##. This is done by defining an open neighborhood around ##x_0## and using the absolute value function.
  • #1
Eclair_de_XII
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TL;DR Summary
The function of continuity is paraphrased as follows.

A function ##f## is said to be continuous at some point ##x_0## in some field if for all positive values of ##\epsilon##, there is an integer ##N## with the property that whenever a natural number ##n## is greater than or equal to ##N##, ##f(x_n)## is within distance ##\epsilon## of ##f(x_0)##, where ##x_n## is a sequence of points converging to ##x_0##.
I argue not. Let ##f:\mathbb{Q}\rightarrow\mathbb{R}## be defined s.t. ##f(r)=r^2##. Consider an increasing sequence of points, to be denoted as ##r_n##, that converges to ##\sqrt2##. It should be clear that ##\sqrt2\equiv\sup\{r_n\}_{n\in\mathbb{N}}##. Continuity defined in terms of sequences of numbers requires that whenever a sequence of points converges to some point, the image of the sequence must converge to the image of that point, also. However, since the concept of the supremum does not even exist in the field of rational numbers, the definition given fails, since the convergence of the sequence given depends on aforementioned concept.
 
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  • #2
Sequences and their convergence can be defined in ##\mathbb Q##. There's no problem there.

I can't see any problem defining continuity of a function whose domain is ##\mathbb Q##.
 
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  • #3
Supremum is the least upper bound. That's well defined for sets of rationals. Although, not every set of rationals has a supremum.
 
  • #4
PeroK said:
Although, not every set of rationals has a supremum.
Are you referring to bounded sets whose supremums are not even contained in the field of rational numbers, such as the sequence I defined above?
 
  • #5
Eclair_de_XII said:
Summary: mbers? The function of continuity is paraphrased as follows.

A function ##f## is said to be continuous at some point ##x_0## in some field if for all positive values of ##\epsilon##, there is an integer ##N## with the property that whenever a natural number ##n## is greater than or equal to ##N##, ##f(x_n)## is within distance ##\epsilon## of ##f(x_0)##, where ##x_n## is a sequence of points converging to ##x_0##.

I argue not. Let ##f:\mathbb{Q}\rightarrow\mathbb{R}## be defined s.t. ##f(r)=r^2##. Consider an increasing sequence of points, to be denoted as ##r_n##, that converges to ##\sqrt2##. It should be clear that ##\sqrt2\equiv\sup\{r_n\}_{n\in\mathbb{N}}##. Continuity defined in terms of sequences of numbers requires that whenever a sequence of points converges to some point, the image of the sequence must converge to the image of that point, also. However, since the concept of the supremum does not even exist in the field of rational numbers, the definition given fails, since the convergence of the sequence given depends on aforementioned concept.
A function ##f## is continuous if every pre-image of an open set is open.

So all it takes are topologies on ##\mathbb{Q}## and ##\mathbb{R}##, i.e. a definition of what sets are called open. We have a distance function given by ##d(x,y)=|x-y|## on both sets, so that we can define an open neighborhood around a point ##x## by ##U_r(x)=\{y\,|\,d(x,y)< r\}##. These sets build our basis for any open set ##O## in both cases, i.e. arbitrary unions of those neighborhoods: ##O=\cup_{r,x}\, U_r(x)##

There is nowhere a distinction between ##\mathbb{Q}## and ##\mathbb{R}## needed. The absolute value works in both cases and therefore continuity can be defined the same way in both cases: ##f^{-1}(O)## is open.

The difference between both sets is, that one is complete and the other one is not. Completeness means that every Cauchy sequence has a limit, i.e. a sequence with members getting closer and closer to each other has a limit. This is true for ##\mathbb{R}## and ##\mathbb{C}## but not for ##\mathbb{Q}.## Completeness is a property of certain topological spaces. Continuity, however, is defined for any function between any topological spaces. Even a function ##f\, : \,\{0,1\} \longrightarrow \mathbb{R}## can be continuous.
 
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  • #6
That is how you get ##\mathbb R##, by adding to ##\mathbb Q## all the limits of sequences of the rational numbers. Some limits are not in ##\mathbb Q##, so we have a strict inclusion ##\mathbb{R} \setminus \mathbb{Q} \neq \emptyset ##.
 

Related to Can the continuity of functions be defined in the field of rational numbers?

1. What is continuity of functions in the field of rational numbers?

Continuity of functions refers to the property of a function being unbroken or uninterrupted at all points within its domain. In the field of rational numbers, this means that the function is defined and has a value at every rational number within its domain.

2. Can a function be continuous on the rational numbers but not on the real numbers?

Yes, it is possible for a function to be continuous on the rational numbers but not on the real numbers. This is because the rational numbers are a subset of the real numbers, so a function can be defined and continuous on the rational numbers but not on the entire set of real numbers.

3. How is continuity of functions on the rational numbers different from continuity on the real numbers?

The main difference is that continuity on the rational numbers only requires the function to be defined and have a value at every rational number within its domain, while continuity on the real numbers requires the function to be defined and have a value at every real number within its domain. This means that a function can be continuous on the rational numbers but not on the real numbers.

4. What are some examples of functions that are continuous on the rational numbers?

Some examples of functions that are continuous on the rational numbers include linear functions, quadratic functions, polynomial functions, and trigonometric functions with rational coefficients. These functions have a defined value at every rational number within their domains.

5. How is the continuity of functions on the rational numbers important in mathematics and science?

The continuity of functions on the rational numbers is important in mathematics and science because it allows us to model and analyze real-world phenomena using rational numbers. It also helps us to understand the behavior of functions and make predictions based on their continuity. In addition, the concept of continuity is fundamental in calculus and plays a crucial role in many mathematical and scientific applications.

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