Find subsequences with specific properties
$begingroup$
Let $(a_n)$ be a bounded sequence such that $inf_{ell} a_{ell}<a_n< sup_{ell} a_{ell}$ for each $n=1,2, dots$
I want to show that there are subsequences $(a_{k_n})$ increasing and $(a_{m_n})$ decreasing such that $a_{k_n} to sup_{ell} a_{ell}$ and $a_{m_n} to inf_{ell} a_{ell}$.
Could you give me a hint how we could find the desired subsequences?
sequences-and-series analysis supremum-and-infimum
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
$endgroup$
add a comment |
$begingroup$
Let $(a_n)$ be a bounded sequence such that $inf_{ell} a_{ell}<a_n< sup_{ell} a_{ell}$ for each $n=1,2, dots$
I want to show that there are subsequences $(a_{k_n})$ increasing and $(a_{m_n})$ decreasing such that $a_{k_n} to sup_{ell} a_{ell}$ and $a_{m_n} to inf_{ell} a_{ell}$.
Could you give me a hint how we could find the desired subsequences?
sequences-and-series analysis supremum-and-infimum
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
$endgroup$
add a comment |
$begingroup$
Let $(a_n)$ be a bounded sequence such that $inf_{ell} a_{ell}<a_n< sup_{ell} a_{ell}$ for each $n=1,2, dots$
I want to show that there are subsequences $(a_{k_n})$ increasing and $(a_{m_n})$ decreasing such that $a_{k_n} to sup_{ell} a_{ell}$ and $a_{m_n} to inf_{ell} a_{ell}$.
Could you give me a hint how we could find the desired subsequences?
sequences-and-series analysis supremum-and-infimum
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
$endgroup$
Let $(a_n)$ be a bounded sequence such that $inf_{ell} a_{ell}<a_n< sup_{ell} a_{ell}$ for each $n=1,2, dots$
I want to show that there are subsequences $(a_{k_n})$ increasing and $(a_{m_n})$ decreasing such that $a_{k_n} to sup_{ell} a_{ell}$ and $a_{m_n} to inf_{ell} a_{ell}$.
Could you give me a hint how we could find the desired subsequences?
sequences-and-series analysis supremum-and-infimum
sequences-and-series analysis supremum-and-infimum
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
asked Dec 1 '18 at 20:58
EvindaEvinda
692513
692513
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint
Suppose the infimum is $0$ and the supremum is $1$. Choose $a_{k_1}=a_1$ as the fist member of the decreasing sequence. Then since $a_1>0$ by hypothesis, there is an element $a_{k_2}$ with $k_2>k_1$ and $0<a_{k_2}<a_{k_1}/2$ and a $k_3>k_2$ and $0<a_{k_3}<a_{k_2}/2$ and so on. (Make sure you see why we can choose $k_{m+1}>k_m.$)
EDIT
In response to the OP's comment.
First of all, since $0 = inf{a_n}$ there are $a_k$ arbitrarily close to $0$ so we can certainly find $k_2$ such that $0<a_{k_2}<a_{k_1}/2$. The only question is whether we can find such a $k_2>k_1$. Suppose not. Then $K={k|a_k<a_{k_1}}$ is finite, and then $inf{a_k}=min_{kin K}a_k>0,$ contradiction.
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
$endgroup$
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint
Suppose the infimum is $0$ and the supremum is $1$. Choose $a_{k_1}=a_1$ as the fist member of the decreasing sequence. Then since $a_1>0$ by hypothesis, there is an element $a_{k_2}$ with $k_2>k_1$ and $0<a_{k_2}<a_{k_1}/2$ and a $k_3>k_2$ and $0<a_{k_3}<a_{k_2}/2$ and so on. (Make sure you see why we can choose $k_{m+1}>k_m.$)
EDIT
In response to the OP's comment.
First of all, since $0 = inf{a_n}$ there are $a_k$ arbitrarily close to $0$ so we can certainly find $k_2$ such that $0<a_{k_2}<a_{k_1}/2$. The only question is whether we can find such a $k_2>k_1$. Suppose not. Then $K={k|a_k<a_{k_1}}$ is finite, and then $inf{a_k}=min_{kin K}a_k>0,$ contradiction.
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
$endgroup$
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
add a comment |
$begingroup$
Hint
Suppose the infimum is $0$ and the supremum is $1$. Choose $a_{k_1}=a_1$ as the fist member of the decreasing sequence. Then since $a_1>0$ by hypothesis, there is an element $a_{k_2}$ with $k_2>k_1$ and $0<a_{k_2}<a_{k_1}/2$ and a $k_3>k_2$ and $0<a_{k_3}<a_{k_2}/2$ and so on. (Make sure you see why we can choose $k_{m+1}>k_m.$)
EDIT
In response to the OP's comment.
First of all, since $0 = inf{a_n}$ there are $a_k$ arbitrarily close to $0$ so we can certainly find $k_2$ such that $0<a_{k_2}<a_{k_1}/2$. The only question is whether we can find such a $k_2>k_1$. Suppose not. Then $K={k|a_k<a_{k_1}}$ is finite, and then $inf{a_k}=min_{kin K}a_k>0,$ contradiction.
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
$endgroup$
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
add a comment |
$begingroup$
Hint
Suppose the infimum is $0$ and the supremum is $1$. Choose $a_{k_1}=a_1$ as the fist member of the decreasing sequence. Then since $a_1>0$ by hypothesis, there is an element $a_{k_2}$ with $k_2>k_1$ and $0<a_{k_2}<a_{k_1}/2$ and a $k_3>k_2$ and $0<a_{k_3}<a_{k_2}/2$ and so on. (Make sure you see why we can choose $k_{m+1}>k_m.$)
EDIT
In response to the OP's comment.
First of all, since $0 = inf{a_n}$ there are $a_k$ arbitrarily close to $0$ so we can certainly find $k_2$ such that $0<a_{k_2}<a_{k_1}/2$. The only question is whether we can find such a $k_2>k_1$. Suppose not. Then $K={k|a_k<a_{k_1}}$ is finite, and then $inf{a_k}=min_{kin K}a_k>0,$ contradiction.
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
$endgroup$
Hint
Suppose the infimum is $0$ and the supremum is $1$. Choose $a_{k_1}=a_1$ as the fist member of the decreasing sequence. Then since $a_1>0$ by hypothesis, there is an element $a_{k_2}$ with $k_2>k_1$ and $0<a_{k_2}<a_{k_1}/2$ and a $k_3>k_2$ and $0<a_{k_3}<a_{k_2}/2$ and so on. (Make sure you see why we can choose $k_{m+1}>k_m.$)
EDIT
In response to the OP's comment.
First of all, since $0 = inf{a_n}$ there are $a_k$ arbitrarily close to $0$ so we can certainly find $k_2$ such that $0<a_{k_2}<a_{k_1}/2$. The only question is whether we can find such a $k_2>k_1$. Suppose not. Then $K={k|a_k<a_{k_1}}$ is finite, and then $inf{a_k}=min_{kin K}a_k>0,$ contradiction.
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
edited Dec 2 '18 at 13:26
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
answered Dec 1 '18 at 21:16
saulspatzsaulspatz
14.2k21329
14.2k21329
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
add a comment |
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
I haven't understood why there is an element $a_{k_2}$ with $k_2> k_1$ and $0< a_{k_2}< frac{a_{k_1}}{2}$ and so on. Could you explain it further to me?
$endgroup$
– Evinda
Dec 2 '18 at 9:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
$begingroup$
@Evinda I added some details.
$endgroup$
– saulspatz
Dec 2 '18 at 13:26
add a comment |
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