For every $xinBbb R$ and $ninBbb Nsetminus {0}$, there exists $r,sinBbb Q$ such that $r< x<s$ and...
For every $xinBbb R$ and $ninBbb Nsetminus {0}$, there exists $r,sinBbb Q$ such that $r< x<s$ and $s-r< 1/n$.
This is a lemma in Chapter 10 from my textbook Introduction to Set Theory by Hrbacek and Jech. Although the authors have provided a proof, I have figured another one. Please help me verify my proof! Thank you so much!
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(s_i mid iinBbb N)$ by $s_i=u_i-l_i$ for all $iinBbb N$. It is easy to verify that $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$.
Let $i_0=min {iinBbb N mid s_i<1/n}$. Since $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$, there exists $s_i$ such that $0<s_i<1/n$ for all $ninBbb N$ and thus such $i_0$ exists.
It is clear that $l_{i_0}<x<u_{i_0}$ and $s_{i_0}=u_{i_0}-l_{i_0}<1/n$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
With a similar reasoning, I have proved the below theorem:
For every $x,kinBbb R$ such that $0<x,1<k$. There exists $r,sinBbb Q$ such that $0<r<x<s$ and $s/r<k$.
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of positive rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(k_i mid iinBbb N)$ by $k_i=u_i/l_i$ for all $iinBbb N$. It is easy to verify that $(k_i mid iinBbb N)$ is a strictly decreasing sequence converging to $1$.
Let $i_0=min {iinBbb N mid k_i<k}$. It is clear that $l_{i_0}<x<u_{i_0}$ and $k_{i_0}=u_{i_0}/l_{i_0}<k$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
real-numbers
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
add a comment |
For every $xinBbb R$ and $ninBbb Nsetminus {0}$, there exists $r,sinBbb Q$ such that $r< x<s$ and $s-r< 1/n$.
This is a lemma in Chapter 10 from my textbook Introduction to Set Theory by Hrbacek and Jech. Although the authors have provided a proof, I have figured another one. Please help me verify my proof! Thank you so much!
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(s_i mid iinBbb N)$ by $s_i=u_i-l_i$ for all $iinBbb N$. It is easy to verify that $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$.
Let $i_0=min {iinBbb N mid s_i<1/n}$. Since $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$, there exists $s_i$ such that $0<s_i<1/n$ for all $ninBbb N$ and thus such $i_0$ exists.
It is clear that $l_{i_0}<x<u_{i_0}$ and $s_{i_0}=u_{i_0}-l_{i_0}<1/n$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
With a similar reasoning, I have proved the below theorem:
For every $x,kinBbb R$ such that $0<x,1<k$. There exists $r,sinBbb Q$ such that $0<r<x<s$ and $s/r<k$.
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of positive rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(k_i mid iinBbb N)$ by $k_i=u_i/l_i$ for all $iinBbb N$. It is easy to verify that $(k_i mid iinBbb N)$ is a strictly decreasing sequence converging to $1$.
Let $i_0=min {iinBbb N mid k_i<k}$. It is clear that $l_{i_0}<x<u_{i_0}$ and $k_{i_0}=u_{i_0}/l_{i_0}<k$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
real-numbers
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
Looks great. I would just add a comment to explain why such an $i_0$ exists.
– mathcounterexamples.net
Jan 4 at 13:17
Hi @JonasDeSchouwer, we have $l_{i+1}-l_i>0$ and $u_{i+1}-u_i<0$. I can not see how $l_{i+1}-l_ile u_{i+1}-u_i$.
– Le Anh Dung
Jan 4 at 13:22
Thank you for your suggestion @mathcounterexamples.net! I have added a argument for that part.
– Le Anh Dung
Jan 4 at 13:27
2
Rather than It is easy to verify that $(s_i)$ is a strictly decreasing sequence converging to $0$., I would say $(s_i)$ is the difference of sequences both converging to $x$, hence it converges to $0$.
– mathcounterexamples.net
Jan 4 at 13:31
add a comment |
For every $xinBbb R$ and $ninBbb Nsetminus {0}$, there exists $r,sinBbb Q$ such that $r< x<s$ and $s-r< 1/n$.
This is a lemma in Chapter 10 from my textbook Introduction to Set Theory by Hrbacek and Jech. Although the authors have provided a proof, I have figured another one. Please help me verify my proof! Thank you so much!
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(s_i mid iinBbb N)$ by $s_i=u_i-l_i$ for all $iinBbb N$. It is easy to verify that $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$.
Let $i_0=min {iinBbb N mid s_i<1/n}$. Since $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$, there exists $s_i$ such that $0<s_i<1/n$ for all $ninBbb N$ and thus such $i_0$ exists.
It is clear that $l_{i_0}<x<u_{i_0}$ and $s_{i_0}=u_{i_0}-l_{i_0}<1/n$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
With a similar reasoning, I have proved the below theorem:
For every $x,kinBbb R$ such that $0<x,1<k$. There exists $r,sinBbb Q$ such that $0<r<x<s$ and $s/r<k$.
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of positive rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(k_i mid iinBbb N)$ by $k_i=u_i/l_i$ for all $iinBbb N$. It is easy to verify that $(k_i mid iinBbb N)$ is a strictly decreasing sequence converging to $1$.
Let $i_0=min {iinBbb N mid k_i<k}$. It is clear that $l_{i_0}<x<u_{i_0}$ and $k_{i_0}=u_{i_0}/l_{i_0}<k$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
real-numbers
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
For every $xinBbb R$ and $ninBbb Nsetminus {0}$, there exists $r,sinBbb Q$ such that $r< x<s$ and $s-r< 1/n$.
This is a lemma in Chapter 10 from my textbook Introduction to Set Theory by Hrbacek and Jech. Although the authors have provided a proof, I have figured another one. Please help me verify my proof! Thank you so much!
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(s_i mid iinBbb N)$ by $s_i=u_i-l_i$ for all $iinBbb N$. It is easy to verify that $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$.
Let $i_0=min {iinBbb N mid s_i<1/n}$. Since $(s_i mid iinBbb N)$ is a strictly decreasing sequence converging to $0$, there exists $s_i$ such that $0<s_i<1/n$ for all $ninBbb N$ and thus such $i_0$ exists.
It is clear that $l_{i_0}<x<u_{i_0}$ and $s_{i_0}=u_{i_0}-l_{i_0}<1/n$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
With a similar reasoning, I have proved the below theorem:
For every $x,kinBbb R$ such that $0<x,1<k$. There exists $r,sinBbb Q$ such that $0<r<x<s$ and $s/r<k$.
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i mid iinBbb N)$ of positive rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i mid iinBbb N)$ of rationals converging to $x$.
We define a sequence $(k_i mid iinBbb N)$ by $k_i=u_i/l_i$ for all $iinBbb N$. It is easy to verify that $(k_i mid iinBbb N)$ is a strictly decreasing sequence converging to $1$.
Let $i_0=min {iinBbb N mid k_i<k}$. It is clear that $l_{i_0}<x<u_{i_0}$ and $k_{i_0}=u_{i_0}/l_{i_0}<k$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
real-numbers
real-numbers
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
edited Jan 4 at 13:26
Le Anh Dung
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
asked Jan 4 at 13:07
Le Anh DungLe Anh Dung
1,0531521
1,0531521
Looks great. I would just add a comment to explain why such an $i_0$ exists.
– mathcounterexamples.net
Jan 4 at 13:17
Hi @JonasDeSchouwer, we have $l_{i+1}-l_i>0$ and $u_{i+1}-u_i<0$. I can not see how $l_{i+1}-l_ile u_{i+1}-u_i$.
– Le Anh Dung
Jan 4 at 13:22
Thank you for your suggestion @mathcounterexamples.net! I have added a argument for that part.
– Le Anh Dung
Jan 4 at 13:27
2
Rather than It is easy to verify that $(s_i)$ is a strictly decreasing sequence converging to $0$., I would say $(s_i)$ is the difference of sequences both converging to $x$, hence it converges to $0$.
– mathcounterexamples.net
Jan 4 at 13:31
add a comment |
Looks great. I would just add a comment to explain why such an $i_0$ exists.
– mathcounterexamples.net
Jan 4 at 13:17
Hi @JonasDeSchouwer, we have $l_{i+1}-l_i>0$ and $u_{i+1}-u_i<0$. I can not see how $l_{i+1}-l_ile u_{i+1}-u_i$.
– Le Anh Dung
Jan 4 at 13:22
Thank you for your suggestion @mathcounterexamples.net! I have added a argument for that part.
– Le Anh Dung
Jan 4 at 13:27
2
Rather than It is easy to verify that $(s_i)$ is a strictly decreasing sequence converging to $0$., I would say $(s_i)$ is the difference of sequences both converging to $x$, hence it converges to $0$.
– mathcounterexamples.net
Jan 4 at 13:31
Looks great. I would just add a comment to explain why such an $i_0$ exists.
– mathcounterexamples.net
Jan 4 at 13:17
Looks great. I would just add a comment to explain why such an $i_0$ exists.
– mathcounterexamples.net
Jan 4 at 13:17
Hi @JonasDeSchouwer, we have $l_{i+1}-l_i>0$ and $u_{i+1}-u_i<0$. I can not see how $l_{i+1}-l_ile u_{i+1}-u_i$.
– Le Anh Dung
Jan 4 at 13:22
Hi @JonasDeSchouwer, we have $l_{i+1}-l_i>0$ and $u_{i+1}-u_i<0$. I can not see how $l_{i+1}-l_ile u_{i+1}-u_i$.
– Le Anh Dung
Jan 4 at 13:22
Thank you for your suggestion @mathcounterexamples.net! I have added a argument for that part.
– Le Anh Dung
Jan 4 at 13:27
Thank you for your suggestion @mathcounterexamples.net! I have added a argument for that part.
– Le Anh Dung
Jan 4 at 13:27
2
2
Rather than It is easy to verify that $(s_i)$ is a strictly decreasing sequence converging to $0$., I would say $(s_i)$ is the difference of sequences both converging to $x$, hence it converges to $0$.
– mathcounterexamples.net
Jan 4 at 13:31
Rather than It is easy to verify that $(s_i)$ is a strictly decreasing sequence converging to $0$., I would say $(s_i)$ is the difference of sequences both converging to $x$, hence it converges to $0$.
– mathcounterexamples.net
Jan 4 at 13:31
add a comment |
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Looks great. I would just add a comment to explain why such an $i_0$ exists.
– mathcounterexamples.net
Jan 4 at 13:17
Hi @JonasDeSchouwer, we have $l_{i+1}-l_i>0$ and $u_{i+1}-u_i<0$. I can not see how $l_{i+1}-l_ile u_{i+1}-u_i$.
– Le Anh Dung
Jan 4 at 13:22
Thank you for your suggestion @mathcounterexamples.net! I have added a argument for that part.
– Le Anh Dung
Jan 4 at 13:27
2
Rather than It is easy to verify that $(s_i)$ is a strictly decreasing sequence converging to $0$., I would say $(s_i)$ is the difference of sequences both converging to $x$, hence it converges to $0$.
– mathcounterexamples.net
Jan 4 at 13:31