A bounded sequence with accumulation points of 2 and 3 must be in the interval $(1,4)$ for large $n$.
Below is the exercise I wish to solve, I have also provided my work. Could I get some feedback? If there is some better approach and this gets me nowhere your help would also very much be appreciated!
Let ${a_n }$ be a bounded sequence that has precisely the accumulation points $2$ and $3$. Show by contradiction that:
$$ exists n_0 in mathbb{N}, forall n geq n_0 : space a_n in (1,4).$$
note: we defined $0 in mathbb{N}$. Hint: argue by contradiction and use Bolzano-Weierstrass.
I will assume on the contrary that $$ forall n_0 in mathbb{N}, exists n < n_0 : space a_n not in (1,4).$$
We know that for some $n_0$ we can always finds points that are outside of ($1,4)$. We also know that $a_n$ is bounded, so we take some arbitrary lower bound $L$ that is less than $1$ and some arbitrary upper bound $U$ that is greater than $4$.
I am not sure about this part. We get whenever $n< n_0$:
Either
$$Lleq a_n leq 1 $$
or,
$$4leq a_n leq U $$
By Bolzano Weierstrass: Every bounded, infinite set of real numbers has at least one accumulation point ( I am not sure I get an infinite amount of points like this.) Anyway,
This would give use that both these interval have at least one accumulation point which contradicts our assumption that $2$ and $3$ were the only ones.
real-analysis proof-verification alternative-proof
asked Nov 24 at 7:19
Wesley Strik
1,494422
|
show 1 more comment
Below is the exercise I wish to solve, I have also provided my work. Could I get some feedback? If there is some better approach and this gets me nowhere your help would also very much be appreciated!
Let ${a_n }$ be a bounded sequence that has precisely the accumulation points $2$ and $3$. Show by contradiction that:
$$ exists n_0 in mathbb{N}, forall n geq n_0 : space a_n in (1,4).$$
note: we defined $0 in mathbb{N}$. Hint: argue by contradiction and use Bolzano-Weierstrass.
I will assume on the contrary that $$ forall n_0 in mathbb{N}, exists n < n_0 : space a_n not in (1,4).$$
We know that for some $n_0$ we can always finds points that are outside of ($1,4)$. We also know that $a_n$ is bounded, so we take some arbitrary lower bound $L$ that is less than $1$ and some arbitrary upper bound $U$ that is greater than $4$.
I am not sure about this part. We get whenever $n< n_0$:
Either
$$Lleq a_n leq 1 $$
or,
$$4leq a_n leq U $$
By Bolzano Weierstrass: Every bounded, infinite set of real numbers has at least one accumulation point ( I am not sure I get an infinite amount of points like this.) Anyway,
This would give use that both these interval have at least one accumulation point which contradicts our assumption that $2$ and $3$ were the only ones.
real-analysis proof-verification alternative-proof
asked Nov 24 at 7:19
Wesley Strik
1,494422
A bounded sequence is necessary, otherwise $2,3,1,2,3,2,2,3,3,...,2,3,n,...$ would be a counter example.
– copper.hat
Nov 24 at 7:35
The sequence is bounded by some $B$ hence the set $[-B,1] cup [4,B]$ can only contain a finite number of points (otherwise, since it is compact...).
– copper.hat
Nov 24 at 7:36
Group shave? ${}{}$
– copper.hat
Nov 24 at 7:37
It is bounded, see the yellow box.
– Wesley Strik
Nov 24 at 7:59
I understand, but the title does not say bounded.
– copper.hat
Nov 24 at 8:00
|
show 1 more comment
Below is the exercise I wish to solve, I have also provided my work. Could I get some feedback? If there is some better approach and this gets me nowhere your help would also very much be appreciated!
Let ${a_n }$ be a bounded sequence that has precisely the accumulation points $2$ and $3$. Show by contradiction that:
$$ exists n_0 in mathbb{N}, forall n geq n_0 : space a_n in (1,4).$$
note: we defined $0 in mathbb{N}$. Hint: argue by contradiction and use Bolzano-Weierstrass.
I will assume on the contrary that $$ forall n_0 in mathbb{N}, exists n < n_0 : space a_n not in (1,4).$$
We know that for some $n_0$ we can always finds points that are outside of ($1,4)$. We also know that $a_n$ is bounded, so we take some arbitrary lower bound $L$ that is less than $1$ and some arbitrary upper bound $U$ that is greater than $4$.
I am not sure about this part. We get whenever $n< n_0$:
Either
$$Lleq a_n leq 1 $$
or,
$$4leq a_n leq U $$
By Bolzano Weierstrass: Every bounded, infinite set of real numbers has at least one accumulation point ( I am not sure I get an infinite amount of points like this.) Anyway,
This would give use that both these interval have at least one accumulation point which contradicts our assumption that $2$ and $3$ were the only ones.
real-analysis proof-verification alternative-proof
asked Nov 24 at 7:19
Wesley Strik
1,494422
Below is the exercise I wish to solve, I have also provided my work. Could I get some feedback? If there is some better approach and this gets me nowhere your help would also very much be appreciated!
Let ${a_n }$ be a bounded sequence that has precisely the accumulation points $2$ and $3$. Show by contradiction that:
$$ exists n_0 in mathbb{N}, forall n geq n_0 : space a_n in (1,4).$$
note: we defined $0 in mathbb{N}$. Hint: argue by contradiction and use Bolzano-Weierstrass.
I will assume on the contrary that $$ forall n_0 in mathbb{N}, exists n < n_0 : space a_n not in (1,4).$$
We know that for some $n_0$ we can always finds points that are outside of ($1,4)$. We also know that $a_n$ is bounded, so we take some arbitrary lower bound $L$ that is less than $1$ and some arbitrary upper bound $U$ that is greater than $4$.
I am not sure about this part. We get whenever $n< n_0$:
Either
$$Lleq a_n leq 1 $$
or,
$$4leq a_n leq U $$
By Bolzano Weierstrass: Every bounded, infinite set of real numbers has at least one accumulation point ( I am not sure I get an infinite amount of points like this.) Anyway,
This would give use that both these interval have at least one accumulation point which contradicts our assumption that $2$ and $3$ were the only ones.
real-analysis proof-verification alternative-proof
real-analysis proof-verification alternative-proof
asked Nov 24 at 7:19
Wesley Strik
1,494422
asked Nov 24 at 7:19
Wesley Strik
1,494422
edited Nov 24 at 8:00
asked Nov 24 at 7:19
Wesley Strik
1,494422
asked Nov 24 at 7:19
Wesley Strik
1,494422
asked Nov 24 at 7:19
Wesley Strik
1,494422
1,494422
A bounded sequence is necessary, otherwise $2,3,1,2,3,2,2,3,3,...,2,3,n,...$ would be a counter example.
– copper.hat
Nov 24 at 7:35
The sequence is bounded by some $B$ hence the set $[-B,1] cup [4,B]$ can only contain a finite number of points (otherwise, since it is compact...).
– copper.hat
Nov 24 at 7:36
Group shave? ${}{}$
– copper.hat
Nov 24 at 7:37
It is bounded, see the yellow box.
– Wesley Strik
Nov 24 at 7:59
I understand, but the title does not say bounded.
– copper.hat
Nov 24 at 8:00
|
show 1 more comment
A bounded sequence is necessary, otherwise $2,3,1,2,3,2,2,3,3,...,2,3,n,...$ would be a counter example.
– copper.hat
Nov 24 at 7:35
The sequence is bounded by some $B$ hence the set $[-B,1] cup [4,B]$ can only contain a finite number of points (otherwise, since it is compact...).
– copper.hat
Nov 24 at 7:36
Group shave? ${}{}$
– copper.hat
Nov 24 at 7:37
It is bounded, see the yellow box.
– Wesley Strik
Nov 24 at 7:59
I understand, but the title does not say bounded.
– copper.hat
Nov 24 at 8:00
A bounded sequence is necessary, otherwise $2,3,1,2,3,2,2,3,3,...,2,3,n,...$ would be a counter example.
– copper.hat
Nov 24 at 7:35
A bounded sequence is necessary, otherwise $2,3,1,2,3,2,2,3,3,...,2,3,n,...$ would be a counter example.
– copper.hat
Nov 24 at 7:35
The sequence is bounded by some $B$ hence the set $[-B,1] cup [4,B]$ can only contain a finite number of points (otherwise, since it is compact...).
– copper.hat
Nov 24 at 7:36
The sequence is bounded by some $B$ hence the set $[-B,1] cup [4,B]$ can only contain a finite number of points (otherwise, since it is compact...).
– copper.hat
Nov 24 at 7:36
Group shave? ${}{}$
– copper.hat
Nov 24 at 7:37
Group shave? ${}{}$
– copper.hat
Nov 24 at 7:37
It is bounded, see the yellow box.
– Wesley Strik
Nov 24 at 7:59
It is bounded, see the yellow box.
– Wesley Strik
Nov 24 at 7:59
I understand, but the title does not say bounded.
– copper.hat
Nov 24 at 8:00
I understand, but the title does not say bounded.
– copper.hat
Nov 24 at 8:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
The negation of the statement is actually $forall n_0, exists ngeq n_0: a_n notin (1,4)$. This implies that there are infinitely many elements $a_n$ of the sequence verifying $L leq a_n leq 1$ or $4 leq a_n leq U$, where $L$ is a lower bound and $U$ is an upper bound of the sequence. Then at least one of the intervals $[L,1]$, $[4,U]$, contains infinitely many elements of the sequence. By Bolzano Weierstrass that interval will contain an accumulation point of the sequence. But by hypotesis the only accumulation points of the sequence are 2 and 3, which are not in $[L,1]$ or $[4,U]$. Contradiction.
answered Nov 24 at 7:49
mlerma54
1,087138
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
add a comment |
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1 Answer
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The negation of the statement is actually $forall n_0, exists ngeq n_0: a_n notin (1,4)$. This implies that there are infinitely many elements $a_n$ of the sequence verifying $L leq a_n leq 1$ or $4 leq a_n leq U$, where $L$ is a lower bound and $U$ is an upper bound of the sequence. Then at least one of the intervals $[L,1]$, $[4,U]$, contains infinitely many elements of the sequence. By Bolzano Weierstrass that interval will contain an accumulation point of the sequence. But by hypotesis the only accumulation points of the sequence are 2 and 3, which are not in $[L,1]$ or $[4,U]$. Contradiction.
answered Nov 24 at 7:49
mlerma54
1,087138
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
add a comment |
The negation of the statement is actually $forall n_0, exists ngeq n_0: a_n notin (1,4)$. This implies that there are infinitely many elements $a_n$ of the sequence verifying $L leq a_n leq 1$ or $4 leq a_n leq U$, where $L$ is a lower bound and $U$ is an upper bound of the sequence. Then at least one of the intervals $[L,1]$, $[4,U]$, contains infinitely many elements of the sequence. By Bolzano Weierstrass that interval will contain an accumulation point of the sequence. But by hypotesis the only accumulation points of the sequence are 2 and 3, which are not in $[L,1]$ or $[4,U]$. Contradiction.
answered Nov 24 at 7:49
mlerma54
1,087138
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
add a comment |
The negation of the statement is actually $forall n_0, exists ngeq n_0: a_n notin (1,4)$. This implies that there are infinitely many elements $a_n$ of the sequence verifying $L leq a_n leq 1$ or $4 leq a_n leq U$, where $L$ is a lower bound and $U$ is an upper bound of the sequence. Then at least one of the intervals $[L,1]$, $[4,U]$, contains infinitely many elements of the sequence. By Bolzano Weierstrass that interval will contain an accumulation point of the sequence. But by hypotesis the only accumulation points of the sequence are 2 and 3, which are not in $[L,1]$ or $[4,U]$. Contradiction.
answered Nov 24 at 7:49
mlerma54
1,087138
The negation of the statement is actually $forall n_0, exists ngeq n_0: a_n notin (1,4)$. This implies that there are infinitely many elements $a_n$ of the sequence verifying $L leq a_n leq 1$ or $4 leq a_n leq U$, where $L$ is a lower bound and $U$ is an upper bound of the sequence. Then at least one of the intervals $[L,1]$, $[4,U]$, contains infinitely many elements of the sequence. By Bolzano Weierstrass that interval will contain an accumulation point of the sequence. But by hypotesis the only accumulation points of the sequence are 2 and 3, which are not in $[L,1]$ or $[4,U]$. Contradiction.
answered Nov 24 at 7:49
mlerma54
1,087138
answered Nov 24 at 7:49
mlerma54
1,087138
answered Nov 24 at 7:49
mlerma54
1,087138
answered Nov 24 at 7:49
mlerma54
1,087138
1,087138
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
add a comment |
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
Thank you, I knew I had made a mistake there somewhere.
– Wesley Strik
Nov 24 at 7:57
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
It also makes more sense, much appreciated.
– Wesley Strik
Nov 24 at 7:58
add a comment |
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A bounded sequence is necessary, otherwise $2,3,1,2,3,2,2,3,3,...,2,3,n,...$ would be a counter example.
– copper.hat
Nov 24 at 7:35
The sequence is bounded by some $B$ hence the set $[-B,1] cup [4,B]$ can only contain a finite number of points (otherwise, since it is compact...).
– copper.hat
Nov 24 at 7:36
Group shave? ${}{}$
– copper.hat
Nov 24 at 7:37
It is bounded, see the yellow box.
– Wesley Strik
Nov 24 at 7:59
I understand, but the title does not say bounded.
– copper.hat
Nov 24 at 8:00