MHB About showing a function : not unbounded on B

[bw0young0math]

Hello, I have a problem. I think this problem about 5 hours. But I couldn't finish this.
In fact, I solve a little. But I don't know whether it is right or not.
If you check this problem, I really appreciate you..

*This is the problem from Bartle 5.1 #14.

Let $A=(0,∞)$, $k:A→\Bbb R$ is given by $$k(x)=\begin{cases}0& (x∈A∩\Bbb Q^c)\\n &(x∈A∩\Bbb Q, x = \frac{m}{n}, (m,n)=1) \end{cases}$$
Let $B$ be an arbitrary open interval in A ($B⊂A$).
Show that function $k$ is unbounded on $B$.* My solution:
There exists $B=(a,b)⊂A=(0,∞)$ such that $k$ is bounded on $B$.
So,there exist $M>0$ such that $\forall x\in B$, $\lvert k(x)\rvert \leq M$
ⅰ)$a,b\in \Bbb R$
Let $C=(a,\min\{a+1,b\})$
Let $D=\{x\in C｜x=m/n ,(m,n)=1, n\leq M\}$. Then $D$ is finite set.
Let $c_1$=the smallest number in D and $c_2$ is next number in $D$ ($c_1<c_2$)
Then by $\Bbb Q$-density, there exist a rational number $c$ in interval ($c_1, c_2$). Then, $c=q/p$, $(p,q)=1$, $p>M$.
So $k(c)=p>M$ and $c \in (c_1,c_2) \subset C = (a,\min\{a+1,b\})\subset B=(a,b)\subset A=(0,∞)$.
Thus this is contradiction to "$k$ is bounded on $B$."
Thus $k$ is not bounded on arbitrary open intervals included in $A$.

ⅱ)$a,b\in \Bbb R \cup \{∞\}$
Let $C=(a,a+1)$. Then Let's do something similar way to ⅰ).



Thus $k$ is not bounded on arbitrary open intervals included in $A$.

How about my inference? Please check above, and if you have good solution, let me know it... Thank you.

 

[Opalg]

Hello, I have a problem. I think this problem about 5 hours. But I couldn't finish this.
In fact, I solve a little. But I don't know whether it is right or not.
If you check this problem, I really appreciate you..

*This is the problem from Bartle 5.1 #14.

A=(0,∞) k:A→R(real numbers) ,k(x)=0 (x∈A∩Q^c) k(x)=n (x∈A∩Q,x=m/n (m,n)=1 )
Let B : arbitrary open interval in A (B⊂A)
Show that function k is unbounded on B.* My solution:
There exists B=(a,b)⊂A=(0,∞) such that k is bounded on B.
So,there exist M>0 such that ∀x∈B=(a,b), ｜k(x)｜≤M
ⅰ)(a,b∈R)
Let C=(a,min{a+1,b})
Let D={x∈C｜x=m/n,(m,n)=1, n≤M}. Then D is finite set.
Let c1=the smallest number in D and c2 is next number in D (c1<c2)
Then by Q-density, there exist a rational number c in interval (c1,c2). Then, c=q/p (p,q)=1, p>M.
So k(c)=p>M & c∈(c1,c2)⊂C=(a,min{a+1,b})⊂B=(a,b)⊂A=(0,∞).
Thus this is contradiction to "k is bounded on B."
Thus k is not bounded on arbitrary open intervals included in A.

ⅱ)(a,b∈R∪{∞})
Let C=(a,a+1). Then Let's do something similar way to ⅰ)(a,b∈R)



Thus k is not bounded on arbitrary open intervals included in A.

How about my inference? Please check above& If you have good solution, let me know it... Thank you.

Your proof is correct. (Yes)

As a matter of personal choice, I would prefer to use a direct proof rather than a proof by contradiction. In fact, the definition of unboundedness says that given a positive number $N$, we must find a point $x$ in $B$ such that $k(x) > N$. To do that, choose a prime number $n > N$ such that $1/n$ is less than half the length of the interval $B$.

Since $n$ is prime, $k(m/n) = n$ whenever $m$ is not a multiple of $n$. But since $1/n$ is less than half the length of $B$, it follows that $B$ contains at least two points of the form $m/n$. For at least one of those points, $m$ will not be a multiple of $n$, and so the function $k$ will take the value $n$ at that point. Since $n>N$, this shows that the definition of unboundedness is satisfied for the function $k$ on the interval $B$.

 

[bw0young0math]

Your proof is correct. (Yes)

As a matter of personal choice, I would prefer to use a direct proof rather than a proof by contradiction. In fact, the definition of unboundedness says that given a positive number $N$, we must find a point $x$ in $B$ such that $k(x) > N$. To do that, choose a prime number $n > N$ such that $1/n$ is less than half the length of the interval $B$.

Since $n$ is prime, $k(m/n) = n$ whenever $m$ is not a multiple of $n$. But since $1/n$ is less than half the length of $B$, it follows that $B$ contains at least two points of the form $m/n$. For at least one of those points, $m$ will not be a multiple of $n$, and so the function $k$ will take the value $n$ at that point. Since $n>N$, this shows that the definition of unboundedness is satisfied for the function $k$ on the interval $B$.

Thank you for your reply.:D
Um.. I have a question. If interval B=(a,b)(b=∞), length of B is infinite. Then how can I prove this?(Because I cannot define an half of the length B when length of B is infinite.)

&

How did you think of an half of length of B? I wonder the inference.

 

[Opalg]

If interval B=(a,b)(b=∞), length of B is infinite. Then how can I prove this?(Because I cannot define an half of the length B when length of B is infinite.)

You essentially dealt with this in your original post. If the length of $B$ is infinite, you can replace $B$ by a subinterval of finite length, say a subinterval of length $1$. If the function is unbounded on that subinterval then it is certainly unbounded on the whole interval.

How did you think of an half of length of B? I wonder the inference.

I wanted to ensure that there would be at least two consecutive multiples of $1/n$ in the interval. For that, it is necessary to choose $n$ so that $1/n$ is less than half the length of the interval.

 

[bw0young0math]

You essentially dealt with this in your original post. If the length of $B$ is infinite, you can replace $B$ by a subinterval of finite length, say a subinterval of length $1$. If the function is unbounded on that subinterval then it is certainly unbounded on the whole interval.I wanted to ensure that there would be at least two consecutive multiples of $1/n$ in the interval. For that, it is necessary to choose $n$ so that $1/n$ is less than half the length of the interval.

Thank you so much!Thank you!