A bounded sequence with accumulation points of 2 and 3 must be in the interval $(1,4)$ for large $n$.











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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.










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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.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:59










  • I understand, but the title does not say bounded.
    – copper.hat
    Nov 24 at 8:00















up vote
0
down vote

favorite












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.










share|cite|improve this question
























  • 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.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:59










  • I understand, but the title does not say bounded.
    – copper.hat
    Nov 24 at 8:00













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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.










share|cite|improve this question















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






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edited Nov 24 at 8:00

























asked Nov 24 at 7:19









WesleyGroupshaveFeelingsToo

1,150321




1,150321












  • 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.
    – WesleyGroupshaveFeelingsToo
    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










  • 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.
    – WesleyGroupshaveFeelingsToo
    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.
– WesleyGroupshaveFeelingsToo
Nov 24 at 7:59




It is bounded, see the yellow box.
– WesleyGroupshaveFeelingsToo
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










1 Answer
1






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1
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accepted










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.






share|cite|improve this answer





















  • Thank you, I knew I had made a mistake there somewhere.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:57










  • It also makes more sense, much appreciated.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:58











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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you, I knew I had made a mistake there somewhere.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:57










  • It also makes more sense, much appreciated.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:58















up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you, I knew I had made a mistake there somewhere.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:57










  • It also makes more sense, much appreciated.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:58













up vote
1
down vote



accepted







up vote
1
down vote



accepted






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 7:49









mlerma54

37416




37416












  • Thank you, I knew I had made a mistake there somewhere.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:57










  • It also makes more sense, much appreciated.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:58


















  • Thank you, I knew I had made a mistake there somewhere.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:57










  • It also makes more sense, much appreciated.
    – WesleyGroupshaveFeelingsToo
    Nov 24 at 7:58
















Thank you, I knew I had made a mistake there somewhere.
– WesleyGroupshaveFeelingsToo
Nov 24 at 7:57




Thank you, I knew I had made a mistake there somewhere.
– WesleyGroupshaveFeelingsToo
Nov 24 at 7:57












It also makes more sense, much appreciated.
– WesleyGroupshaveFeelingsToo
Nov 24 at 7:58




It also makes more sense, much appreciated.
– WesleyGroupshaveFeelingsToo
Nov 24 at 7:58


















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