Jtdylktuy

Between any two distinct rational numbers, there is an irrational number, and vice versa.

hint

Let $bne a$ be a rationnal.

then $x_n=b+frac{pi}{n}$ is a sequence of irrationals which converges to $b$.

$$f(x_n)=x_nto bne f(b)$$

By the same if $bne a$ is irrational, put $y_n=frac{lfloor 10^n brfloor}{10^n}$ is a sequence of rationals which converges to $b$. but

$$f(y_n)=ane f(b).$$

Continuity at $x=a$.

let $(z_n)$ a sequence which goes to $a$.
then

$$f(z_n)=a text{ or } z_n$$
thus
$$f(z_n)to a=f(a)$$

Since $yne a$, we know $x$ is irrational and $xne a$.

Let $epsilon_0=|x-a|>0$. For any $delta>0$, there exists a rational $r$ such that $|x-r|<delta$, but $|f(x)-f(r)|=|x-a|=epsilon_0$. That is, $f(x)$ is discontinuous at $x$.

Let's assume that a limit exists at $x_0$(call it $l$).

In any neighborhood of $x_0$ we can find rational number, so $f(x) = a$, for some $x$.

Choose $epsilon$ smaller than $|l - a|$.

This argument doesn't work only if $l$ is $a$. The the limit if exists at any point, it should be $a$.

The only point where it can be the limit is $x_0 = a (= f(x_0))$.

At other points choose $epsilon$ less than $|frac{x_0 - a}{2}|$.
Now choose an irrational $x$ sufficiently close to $x_0$ so that $|x-a|$ is greater than $epsilon$.

So that proves not only that it is continuous at only one point, but also that limit doesn't exist at others.

Without loss of generality, fix $a=1$. For $varepsilon=dfrac12$ consider the cases:

• 
  $x,ynotin(1-varepsilon,1+varepsilon)$. For every $delta>0$, $|x-y|<deltaimplies |f(x)-f(y)|=|1-y|geqdfrac12$

(This is true if either $xinmathbb{Q}$ and $yinmathbb{I}$ or otherwise.)

• If either $x$ or $yin(1-varepsilon,1+varepsilon)$, it follows as above.

For continuity at $x=a$, choose $delta=varepsilon$.

In Particular

Suppose $a = 1$

Suppose $f(x)$ continuous
begin{align}
forall delta hspace{0.5cm} |x - y| < delta & quad Rightarrow | 1 - f(y) | < epsilon quad &&wedge quad x_i - f(y) < epsilon
end{align}
Suppose irrational $y_i$, where $y_i > 1 + epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow |1 - y_i| < epsilon \
&Rightarrow |y_i| < epsilon + 1 \
&Rightarrow y_i < 1 + epsilon \
end{split}
end{equation*}
a contradiction

Consider rational $y_r$, where $y_i > y_r > 1 + epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow y_i < 1 + epsilon \
&Rightarrow y_r < 1 + epsilon \
end{split}
end{equation*}
a contradiction

In other words, $f(x)$ is discontinuous for all $y > 1 + epsilon$

Suppose $y_i$, where $y_i < 1 - epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow |1 - y_i| < epsilon \
&Rightarrow |y_i - 1| < epsilon \
&Rightarrow 1 - epsilon < y_i
end{split}
end{equation*}
a contradiction

Consider $y_r$, where $y_i < y_r < 1 - epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow 1 - epsilon < y_i \
&Rightarrow 1 - epsilon < y_r
end{split}
end{equation*}
a contradiction

In other words, $f(x)$ is discontinuous for all $y < 1 - epsilon$

Generally

Suppose $f(x)$ continuous
begin{align}
forall delta hspace{0.5cm} |x - y| < delta & quad Rightarrow | a - f(y) | < epsilon quad &&wedge quad x_i - f(y) < epsilon
end{align}
Suppose $y_i$, where $y_i > |a| + epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow |a - y_i| < epsilon \
&Rightarrow |y_i| < epsilon + |a| \
&Rightarrow y_i < epsilon + |a| \
end{split}
end{equation*}
a contradiction

Consider $y_r$, where $y_i > y_r > |a| + epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow y_i < |a| + epsilon \
&Rightarrow y_r < |a| + epsilon \
end{split}
end{equation*}
a contradiction

In other words, $f(x)$ is discontinuous for all $y > |a| + epsilon$,
and therefore discontinuous for all $y > a + epsilon$

Suppose $y_i$, where $y_i < a - epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow |a - y_i| < epsilon \
&Rightarrow |y_i - a| < epsilon \
&Rightarrow a - epsilon < y_i
end{split}
end{equation*}
a contradiction

Consider $y_r$, where $y_i < y_r < a - epsilon$
begin{equation*}
begin{split}
forall delta hspace{0.5cm} |x - y| < delta &Rightarrow | 1 - f(y_i) | < epsilon \
&Rightarrow a - epsilon < y_i \
&Rightarrow a - epsilon < y_r
end{split}
end{equation*}
a contradiction

In other words, $f(x)$ is discontinuous for all $y < 1 - epsilon$