Mixing
Proving that $mathbb{N}$ isn't bounded from above using Bolzano-Weierstrass
$begingroup$
Here's my attempt but I don't think it's good enough:
We suppose thar $mathbb{N}$ is bounded from above (we already know it's bounded from below). Let $a_n$ be the sequence of natural numbers, $nin mathbb{N}$
Since $mathbb{N}$ is bounded then $a_n$ is also bounded so there exists a $M>0$ such that
$|a_n|<M$
From the Bolzano-Weierstrass theorem there exists at least one subsequence of $a_n$ that converges to a real number $a$ $(1)$ and since $a_n in mathbb{N}$ $Rightarrow $ $ain mathbb{N}$
Let $K$ be the set that contains the limits of the subsequences of $a_n$
$K$ has at least one element from $(1)$ and is also bounded since $a_n$ is bounded
Thus there exists the $supmathbb{N}=s$ and in fact $sin mathbb{N}$
Then $s-1$ is not an upper bound of $mathbb{N}$ and thus there is $ninmathbb{N}$:
$s-1<n<sRightarrow n+1>s$ though $n+1inmathbb{N}$ which means that $s$ isn't an upper bound of $mathbb{N}$
Contradiction
What do you think any ideas??
real-analysis calculus sequences-and-series proof-verification natural-numbers
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
$endgroup$
add a comment |
$begingroup$
Here's my attempt but I don't think it's good enough:
We suppose thar $mathbb{N}$ is bounded from above (we already know it's bounded from below). Let $a_n$ be the sequence of natural numbers, $nin mathbb{N}$
Since $mathbb{N}$ is bounded then $a_n$ is also bounded so there exists a $M>0$ such that
$|a_n|<M$
From the Bolzano-Weierstrass theorem there exists at least one subsequence of $a_n$ that converges to a real number $a$ $(1)$ and since $a_n in mathbb{N}$ $Rightarrow $ $ain mathbb{N}$
Let $K$ be the set that contains the limits of the subsequences of $a_n$
$K$ has at least one element from $(1)$ and is also bounded since $a_n$ is bounded
Thus there exists the $supmathbb{N}=s$ and in fact $sin mathbb{N}$
Then $s-1$ is not an upper bound of $mathbb{N}$ and thus there is $ninmathbb{N}$:
$s-1<n<sRightarrow n+1>s$ though $n+1inmathbb{N}$ which means that $s$ isn't an upper bound of $mathbb{N}$
Contradiction
What do you think any ideas??
real-analysis calculus sequences-and-series proof-verification natural-numbers
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
$endgroup$
1
$begingroup$
This argument is basically correct, although there is a (cleaner) direct approach by considering the sequence $(a_n) = (n)$.
$endgroup$
– David Kraemer
Jan 21 at 19:51
add a comment |
$begingroup$
Here's my attempt but I don't think it's good enough:
We suppose thar $mathbb{N}$ is bounded from above (we already know it's bounded from below). Let $a_n$ be the sequence of natural numbers, $nin mathbb{N}$
Since $mathbb{N}$ is bounded then $a_n$ is also bounded so there exists a $M>0$ such that
$|a_n|<M$
From the Bolzano-Weierstrass theorem there exists at least one subsequence of $a_n$ that converges to a real number $a$ $(1)$ and since $a_n in mathbb{N}$ $Rightarrow $ $ain mathbb{N}$
Let $K$ be the set that contains the limits of the subsequences of $a_n$
$K$ has at least one element from $(1)$ and is also bounded since $a_n$ is bounded
Thus there exists the $supmathbb{N}=s$ and in fact $sin mathbb{N}$
Then $s-1$ is not an upper bound of $mathbb{N}$ and thus there is $ninmathbb{N}$:
$s-1<n<sRightarrow n+1>s$ though $n+1inmathbb{N}$ which means that $s$ isn't an upper bound of $mathbb{N}$
Contradiction
What do you think any ideas??
real-analysis calculus sequences-and-series proof-verification natural-numbers
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
$endgroup$
Here's my attempt but I don't think it's good enough:
We suppose thar $mathbb{N}$ is bounded from above (we already know it's bounded from below). Let $a_n$ be the sequence of natural numbers, $nin mathbb{N}$
Since $mathbb{N}$ is bounded then $a_n$ is also bounded so there exists a $M>0$ such that
$|a_n|<M$
From the Bolzano-Weierstrass theorem there exists at least one subsequence of $a_n$ that converges to a real number $a$ $(1)$ and since $a_n in mathbb{N}$ $Rightarrow $ $ain mathbb{N}$
Let $K$ be the set that contains the limits of the subsequences of $a_n$
$K$ has at least one element from $(1)$ and is also bounded since $a_n$ is bounded
Thus there exists the $supmathbb{N}=s$ and in fact $sin mathbb{N}$
Then $s-1$ is not an upper bound of $mathbb{N}$ and thus there is $ninmathbb{N}$:
$s-1<n<sRightarrow n+1>s$ though $n+1inmathbb{N}$ which means that $s$ isn't an upper bound of $mathbb{N}$
Contradiction
What do you think any ideas??
real-analysis calculus sequences-and-series proof-verification natural-numbers
real-analysis calculus sequences-and-series proof-verification natural-numbers
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
edited Jan 21 at 19:45
VakiPitsi
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
asked Jan 21 at 19:33
VakiPitsiVakiPitsi
1978
1978
1
$begingroup$
This argument is basically correct, although there is a (cleaner) direct approach by considering the sequence $(a_n) = (n)$.
$endgroup$
– David Kraemer
Jan 21 at 19:51
add a comment |
1
$begingroup$
This argument is basically correct, although there is a (cleaner) direct approach by considering the sequence $(a_n) = (n)$.
$endgroup$
– David Kraemer
Jan 21 at 19:51
1
1
$begingroup$
This argument is basically correct, although there is a (cleaner) direct approach by considering the sequence $(a_n) = (n)$.
$endgroup$
– David Kraemer
Jan 21 at 19:51
$begingroup$
This argument is basically correct, although there is a (cleaner) direct approach by considering the sequence $(a_n) = (n)$.
$endgroup$
– David Kraemer
Jan 21 at 19:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you want to prove that $mathbb{N}$ is bounded using BW, here is a way that uses it more directly.
Define $a_n$ = n.
Lets assume $mathbb{N}$ is bounded. So $a_n$ is bounded. So according to BW, there exists a convergent sub sequence $a_{n_k}$ such that $a_{n_k} rightarrow L$ where L is a real number. So $a_{n_k}$ is a Cauchy sequence, which gives us a contradiction, since $|a_{n_{k+1}}-a_{n_{k}}|>frac{1}{2}$. So $mathbb{N}$ isn't bounded.
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
$endgroup$
1
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
add a comment |
$begingroup$
The BW theorem is an awkward way to prove N is unbounded.
Your proof uses sup N, which is all that is needed.
If sup N exists, then (sup N) - 1 is an upper bound of N.
That's the contradiction you pointed out and all that requires is N is bounded and the completeness of R.
Seemingly you did not use the BW theorem except to beat around the bush with it.
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
$endgroup$
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want to prove that $mathbb{N}$ is bounded using BW, here is a way that uses it more directly.
Define $a_n$ = n.
Lets assume $mathbb{N}$ is bounded. So $a_n$ is bounded. So according to BW, there exists a convergent sub sequence $a_{n_k}$ such that $a_{n_k} rightarrow L$ where L is a real number. So $a_{n_k}$ is a Cauchy sequence, which gives us a contradiction, since $|a_{n_{k+1}}-a_{n_{k}}|>frac{1}{2}$. So $mathbb{N}$ isn't bounded.
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
$endgroup$
1
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
add a comment |
$begingroup$
If you want to prove that $mathbb{N}$ is bounded using BW, here is a way that uses it more directly.
Define $a_n$ = n.
Lets assume $mathbb{N}$ is bounded. So $a_n$ is bounded. So according to BW, there exists a convergent sub sequence $a_{n_k}$ such that $a_{n_k} rightarrow L$ where L is a real number. So $a_{n_k}$ is a Cauchy sequence, which gives us a contradiction, since $|a_{n_{k+1}}-a_{n_{k}}|>frac{1}{2}$. So $mathbb{N}$ isn't bounded.
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
$endgroup$
1
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
add a comment |
$begingroup$
If you want to prove that $mathbb{N}$ is bounded using BW, here is a way that uses it more directly.
Define $a_n$ = n.
Lets assume $mathbb{N}$ is bounded. So $a_n$ is bounded. So according to BW, there exists a convergent sub sequence $a_{n_k}$ such that $a_{n_k} rightarrow L$ where L is a real number. So $a_{n_k}$ is a Cauchy sequence, which gives us a contradiction, since $|a_{n_{k+1}}-a_{n_{k}}|>frac{1}{2}$. So $mathbb{N}$ isn't bounded.
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
$endgroup$
If you want to prove that $mathbb{N}$ is bounded using BW, here is a way that uses it more directly.
Define $a_n$ = n.
Lets assume $mathbb{N}$ is bounded. So $a_n$ is bounded. So according to BW, there exists a convergent sub sequence $a_{n_k}$ such that $a_{n_k} rightarrow L$ where L is a real number. So $a_{n_k}$ is a Cauchy sequence, which gives us a contradiction, since $|a_{n_{k+1}}-a_{n_{k}}|>frac{1}{2}$. So $mathbb{N}$ isn't bounded.
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
answered Jan 24 at 8:41
Amit LevyAmit Levy
747
747
1
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
add a comment |
1
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
1
1
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
$begingroup$
Excellent! very direct proof much better than mine thank you a ton!
$endgroup$
– VakiPitsi
Jan 25 at 11:06
add a comment |
$begingroup$
The BW theorem is an awkward way to prove N is unbounded.
Your proof uses sup N, which is all that is needed.
If sup N exists, then (sup N) - 1 is an upper bound of N.
That's the contradiction you pointed out and all that requires is N is bounded and the completeness of R.
Seemingly you did not use the BW theorem except to beat around the bush with it.
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
$endgroup$
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
add a comment |
$begingroup$
The BW theorem is an awkward way to prove N is unbounded.
Your proof uses sup N, which is all that is needed.
If sup N exists, then (sup N) - 1 is an upper bound of N.
That's the contradiction you pointed out and all that requires is N is bounded and the completeness of R.
Seemingly you did not use the BW theorem except to beat around the bush with it.
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
$endgroup$
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
add a comment |
$begingroup$
The BW theorem is an awkward way to prove N is unbounded.
Your proof uses sup N, which is all that is needed.
If sup N exists, then (sup N) - 1 is an upper bound of N.
That's the contradiction you pointed out and all that requires is N is bounded and the completeness of R.
Seemingly you did not use the BW theorem except to beat around the bush with it.
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
$endgroup$
The BW theorem is an awkward way to prove N is unbounded.
Your proof uses sup N, which is all that is needed.
If sup N exists, then (sup N) - 1 is an upper bound of N.
That's the contradiction you pointed out and all that requires is N is bounded and the completeness of R.
Seemingly you did not use the BW theorem except to beat around the bush with it.
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
answered Jan 21 at 23:14
William ElliotWilliam Elliot
8,5672720
8,5672720
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
add a comment |
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
$begingroup$
I'm aware of the better ways I can go to prove this. But it was a suggested excercise and I was unsure if my use of BW was usefull at all
$endgroup$
– VakiPitsi
Jan 22 at 19:05
add a comment |
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$begingroup$
This argument is basically correct, although there is a (cleaner) direct approach by considering the sequence $(a_n) = (n)$.
$endgroup$
– David Kraemer
Jan 21 at 19:51