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

[Eclair_de_XII]

TL;DR

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.

 

[PeroK]

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 $\mathbb Q$.

 

[PeroK]

Supremum is the least upper bound. That's well defined for sets of rationals. Although, not every set of rationals has a supremum.

 

[Eclair_de_XII]

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?

 

[fresh_42]

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.

 

[dextercioby]

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 $\mathbb{R} \setminus \mathbb{Q} \neq \emptyset$.