Prove that ${x_n}$ is a Cauchy sequence iff $forall epsilon > 0 exists N: forall n > N implies |x_n -...
$begingroup$
Let ${x_n}$ denote a sequence. Prove that:
$$
{x_n} text{is fundamental} iff forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilon
$$
Let $P$ be a statement that:
$$
P = forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilontag1
$$
Let $Q$ be a statement that $x_n$ is a Cauchy sequence:
$$
Q = forall epsilon > 0 exists N in Bbb N:forall n,m > N implies |x_n - x_m| < epsilon tag2
$$
We want to show two things: $P implies Q$ and $Q implies P$.
$Box$ Start with $P implies Q$. Then by $(1)$:
$$
forall epsilon>0 exists N_1 inBbb N : forall n > N_1 implies |x_n - x_{N_1}| < epsilon tag3
$$
At the same time:
$$
forall epsilon>0 exists N_2 inBbb N : forall m > N_2 implies |x_m - x_{N_2}| < epsilon tag4
$$
If we now choose $N = max{N_1, N_2}$ both statements are true and we obtain:
$$
forallepsilon > 0 exists N =max{N_1, N_2}: forall n, m> N implies
begin{cases}
|x_n - x_N| < epsilon \
|x_m - x_N| < epsilon \
end{cases}
$$
Consider the sum of the inequalities:
$$
|x_n - x_N| + |x_m - x_N| < 2epsilon \
|(x_n - x_N) - (x_m - x_N)| le |x_n - x_N| + |x_m - x_N| \
|x_n - x_m| < 2epsilon
$$
Now if we choose $epsilon = {epsilon_0 over 2}$ in $(3)$ and $(4)$ we get:
$$
forall epsilon_0>0 exists N =max{N_1, N_2} : forall n, m > N implies |x_n - x_m| < epsilon_0 Box
$$
$Box$ Proceed to $Qimplies P$. This is where I got stuck. So far I've got that if a sequence is Cauchy then it must be bounded.
What should be my steps to show the second implication? Also I'm not quite sure about the first part is correct, could you please confirm/reject it as well?
real-analysis calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 18:42
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
Let ${x_n}$ denote a sequence. Prove that:
$$
{x_n} text{is fundamental} iff forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilon
$$
Let $P$ be a statement that:
$$
P = forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilontag1
$$
Let $Q$ be a statement that $x_n$ is a Cauchy sequence:
$$
Q = forall epsilon > 0 exists N in Bbb N:forall n,m > N implies |x_n - x_m| < epsilon tag2
$$
We want to show two things: $P implies Q$ and $Q implies P$.
$Box$ Start with $P implies Q$. Then by $(1)$:
$$
forall epsilon>0 exists N_1 inBbb N : forall n > N_1 implies |x_n - x_{N_1}| < epsilon tag3
$$
At the same time:
$$
forall epsilon>0 exists N_2 inBbb N : forall m > N_2 implies |x_m - x_{N_2}| < epsilon tag4
$$
If we now choose $N = max{N_1, N_2}$ both statements are true and we obtain:
$$
forallepsilon > 0 exists N =max{N_1, N_2}: forall n, m> N implies
begin{cases}
|x_n - x_N| < epsilon \
|x_m - x_N| < epsilon \
end{cases}
$$
Consider the sum of the inequalities:
$$
|x_n - x_N| + |x_m - x_N| < 2epsilon \
|(x_n - x_N) - (x_m - x_N)| le |x_n - x_N| + |x_m - x_N| \
|x_n - x_m| < 2epsilon
$$
Now if we choose $epsilon = {epsilon_0 over 2}$ in $(3)$ and $(4)$ we get:
$$
forall epsilon_0>0 exists N =max{N_1, N_2} : forall n, m > N implies |x_n - x_m| < epsilon_0 Box
$$
$Box$ Proceed to $Qimplies P$. This is where I got stuck. So far I've got that if a sequence is Cauchy then it must be bounded.
What should be my steps to show the second implication? Also I'm not quite sure about the first part is correct, could you please confirm/reject it as well?
real-analysis calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 18:42
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
Let ${x_n}$ denote a sequence. Prove that:
$$
{x_n} text{is fundamental} iff forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilon
$$
Let $P$ be a statement that:
$$
P = forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilontag1
$$
Let $Q$ be a statement that $x_n$ is a Cauchy sequence:
$$
Q = forall epsilon > 0 exists N in Bbb N:forall n,m > N implies |x_n - x_m| < epsilon tag2
$$
We want to show two things: $P implies Q$ and $Q implies P$.
$Box$ Start with $P implies Q$. Then by $(1)$:
$$
forall epsilon>0 exists N_1 inBbb N : forall n > N_1 implies |x_n - x_{N_1}| < epsilon tag3
$$
At the same time:
$$
forall epsilon>0 exists N_2 inBbb N : forall m > N_2 implies |x_m - x_{N_2}| < epsilon tag4
$$
If we now choose $N = max{N_1, N_2}$ both statements are true and we obtain:
$$
forallepsilon > 0 exists N =max{N_1, N_2}: forall n, m> N implies
begin{cases}
|x_n - x_N| < epsilon \
|x_m - x_N| < epsilon \
end{cases}
$$
Consider the sum of the inequalities:
$$
|x_n - x_N| + |x_m - x_N| < 2epsilon \
|(x_n - x_N) - (x_m - x_N)| le |x_n - x_N| + |x_m - x_N| \
|x_n - x_m| < 2epsilon
$$
Now if we choose $epsilon = {epsilon_0 over 2}$ in $(3)$ and $(4)$ we get:
$$
forall epsilon_0>0 exists N =max{N_1, N_2} : forall n, m > N implies |x_n - x_m| < epsilon_0 Box
$$
$Box$ Proceed to $Qimplies P$. This is where I got stuck. So far I've got that if a sequence is Cauchy then it must be bounded.
What should be my steps to show the second implication? Also I'm not quite sure about the first part is correct, could you please confirm/reject it as well?
real-analysis calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 18:42
romanroman
2,43721226
$endgroup$
Let ${x_n}$ denote a sequence. Prove that:
$$
{x_n} text{is fundamental} iff forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilon
$$
Let $P$ be a statement that:
$$
P = forall epsilon > 0 exists N: forall n > N implies |x_n - x_N| < epsilontag1
$$
Let $Q$ be a statement that $x_n$ is a Cauchy sequence:
$$
Q = forall epsilon > 0 exists N in Bbb N:forall n,m > N implies |x_n - x_m| < epsilon tag2
$$
We want to show two things: $P implies Q$ and $Q implies P$.
$Box$ Start with $P implies Q$. Then by $(1)$:
$$
forall epsilon>0 exists N_1 inBbb N : forall n > N_1 implies |x_n - x_{N_1}| < epsilon tag3
$$
At the same time:
$$
forall epsilon>0 exists N_2 inBbb N : forall m > N_2 implies |x_m - x_{N_2}| < epsilon tag4
$$
If we now choose $N = max{N_1, N_2}$ both statements are true and we obtain:
$$
forallepsilon > 0 exists N =max{N_1, N_2}: forall n, m> N implies
begin{cases}
|x_n - x_N| < epsilon \
|x_m - x_N| < epsilon \
end{cases}
$$
Consider the sum of the inequalities:
$$
|x_n - x_N| + |x_m - x_N| < 2epsilon \
|(x_n - x_N) - (x_m - x_N)| le |x_n - x_N| + |x_m - x_N| \
|x_n - x_m| < 2epsilon
$$
Now if we choose $epsilon = {epsilon_0 over 2}$ in $(3)$ and $(4)$ we get:
$$
forall epsilon_0>0 exists N =max{N_1, N_2} : forall n, m > N implies |x_n - x_m| < epsilon_0 Box
$$
$Box$ Proceed to $Qimplies P$. This is where I got stuck. So far I've got that if a sequence is Cauchy then it must be bounded.
What should be my steps to show the second implication? Also I'm not quite sure about the first part is correct, could you please confirm/reject it as well?
real-analysis calculus sequences-and-series proof-verification cauchy-sequences
real-analysis calculus sequences-and-series proof-verification cauchy-sequences
asked Dec 19 '18 at 18:42
romanroman
2,43721226
asked Dec 19 '18 at 18:42
romanroman
2,43721226
asked Dec 19 '18 at 18:42
romanroman
2,43721226
asked Dec 19 '18 at 18:42
romanroman
2,43721226
asked Dec 19 '18 at 18:42
romanroman
2,43721226
2,43721226
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$Qimplies P$ is evident from definition of Cauchy sequence:
$$ forall epsilon > 0 exists N in Bbb N:forall n,m ge N implies |x_n - x_m| < epsilon $$
if we let $m=N$, we get $P$.
while for $Pimplies Q$, by carefully picking $N$ so that for given $epsilon>0$, $n>N$ implies $|x_n-X_N|<cfrac {epsilon}2$, we have:
$|x_n-x_m|le|x_n-x_N|+|x_m-x_N|<epsilon$, when both $n, m>N$
hence $(x_n)$ is Cauchy or Q is true.
answered Dec 19 '18 at 19:01
LanceLance
64239
$endgroup$
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
$Qimplies P$ is evident from definition of Cauchy sequence:
$$ forall epsilon > 0 exists N in Bbb N:forall n,m ge N implies |x_n - x_m| < epsilon $$
if we let $m=N$, we get $P$.
while for $Pimplies Q$, by carefully picking $N$ so that for given $epsilon>0$, $n>N$ implies $|x_n-X_N|<cfrac {epsilon}2$, we have:
$|x_n-x_m|le|x_n-x_N|+|x_m-x_N|<epsilon$, when both $n, m>N$
hence $(x_n)$ is Cauchy or Q is true.
answered Dec 19 '18 at 19:01
LanceLance
64239
$endgroup$
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
add a comment |
$begingroup$
$Qimplies P$ is evident from definition of Cauchy sequence:
$$ forall epsilon > 0 exists N in Bbb N:forall n,m ge N implies |x_n - x_m| < epsilon $$
if we let $m=N$, we get $P$.
while for $Pimplies Q$, by carefully picking $N$ so that for given $epsilon>0$, $n>N$ implies $|x_n-X_N|<cfrac {epsilon}2$, we have:
$|x_n-x_m|le|x_n-x_N|+|x_m-x_N|<epsilon$, when both $n, m>N$
hence $(x_n)$ is Cauchy or Q is true.
answered Dec 19 '18 at 19:01
LanceLance
64239
$endgroup$
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
add a comment |
$begingroup$
$Qimplies P$ is evident from definition of Cauchy sequence:
$$ forall epsilon > 0 exists N in Bbb N:forall n,m ge N implies |x_n - x_m| < epsilon $$
if we let $m=N$, we get $P$.
while for $Pimplies Q$, by carefully picking $N$ so that for given $epsilon>0$, $n>N$ implies $|x_n-X_N|<cfrac {epsilon}2$, we have:
$|x_n-x_m|le|x_n-x_N|+|x_m-x_N|<epsilon$, when both $n, m>N$
hence $(x_n)$ is Cauchy or Q is true.
answered Dec 19 '18 at 19:01
LanceLance
64239
$endgroup$
$Qimplies P$ is evident from definition of Cauchy sequence:
$$ forall epsilon > 0 exists N in Bbb N:forall n,m ge N implies |x_n - x_m| < epsilon $$
if we let $m=N$, we get $P$.
while for $Pimplies Q$, by carefully picking $N$ so that for given $epsilon>0$, $n>N$ implies $|x_n-X_N|<cfrac {epsilon}2$, we have:
$|x_n-x_m|le|x_n-x_N|+|x_m-x_N|<epsilon$, when both $n, m>N$
hence $(x_n)$ is Cauchy or Q is true.
answered Dec 19 '18 at 19:01
LanceLance
64239
edited Dec 19 '18 at 19:06
answered Dec 19 '18 at 19:01
LanceLance
64239
answered Dec 19 '18 at 19:01
LanceLance
64239
answered Dec 19 '18 at 19:01
LanceLance
64239
64239
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
add a comment |
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
$begingroup$
thank you for the answer, looks i've overcomplicated things a lot
$endgroup$
– roman
Dec 20 '18 at 9:20
add a comment |
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