Prove any number $c in [a, b]$ is a subsequential limit if $liminf x_n = a$, $lim sup x_n = b$, $ane b$,...
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I'm trying to solve the following problem:
Let ${x_n}$ denote a bounded sequence. Prove that any number $c in [a, b]$ is a subsequential limit of ${x_n}$ if:
$$
begin{cases}
lim_{ntoinfty} (x_n - x_{n+1})=0\
liminf x_n = a\
lim sup x_n = b\
ane b
end{cases}
$$
Here are some of my thoughts. We know that $x_n$ is bounded. Then by Bolzano-Weierstrass we may choose some subsequence such that it has a finite limit:
$$
exists c in [a, b] : lim x_{n_k} = c iff forall epsilon_1 > 0 exists N_1inBbb N: forall n_k > N_1 implies |x_{n_k} - c| < epsilon_1
$$
We are also given that limsup and liminf exist and therefore:
$$
exists N_2 in Bbb N : forall n_k > N_2 implies x_{n_k} ge a \
exists N_3 in Bbb N : forall n_k > N_3 implies x_{n_k} le b
$$
If we now choose $N$ to be $max{N_2, N_3}$ we obtain:
$$
exists N = max{N_2, N_3}: forall n_k > N implies a le x_{n_k} le b tag1
$$
Also we are given the fact that $lim (x_n - x_{n+1}) = 0$:
$$
forall epsilon_2 > 0, exists N_4 in Bbb N: forall n_k > N_4implies |x_n - x_{n+1}| < epsilon_2
$$
But if $lim (x_n - x_{n+1}) = 0$, then it is also true for the subsequences:
$$
forall epsilon_3 > 0, exists N_5 in Bbb N: forall n_k > N_5implies |x_{n_k} - x_{n_k+1}| < epsilon_3 tag 2
$$
Now I'm struggling to combine that facts in order to show that any $c in [a, b]$ is a subsequential limit of ${x_n}$, how do I proceed? Feels like i have to consider $(1)$ and $(2)$ in tandem in order to finish the proof.
calculus limits limsup-and-liminf
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add a comment |
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I'm trying to solve the following problem:
Let ${x_n}$ denote a bounded sequence. Prove that any number $c in [a, b]$ is a subsequential limit of ${x_n}$ if:
$$
begin{cases}
lim_{ntoinfty} (x_n - x_{n+1})=0\
liminf x_n = a\
lim sup x_n = b\
ane b
end{cases}
$$
Here are some of my thoughts. We know that $x_n$ is bounded. Then by Bolzano-Weierstrass we may choose some subsequence such that it has a finite limit:
$$
exists c in [a, b] : lim x_{n_k} = c iff forall epsilon_1 > 0 exists N_1inBbb N: forall n_k > N_1 implies |x_{n_k} - c| < epsilon_1
$$
We are also given that limsup and liminf exist and therefore:
$$
exists N_2 in Bbb N : forall n_k > N_2 implies x_{n_k} ge a \
exists N_3 in Bbb N : forall n_k > N_3 implies x_{n_k} le b
$$
If we now choose $N$ to be $max{N_2, N_3}$ we obtain:
$$
exists N = max{N_2, N_3}: forall n_k > N implies a le x_{n_k} le b tag1
$$
Also we are given the fact that $lim (x_n - x_{n+1}) = 0$:
$$
forall epsilon_2 > 0, exists N_4 in Bbb N: forall n_k > N_4implies |x_n - x_{n+1}| < epsilon_2
$$
But if $lim (x_n - x_{n+1}) = 0$, then it is also true for the subsequences:
$$
forall epsilon_3 > 0, exists N_5 in Bbb N: forall n_k > N_5implies |x_{n_k} - x_{n_k+1}| < epsilon_3 tag 2
$$
Now I'm struggling to combine that facts in order to show that any $c in [a, b]$ is a subsequential limit of ${x_n}$, how do I proceed? Feels like i have to consider $(1)$ and $(2)$ in tandem in order to finish the proof.
calculus limits limsup-and-liminf
$endgroup$
add a comment |
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I'm trying to solve the following problem:
Let ${x_n}$ denote a bounded sequence. Prove that any number $c in [a, b]$ is a subsequential limit of ${x_n}$ if:
$$
begin{cases}
lim_{ntoinfty} (x_n - x_{n+1})=0\
liminf x_n = a\
lim sup x_n = b\
ane b
end{cases}
$$
Here are some of my thoughts. We know that $x_n$ is bounded. Then by Bolzano-Weierstrass we may choose some subsequence such that it has a finite limit:
$$
exists c in [a, b] : lim x_{n_k} = c iff forall epsilon_1 > 0 exists N_1inBbb N: forall n_k > N_1 implies |x_{n_k} - c| < epsilon_1
$$
We are also given that limsup and liminf exist and therefore:
$$
exists N_2 in Bbb N : forall n_k > N_2 implies x_{n_k} ge a \
exists N_3 in Bbb N : forall n_k > N_3 implies x_{n_k} le b
$$
If we now choose $N$ to be $max{N_2, N_3}$ we obtain:
$$
exists N = max{N_2, N_3}: forall n_k > N implies a le x_{n_k} le b tag1
$$
Also we are given the fact that $lim (x_n - x_{n+1}) = 0$:
$$
forall epsilon_2 > 0, exists N_4 in Bbb N: forall n_k > N_4implies |x_n - x_{n+1}| < epsilon_2
$$
But if $lim (x_n - x_{n+1}) = 0$, then it is also true for the subsequences:
$$
forall epsilon_3 > 0, exists N_5 in Bbb N: forall n_k > N_5implies |x_{n_k} - x_{n_k+1}| < epsilon_3 tag 2
$$
Now I'm struggling to combine that facts in order to show that any $c in [a, b]$ is a subsequential limit of ${x_n}$, how do I proceed? Feels like i have to consider $(1)$ and $(2)$ in tandem in order to finish the proof.
calculus limits limsup-and-liminf
$endgroup$
I'm trying to solve the following problem:
Let ${x_n}$ denote a bounded sequence. Prove that any number $c in [a, b]$ is a subsequential limit of ${x_n}$ if:
$$
begin{cases}
lim_{ntoinfty} (x_n - x_{n+1})=0\
liminf x_n = a\
lim sup x_n = b\
ane b
end{cases}
$$
Here are some of my thoughts. We know that $x_n$ is bounded. Then by Bolzano-Weierstrass we may choose some subsequence such that it has a finite limit:
$$
exists c in [a, b] : lim x_{n_k} = c iff forall epsilon_1 > 0 exists N_1inBbb N: forall n_k > N_1 implies |x_{n_k} - c| < epsilon_1
$$
We are also given that limsup and liminf exist and therefore:
$$
exists N_2 in Bbb N : forall n_k > N_2 implies x_{n_k} ge a \
exists N_3 in Bbb N : forall n_k > N_3 implies x_{n_k} le b
$$
If we now choose $N$ to be $max{N_2, N_3}$ we obtain:
$$
exists N = max{N_2, N_3}: forall n_k > N implies a le x_{n_k} le b tag1
$$
Also we are given the fact that $lim (x_n - x_{n+1}) = 0$:
$$
forall epsilon_2 > 0, exists N_4 in Bbb N: forall n_k > N_4implies |x_n - x_{n+1}| < epsilon_2
$$
But if $lim (x_n - x_{n+1}) = 0$, then it is also true for the subsequences:
$$
forall epsilon_3 > 0, exists N_5 in Bbb N: forall n_k > N_5implies |x_{n_k} - x_{n_k+1}| < epsilon_3 tag 2
$$
Now I'm struggling to combine that facts in order to show that any $c in [a, b]$ is a subsequential limit of ${x_n}$, how do I proceed? Feels like i have to consider $(1)$ and $(2)$ in tandem in order to finish the proof.
calculus limits limsup-and-liminf
calculus limits limsup-and-liminf
asked Dec 17 '18 at 13:50
romanroman
2,28421224
2,28421224
add a comment |
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4 Answers
4
active
oldest
votes
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Pick $cin (a,b)$ and enumerate $L={x_n:x_nle c}$ by ${y_n}$ and $U={x_n:x_nge c}$ by ${z_n}$. Notice ${x_n}={y_n}cup{z_n}$ and that both $y_n$ and $z_n$ are infinite because $liminf x_n=a$ and $liminf x_n=b$. If either $limsup y_n =c$ or $liminf z_n=c$ we are done, so suppose neither holds. Then there exists some $N$ s.t. for $m,nge N$, $y_m < c-epsilon<c<c+epsilon<z_n$ for some
$epsilon>0$. If there were only finitely many $k$ s.t. $x_kle cle x_{k+1}$, we could not have $liminf x_n=a$ and $limsup x_n=b$. Therefore there are infinitely many $k$ s.t. $x_k=y_{n_k}$ and $x_{k+1}=z_{m_k}$. For $k$ large enough, $x_{k+1}-x_kge 2epsilon$, so we can not have $lim( x_n-x_{n+1})=0$.
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Could you please elaborate on the last statement? I can't see where the contradiction comes from
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– roman
Dec 17 '18 at 15:13
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@roman I edited the answer to be more explicit.
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– Guacho Perez
Dec 17 '18 at 15:29
add a comment |
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Claim: Let $A,B$ be subsets of $Bbb R$ such that $delta:=inf B-sup A>0$. Let $x_nin Acup B$ be a sequence such that $lim_{ntoinfty} (x_n-x_{n+1})=0$, then $exists MinBbb N$ such that either $x_nin A$ for all $nge M$ or $x_nin B$ for all $nge M$.
Proof: Let $n_0inBbb N$ be such that $|x_n-x_{n+1}|<delta/2$ for all $nge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $exists N>n_0$ such that $x_Nin A$ but $x_{N+1}in B$. This, however, implies that
$$
deltale |x_N-x_{N+1}|<delta/2,
$$
a contradiction.
Let's apply the above to your problem:
Assume that $exists cin (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $varepsilon>0$ and $NinBbb N$ such that for all $nge N$, $x_nnotin (c-varepsilon,c+varepsilon)$. This splits the set in which the sequence $x_n$ (for $nge N$) lies into two parts, i.e.
$$
A=[a,c-varepsilon], B=[c+varepsilon,b].
$$
The sets $A$ and $B$ satisfy the conditions above.
Without loss of generality, let's say $x_nin A$ for all $nge M$. Then we have
$$
limsup_{ntoinfty} x_n le sup_{nge M} x_n le c-varepsilon < b,
$$
which contradicts $limsup_{ntoinfty} x_n = b$.
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Here's my very informal answer.
The lim sup and lim inf info
says that the sequence
gets infinitely often
arbitrarily close to
a and b.
The difference info
says that
successive terms
get eventually arbitrarily close.
So,
on the way between a and b,
the successive terms
are arbitrarily close
and so will be,
somewhere,
arbitrarily close to c.
Since this happens
infinitely often,
choose these terms close to c
as the subsequence.
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add a comment |
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The answer has already been given, but I will try to add my own thoughts on this, hopefully valid. Please let me know whether the reasoning below is wrong and I will either edit it or delete it completely.
Choose some $c in [a, b]$. If $c = a$ or $c = b$ then we are done since:
$$
limsup x_n = b = c \
text{or}\
liminf x_n = a = c
$$
Suppose $ c notin {a, b}$. Then by Bolzano-Weierstrass (and the fact that $x_n$ is bounded and hence $x_{n_k}$ is also) we may choose a convergent subsequence ${x_{n_k}}$ from ${x_n}$. Suppose its limit is $c$:
$$
lim x_{n_k} = c
$$
Using the definition of a limit:
$$
forall epsilon_1 > 0 exists N in Bbb N: forall n_k > N implies |x_{n_k} - c| < epsilon_1 tag1
$$
On the other hand we have the following property:
$$
lim_{ntoinfty} (x_n - x_{n+1}) = 0
$$
This property also applies to subsequences so for ${x_{n_k}}$:
$$
lim_{ktoinfty} (x_{n_k} - x_{n_k + 1}) = 0
$$
Rewriting it by definition we obtain:
$$
forall epsilon_2 > 0 exists N inBbb N: forall n_k > N implies |x_{n_k} - x_{n_k + 1}| < epsilon_2 tag2
$$
Based on $(1)$ and $(2)$ we may choose $N$ such that both statements are satisfied. Since both inequalities in $(1)$ and $(2)$ become valid we may sum them up:
$$
forall epsilon = max{epsilon_1, epsilon_2} > 0 exists N = max{N_1, N_2}: n_k > N implies |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| < epsilon
$$
By trialnge inequality:
$$
|x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| ge |x_{n_k} - c + x_{n_k} - x_{n_k + 1}| = |2x_{n_k} - x_{n_k + 1} - c|
$$
Going from definition to a limit we get:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = c
$$
We've chosen ${x_{n_k}}$ such that $lim x_{n_k} = c$ exists and we know that $lim({x_{n_k} - x_{n_k + 1}}) = 0$ also exists, so:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}x_{n_k} + lim_{ktoinfty}(x_{n_k} - x_{n_k + 1}) = \
= lim_{ktoinfty}x_{n_k} + 0 = \
= lim_{ktoinfty}x_{n_k} = c
$$
Since $c$ is arbitrary in $[a, b]$ and $x_{n_k}$ is a subsequence of $x_n$ this shows that every $cin[a, b]$ appears to be a subsequential limit.
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Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
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– Guacho Perez
Dec 17 '18 at 20:58
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@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
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– roman
Dec 18 '18 at 7:50
add a comment |
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4 Answers
4
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4 Answers
4
active
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Pick $cin (a,b)$ and enumerate $L={x_n:x_nle c}$ by ${y_n}$ and $U={x_n:x_nge c}$ by ${z_n}$. Notice ${x_n}={y_n}cup{z_n}$ and that both $y_n$ and $z_n$ are infinite because $liminf x_n=a$ and $liminf x_n=b$. If either $limsup y_n =c$ or $liminf z_n=c$ we are done, so suppose neither holds. Then there exists some $N$ s.t. for $m,nge N$, $y_m < c-epsilon<c<c+epsilon<z_n$ for some
$epsilon>0$. If there were only finitely many $k$ s.t. $x_kle cle x_{k+1}$, we could not have $liminf x_n=a$ and $limsup x_n=b$. Therefore there are infinitely many $k$ s.t. $x_k=y_{n_k}$ and $x_{k+1}=z_{m_k}$. For $k$ large enough, $x_{k+1}-x_kge 2epsilon$, so we can not have $lim( x_n-x_{n+1})=0$.
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Could you please elaborate on the last statement? I can't see where the contradiction comes from
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– roman
Dec 17 '18 at 15:13
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@roman I edited the answer to be more explicit.
$endgroup$
– Guacho Perez
Dec 17 '18 at 15:29
add a comment |
$begingroup$
Pick $cin (a,b)$ and enumerate $L={x_n:x_nle c}$ by ${y_n}$ and $U={x_n:x_nge c}$ by ${z_n}$. Notice ${x_n}={y_n}cup{z_n}$ and that both $y_n$ and $z_n$ are infinite because $liminf x_n=a$ and $liminf x_n=b$. If either $limsup y_n =c$ or $liminf z_n=c$ we are done, so suppose neither holds. Then there exists some $N$ s.t. for $m,nge N$, $y_m < c-epsilon<c<c+epsilon<z_n$ for some
$epsilon>0$. If there were only finitely many $k$ s.t. $x_kle cle x_{k+1}$, we could not have $liminf x_n=a$ and $limsup x_n=b$. Therefore there are infinitely many $k$ s.t. $x_k=y_{n_k}$ and $x_{k+1}=z_{m_k}$. For $k$ large enough, $x_{k+1}-x_kge 2epsilon$, so we can not have $lim( x_n-x_{n+1})=0$.
$endgroup$
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Could you please elaborate on the last statement? I can't see where the contradiction comes from
$endgroup$
– roman
Dec 17 '18 at 15:13
$begingroup$
@roman I edited the answer to be more explicit.
$endgroup$
– Guacho Perez
Dec 17 '18 at 15:29
add a comment |
$begingroup$
Pick $cin (a,b)$ and enumerate $L={x_n:x_nle c}$ by ${y_n}$ and $U={x_n:x_nge c}$ by ${z_n}$. Notice ${x_n}={y_n}cup{z_n}$ and that both $y_n$ and $z_n$ are infinite because $liminf x_n=a$ and $liminf x_n=b$. If either $limsup y_n =c$ or $liminf z_n=c$ we are done, so suppose neither holds. Then there exists some $N$ s.t. for $m,nge N$, $y_m < c-epsilon<c<c+epsilon<z_n$ for some
$epsilon>0$. If there were only finitely many $k$ s.t. $x_kle cle x_{k+1}$, we could not have $liminf x_n=a$ and $limsup x_n=b$. Therefore there are infinitely many $k$ s.t. $x_k=y_{n_k}$ and $x_{k+1}=z_{m_k}$. For $k$ large enough, $x_{k+1}-x_kge 2epsilon$, so we can not have $lim( x_n-x_{n+1})=0$.
$endgroup$
Pick $cin (a,b)$ and enumerate $L={x_n:x_nle c}$ by ${y_n}$ and $U={x_n:x_nge c}$ by ${z_n}$. Notice ${x_n}={y_n}cup{z_n}$ and that both $y_n$ and $z_n$ are infinite because $liminf x_n=a$ and $liminf x_n=b$. If either $limsup y_n =c$ or $liminf z_n=c$ we are done, so suppose neither holds. Then there exists some $N$ s.t. for $m,nge N$, $y_m < c-epsilon<c<c+epsilon<z_n$ for some
$epsilon>0$. If there were only finitely many $k$ s.t. $x_kle cle x_{k+1}$, we could not have $liminf x_n=a$ and $limsup x_n=b$. Therefore there are infinitely many $k$ s.t. $x_k=y_{n_k}$ and $x_{k+1}=z_{m_k}$. For $k$ large enough, $x_{k+1}-x_kge 2epsilon$, so we can not have $lim( x_n-x_{n+1})=0$.
edited Dec 17 '18 at 15:27
answered Dec 17 '18 at 14:33
Guacho PerezGuacho Perez
3,92911132
3,92911132
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Could you please elaborate on the last statement? I can't see where the contradiction comes from
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– roman
Dec 17 '18 at 15:13
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@roman I edited the answer to be more explicit.
$endgroup$
– Guacho Perez
Dec 17 '18 at 15:29
add a comment |
$begingroup$
Could you please elaborate on the last statement? I can't see where the contradiction comes from
$endgroup$
– roman
Dec 17 '18 at 15:13
$begingroup$
@roman I edited the answer to be more explicit.
$endgroup$
– Guacho Perez
Dec 17 '18 at 15:29
$begingroup$
Could you please elaborate on the last statement? I can't see where the contradiction comes from
$endgroup$
– roman
Dec 17 '18 at 15:13
$begingroup$
Could you please elaborate on the last statement? I can't see where the contradiction comes from
$endgroup$
– roman
Dec 17 '18 at 15:13
$begingroup$
@roman I edited the answer to be more explicit.
$endgroup$
– Guacho Perez
Dec 17 '18 at 15:29
$begingroup$
@roman I edited the answer to be more explicit.
$endgroup$
– Guacho Perez
Dec 17 '18 at 15:29
add a comment |
$begingroup$
Claim: Let $A,B$ be subsets of $Bbb R$ such that $delta:=inf B-sup A>0$. Let $x_nin Acup B$ be a sequence such that $lim_{ntoinfty} (x_n-x_{n+1})=0$, then $exists MinBbb N$ such that either $x_nin A$ for all $nge M$ or $x_nin B$ for all $nge M$.
Proof: Let $n_0inBbb N$ be such that $|x_n-x_{n+1}|<delta/2$ for all $nge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $exists N>n_0$ such that $x_Nin A$ but $x_{N+1}in B$. This, however, implies that
$$
deltale |x_N-x_{N+1}|<delta/2,
$$
a contradiction.
Let's apply the above to your problem:
Assume that $exists cin (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $varepsilon>0$ and $NinBbb N$ such that for all $nge N$, $x_nnotin (c-varepsilon,c+varepsilon)$. This splits the set in which the sequence $x_n$ (for $nge N$) lies into two parts, i.e.
$$
A=[a,c-varepsilon], B=[c+varepsilon,b].
$$
The sets $A$ and $B$ satisfy the conditions above.
Without loss of generality, let's say $x_nin A$ for all $nge M$. Then we have
$$
limsup_{ntoinfty} x_n le sup_{nge M} x_n le c-varepsilon < b,
$$
which contradicts $limsup_{ntoinfty} x_n = b$.
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add a comment |
$begingroup$
Claim: Let $A,B$ be subsets of $Bbb R$ such that $delta:=inf B-sup A>0$. Let $x_nin Acup B$ be a sequence such that $lim_{ntoinfty} (x_n-x_{n+1})=0$, then $exists MinBbb N$ such that either $x_nin A$ for all $nge M$ or $x_nin B$ for all $nge M$.
Proof: Let $n_0inBbb N$ be such that $|x_n-x_{n+1}|<delta/2$ for all $nge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $exists N>n_0$ such that $x_Nin A$ but $x_{N+1}in B$. This, however, implies that
$$
deltale |x_N-x_{N+1}|<delta/2,
$$
a contradiction.
Let's apply the above to your problem:
Assume that $exists cin (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $varepsilon>0$ and $NinBbb N$ such that for all $nge N$, $x_nnotin (c-varepsilon,c+varepsilon)$. This splits the set in which the sequence $x_n$ (for $nge N$) lies into two parts, i.e.
$$
A=[a,c-varepsilon], B=[c+varepsilon,b].
$$
The sets $A$ and $B$ satisfy the conditions above.
Without loss of generality, let's say $x_nin A$ for all $nge M$. Then we have
$$
limsup_{ntoinfty} x_n le sup_{nge M} x_n le c-varepsilon < b,
$$
which contradicts $limsup_{ntoinfty} x_n = b$.
$endgroup$
add a comment |
$begingroup$
Claim: Let $A,B$ be subsets of $Bbb R$ such that $delta:=inf B-sup A>0$. Let $x_nin Acup B$ be a sequence such that $lim_{ntoinfty} (x_n-x_{n+1})=0$, then $exists MinBbb N$ such that either $x_nin A$ for all $nge M$ or $x_nin B$ for all $nge M$.
Proof: Let $n_0inBbb N$ be such that $|x_n-x_{n+1}|<delta/2$ for all $nge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $exists N>n_0$ such that $x_Nin A$ but $x_{N+1}in B$. This, however, implies that
$$
deltale |x_N-x_{N+1}|<delta/2,
$$
a contradiction.
Let's apply the above to your problem:
Assume that $exists cin (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $varepsilon>0$ and $NinBbb N$ such that for all $nge N$, $x_nnotin (c-varepsilon,c+varepsilon)$. This splits the set in which the sequence $x_n$ (for $nge N$) lies into two parts, i.e.
$$
A=[a,c-varepsilon], B=[c+varepsilon,b].
$$
The sets $A$ and $B$ satisfy the conditions above.
Without loss of generality, let's say $x_nin A$ for all $nge M$. Then we have
$$
limsup_{ntoinfty} x_n le sup_{nge M} x_n le c-varepsilon < b,
$$
which contradicts $limsup_{ntoinfty} x_n = b$.
$endgroup$
Claim: Let $A,B$ be subsets of $Bbb R$ such that $delta:=inf B-sup A>0$. Let $x_nin Acup B$ be a sequence such that $lim_{ntoinfty} (x_n-x_{n+1})=0$, then $exists MinBbb N$ such that either $x_nin A$ for all $nge M$ or $x_nin B$ for all $nge M$.
Proof: Let $n_0inBbb N$ be such that $|x_n-x_{n+1}|<delta/2$ for all $nge n_0$. Suppose for contradiction that such an $M$ cannot be found, then $exists N>n_0$ such that $x_Nin A$ but $x_{N+1}in B$. This, however, implies that
$$
deltale |x_N-x_{N+1}|<delta/2,
$$
a contradiction.
Let's apply the above to your problem:
Assume that $exists cin (a,b)$ such that $c$ is not a subsequential limit of $x_n$, then there is an $varepsilon>0$ and $NinBbb N$ such that for all $nge N$, $x_nnotin (c-varepsilon,c+varepsilon)$. This splits the set in which the sequence $x_n$ (for $nge N$) lies into two parts, i.e.
$$
A=[a,c-varepsilon], B=[c+varepsilon,b].
$$
The sets $A$ and $B$ satisfy the conditions above.
Without loss of generality, let's say $x_nin A$ for all $nge M$. Then we have
$$
limsup_{ntoinfty} x_n le sup_{nge M} x_n le c-varepsilon < b,
$$
which contradicts $limsup_{ntoinfty} x_n = b$.
edited Dec 18 '18 at 7:58
answered Dec 17 '18 at 17:12
BigbearZzzBigbearZzz
8,79421652
8,79421652
add a comment |
add a comment |
$begingroup$
Here's my very informal answer.
The lim sup and lim inf info
says that the sequence
gets infinitely often
arbitrarily close to
a and b.
The difference info
says that
successive terms
get eventually arbitrarily close.
So,
on the way between a and b,
the successive terms
are arbitrarily close
and so will be,
somewhere,
arbitrarily close to c.
Since this happens
infinitely often,
choose these terms close to c
as the subsequence.
$endgroup$
add a comment |
$begingroup$
Here's my very informal answer.
The lim sup and lim inf info
says that the sequence
gets infinitely often
arbitrarily close to
a and b.
The difference info
says that
successive terms
get eventually arbitrarily close.
So,
on the way between a and b,
the successive terms
are arbitrarily close
and so will be,
somewhere,
arbitrarily close to c.
Since this happens
infinitely often,
choose these terms close to c
as the subsequence.
$endgroup$
add a comment |
$begingroup$
Here's my very informal answer.
The lim sup and lim inf info
says that the sequence
gets infinitely often
arbitrarily close to
a and b.
The difference info
says that
successive terms
get eventually arbitrarily close.
So,
on the way between a and b,
the successive terms
are arbitrarily close
and so will be,
somewhere,
arbitrarily close to c.
Since this happens
infinitely often,
choose these terms close to c
as the subsequence.
$endgroup$
Here's my very informal answer.
The lim sup and lim inf info
says that the sequence
gets infinitely often
arbitrarily close to
a and b.
The difference info
says that
successive terms
get eventually arbitrarily close.
So,
on the way between a and b,
the successive terms
are arbitrarily close
and so will be,
somewhere,
arbitrarily close to c.
Since this happens
infinitely often,
choose these terms close to c
as the subsequence.
answered Dec 17 '18 at 14:27
marty cohenmarty cohen
73.9k549128
73.9k549128
add a comment |
add a comment |
$begingroup$
The answer has already been given, but I will try to add my own thoughts on this, hopefully valid. Please let me know whether the reasoning below is wrong and I will either edit it or delete it completely.
Choose some $c in [a, b]$. If $c = a$ or $c = b$ then we are done since:
$$
limsup x_n = b = c \
text{or}\
liminf x_n = a = c
$$
Suppose $ c notin {a, b}$. Then by Bolzano-Weierstrass (and the fact that $x_n$ is bounded and hence $x_{n_k}$ is also) we may choose a convergent subsequence ${x_{n_k}}$ from ${x_n}$. Suppose its limit is $c$:
$$
lim x_{n_k} = c
$$
Using the definition of a limit:
$$
forall epsilon_1 > 0 exists N in Bbb N: forall n_k > N implies |x_{n_k} - c| < epsilon_1 tag1
$$
On the other hand we have the following property:
$$
lim_{ntoinfty} (x_n - x_{n+1}) = 0
$$
This property also applies to subsequences so for ${x_{n_k}}$:
$$
lim_{ktoinfty} (x_{n_k} - x_{n_k + 1}) = 0
$$
Rewriting it by definition we obtain:
$$
forall epsilon_2 > 0 exists N inBbb N: forall n_k > N implies |x_{n_k} - x_{n_k + 1}| < epsilon_2 tag2
$$
Based on $(1)$ and $(2)$ we may choose $N$ such that both statements are satisfied. Since both inequalities in $(1)$ and $(2)$ become valid we may sum them up:
$$
forall epsilon = max{epsilon_1, epsilon_2} > 0 exists N = max{N_1, N_2}: n_k > N implies |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| < epsilon
$$
By trialnge inequality:
$$
|x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| ge |x_{n_k} - c + x_{n_k} - x_{n_k + 1}| = |2x_{n_k} - x_{n_k + 1} - c|
$$
Going from definition to a limit we get:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = c
$$
We've chosen ${x_{n_k}}$ such that $lim x_{n_k} = c$ exists and we know that $lim({x_{n_k} - x_{n_k + 1}}) = 0$ also exists, so:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}x_{n_k} + lim_{ktoinfty}(x_{n_k} - x_{n_k + 1}) = \
= lim_{ktoinfty}x_{n_k} + 0 = \
= lim_{ktoinfty}x_{n_k} = c
$$
Since $c$ is arbitrary in $[a, b]$ and $x_{n_k}$ is a subsequence of $x_n$ this shows that every $cin[a, b]$ appears to be a subsequential limit.
$endgroup$
$begingroup$
Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
$endgroup$
– Guacho Perez
Dec 17 '18 at 20:58
$begingroup$
@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
$endgroup$
– roman
Dec 18 '18 at 7:50
add a comment |
$begingroup$
The answer has already been given, but I will try to add my own thoughts on this, hopefully valid. Please let me know whether the reasoning below is wrong and I will either edit it or delete it completely.
Choose some $c in [a, b]$. If $c = a$ or $c = b$ then we are done since:
$$
limsup x_n = b = c \
text{or}\
liminf x_n = a = c
$$
Suppose $ c notin {a, b}$. Then by Bolzano-Weierstrass (and the fact that $x_n$ is bounded and hence $x_{n_k}$ is also) we may choose a convergent subsequence ${x_{n_k}}$ from ${x_n}$. Suppose its limit is $c$:
$$
lim x_{n_k} = c
$$
Using the definition of a limit:
$$
forall epsilon_1 > 0 exists N in Bbb N: forall n_k > N implies |x_{n_k} - c| < epsilon_1 tag1
$$
On the other hand we have the following property:
$$
lim_{ntoinfty} (x_n - x_{n+1}) = 0
$$
This property also applies to subsequences so for ${x_{n_k}}$:
$$
lim_{ktoinfty} (x_{n_k} - x_{n_k + 1}) = 0
$$
Rewriting it by definition we obtain:
$$
forall epsilon_2 > 0 exists N inBbb N: forall n_k > N implies |x_{n_k} - x_{n_k + 1}| < epsilon_2 tag2
$$
Based on $(1)$ and $(2)$ we may choose $N$ such that both statements are satisfied. Since both inequalities in $(1)$ and $(2)$ become valid we may sum them up:
$$
forall epsilon = max{epsilon_1, epsilon_2} > 0 exists N = max{N_1, N_2}: n_k > N implies |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| < epsilon
$$
By trialnge inequality:
$$
|x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| ge |x_{n_k} - c + x_{n_k} - x_{n_k + 1}| = |2x_{n_k} - x_{n_k + 1} - c|
$$
Going from definition to a limit we get:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = c
$$
We've chosen ${x_{n_k}}$ such that $lim x_{n_k} = c$ exists and we know that $lim({x_{n_k} - x_{n_k + 1}}) = 0$ also exists, so:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}x_{n_k} + lim_{ktoinfty}(x_{n_k} - x_{n_k + 1}) = \
= lim_{ktoinfty}x_{n_k} + 0 = \
= lim_{ktoinfty}x_{n_k} = c
$$
Since $c$ is arbitrary in $[a, b]$ and $x_{n_k}$ is a subsequence of $x_n$ this shows that every $cin[a, b]$ appears to be a subsequential limit.
$endgroup$
$begingroup$
Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
$endgroup$
– Guacho Perez
Dec 17 '18 at 20:58
$begingroup$
@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
$endgroup$
– roman
Dec 18 '18 at 7:50
add a comment |
$begingroup$
The answer has already been given, but I will try to add my own thoughts on this, hopefully valid. Please let me know whether the reasoning below is wrong and I will either edit it or delete it completely.
Choose some $c in [a, b]$. If $c = a$ or $c = b$ then we are done since:
$$
limsup x_n = b = c \
text{or}\
liminf x_n = a = c
$$
Suppose $ c notin {a, b}$. Then by Bolzano-Weierstrass (and the fact that $x_n$ is bounded and hence $x_{n_k}$ is also) we may choose a convergent subsequence ${x_{n_k}}$ from ${x_n}$. Suppose its limit is $c$:
$$
lim x_{n_k} = c
$$
Using the definition of a limit:
$$
forall epsilon_1 > 0 exists N in Bbb N: forall n_k > N implies |x_{n_k} - c| < epsilon_1 tag1
$$
On the other hand we have the following property:
$$
lim_{ntoinfty} (x_n - x_{n+1}) = 0
$$
This property also applies to subsequences so for ${x_{n_k}}$:
$$
lim_{ktoinfty} (x_{n_k} - x_{n_k + 1}) = 0
$$
Rewriting it by definition we obtain:
$$
forall epsilon_2 > 0 exists N inBbb N: forall n_k > N implies |x_{n_k} - x_{n_k + 1}| < epsilon_2 tag2
$$
Based on $(1)$ and $(2)$ we may choose $N$ such that both statements are satisfied. Since both inequalities in $(1)$ and $(2)$ become valid we may sum them up:
$$
forall epsilon = max{epsilon_1, epsilon_2} > 0 exists N = max{N_1, N_2}: n_k > N implies |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| < epsilon
$$
By trialnge inequality:
$$
|x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| ge |x_{n_k} - c + x_{n_k} - x_{n_k + 1}| = |2x_{n_k} - x_{n_k + 1} - c|
$$
Going from definition to a limit we get:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = c
$$
We've chosen ${x_{n_k}}$ such that $lim x_{n_k} = c$ exists and we know that $lim({x_{n_k} - x_{n_k + 1}}) = 0$ also exists, so:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}x_{n_k} + lim_{ktoinfty}(x_{n_k} - x_{n_k + 1}) = \
= lim_{ktoinfty}x_{n_k} + 0 = \
= lim_{ktoinfty}x_{n_k} = c
$$
Since $c$ is arbitrary in $[a, b]$ and $x_{n_k}$ is a subsequence of $x_n$ this shows that every $cin[a, b]$ appears to be a subsequential limit.
$endgroup$
The answer has already been given, but I will try to add my own thoughts on this, hopefully valid. Please let me know whether the reasoning below is wrong and I will either edit it or delete it completely.
Choose some $c in [a, b]$. If $c = a$ or $c = b$ then we are done since:
$$
limsup x_n = b = c \
text{or}\
liminf x_n = a = c
$$
Suppose $ c notin {a, b}$. Then by Bolzano-Weierstrass (and the fact that $x_n$ is bounded and hence $x_{n_k}$ is also) we may choose a convergent subsequence ${x_{n_k}}$ from ${x_n}$. Suppose its limit is $c$:
$$
lim x_{n_k} = c
$$
Using the definition of a limit:
$$
forall epsilon_1 > 0 exists N in Bbb N: forall n_k > N implies |x_{n_k} - c| < epsilon_1 tag1
$$
On the other hand we have the following property:
$$
lim_{ntoinfty} (x_n - x_{n+1}) = 0
$$
This property also applies to subsequences so for ${x_{n_k}}$:
$$
lim_{ktoinfty} (x_{n_k} - x_{n_k + 1}) = 0
$$
Rewriting it by definition we obtain:
$$
forall epsilon_2 > 0 exists N inBbb N: forall n_k > N implies |x_{n_k} - x_{n_k + 1}| < epsilon_2 tag2
$$
Based on $(1)$ and $(2)$ we may choose $N$ such that both statements are satisfied. Since both inequalities in $(1)$ and $(2)$ become valid we may sum them up:
$$
forall epsilon = max{epsilon_1, epsilon_2} > 0 exists N = max{N_1, N_2}: n_k > N implies |x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| < epsilon
$$
By trialnge inequality:
$$
|x_{n_k} - c| + |x_{n_k} - x_{n_k + 1}| ge |x_{n_k} - c + x_{n_k} - x_{n_k + 1}| = |2x_{n_k} - x_{n_k + 1} - c|
$$
Going from definition to a limit we get:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = c
$$
We've chosen ${x_{n_k}}$ such that $lim x_{n_k} = c$ exists and we know that $lim({x_{n_k} - x_{n_k + 1}}) = 0$ also exists, so:
$$
lim_{ktoinfty}(2x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}(x_{n_k} + x_{n_k} - x_{n_k + 1}) = \
lim_{ktoinfty}x_{n_k} + lim_{ktoinfty}(x_{n_k} - x_{n_k + 1}) = \
= lim_{ktoinfty}x_{n_k} + 0 = \
= lim_{ktoinfty}x_{n_k} = c
$$
Since $c$ is arbitrary in $[a, b]$ and $x_{n_k}$ is a subsequence of $x_n$ this shows that every $cin[a, b]$ appears to be a subsequential limit.
edited Dec 17 '18 at 16:52
answered Dec 17 '18 at 16:39
romanroman
2,28421224
2,28421224
$begingroup$
Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
$endgroup$
– Guacho Perez
Dec 17 '18 at 20:58
$begingroup$
@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
$endgroup$
– roman
Dec 18 '18 at 7:50
add a comment |
$begingroup$
Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
$endgroup$
– Guacho Perez
Dec 17 '18 at 20:58
$begingroup$
@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
$endgroup$
– roman
Dec 18 '18 at 7:50
$begingroup$
Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
$endgroup$
– Guacho Perez
Dec 17 '18 at 20:58
$begingroup$
Remember you are trying to prove the existence of a subsequence that converges to $c$, so when you say "suppose its limit is $c$" after invoking Bolzano-Weierstrass, you are assuming what you are trying to prove. Here, you have only that $x_{n_k}$ converges to some $c$ in $[a,b]$, but you are trying to prove this happens for all $c$.
$endgroup$
– Guacho Perez
Dec 17 '18 at 20:58
$begingroup$
@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
$endgroup$
– roman
Dec 18 '18 at 7:50
$begingroup$
@GuachoPerez Well, after going though this answer once again, you are right, thank you for the notices. That problem is driving me mad. I'm going to accept you answer while it seems i'm too dumb to prove it myself
$endgroup$
– roman
Dec 18 '18 at 7:50
add a comment |
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