Convergent subsequences of sequences whose difference tends to zero
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Suppose we have two sequences $p, q:mathbb{N} rightarrow K$, where $K$ is a compact set and $K subset mathbb{R}$, such that $|p_n - q_n|rightarrow 0$. Since $K$ is compact, there are $p'_n, q'
_n$ convergent subsequences of $p, q$ such that their limit are in $K$. Is it true that $|p'_n-q'_n|rightarrow 0$, or in other words, that $lim_{ntoinfty}p'_n = lim_{ntoinfty}q'_n$?
sequences-and-series limits
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
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add a comment |
$begingroup$
Suppose we have two sequences $p, q:mathbb{N} rightarrow K$, where $K$ is a compact set and $K subset mathbb{R}$, such that $|p_n - q_n|rightarrow 0$. Since $K$ is compact, there are $p'_n, q'
_n$ convergent subsequences of $p, q$ such that their limit are in $K$. Is it true that $|p'_n-q'_n|rightarrow 0$, or in other words, that $lim_{ntoinfty}p'_n = lim_{ntoinfty}q'_n$?
sequences-and-series limits
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
$endgroup$
add a comment |
$begingroup$
Suppose we have two sequences $p, q:mathbb{N} rightarrow K$, where $K$ is a compact set and $K subset mathbb{R}$, such that $|p_n - q_n|rightarrow 0$. Since $K$ is compact, there are $p'_n, q'
_n$ convergent subsequences of $p, q$ such that their limit are in $K$. Is it true that $|p'_n-q'_n|rightarrow 0$, or in other words, that $lim_{ntoinfty}p'_n = lim_{ntoinfty}q'_n$?
sequences-and-series limits
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
$endgroup$
Suppose we have two sequences $p, q:mathbb{N} rightarrow K$, where $K$ is a compact set and $K subset mathbb{R}$, such that $|p_n - q_n|rightarrow 0$. Since $K$ is compact, there are $p'_n, q'
_n$ convergent subsequences of $p, q$ such that their limit are in $K$. Is it true that $|p'_n-q'_n|rightarrow 0$, or in other words, that $lim_{ntoinfty}p'_n = lim_{ntoinfty}q'_n$?
sequences-and-series limits
sequences-and-series limits
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
asked Dec 18 '18 at 20:24
Sz. AkosSz. Akos
165
165
add a comment |
add a comment |
1 Answer
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Since $K$ is a compact set $K subset mathbb R$ that means that $K$ is closed and bounded. Now, since $|p_n - q_n| to 0$ this essentialy means that $p_n$ can approach $q_n$ infinitely closed and since they are $p,q : mathbb N to K$, then we can imply that they are bounded. But by the Bolzano-Weierstrass Theorem, each bounded sequence has a convergent subsequence, thus it is true that $|p'_n - q'_n| to 0$.
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since $K$ is a compact set $K subset mathbb R$ that means that $K$ is closed and bounded. Now, since $|p_n - q_n| to 0$ this essentialy means that $p_n$ can approach $q_n$ infinitely closed and since they are $p,q : mathbb N to K$, then we can imply that they are bounded. But by the Bolzano-Weierstrass Theorem, each bounded sequence has a convergent subsequence, thus it is true that $|p'_n - q'_n| to 0$.
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
add a comment |
$begingroup$
Since $K$ is a compact set $K subset mathbb R$ that means that $K$ is closed and bounded. Now, since $|p_n - q_n| to 0$ this essentialy means that $p_n$ can approach $q_n$ infinitely closed and since they are $p,q : mathbb N to K$, then we can imply that they are bounded. But by the Bolzano-Weierstrass Theorem, each bounded sequence has a convergent subsequence, thus it is true that $|p'_n - q'_n| to 0$.
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
add a comment |
$begingroup$
Since $K$ is a compact set $K subset mathbb R$ that means that $K$ is closed and bounded. Now, since $|p_n - q_n| to 0$ this essentialy means that $p_n$ can approach $q_n$ infinitely closed and since they are $p,q : mathbb N to K$, then we can imply that they are bounded. But by the Bolzano-Weierstrass Theorem, each bounded sequence has a convergent subsequence, thus it is true that $|p'_n - q'_n| to 0$.
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
$endgroup$
Since $K$ is a compact set $K subset mathbb R$ that means that $K$ is closed and bounded. Now, since $|p_n - q_n| to 0$ this essentialy means that $p_n$ can approach $q_n$ infinitely closed and since they are $p,q : mathbb N to K$, then we can imply that they are bounded. But by the Bolzano-Weierstrass Theorem, each bounded sequence has a convergent subsequence, thus it is true that $|p'_n - q'_n| to 0$.
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
answered Dec 18 '18 at 20:29
RebellosRebellos
14.8k31248
14.8k31248
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
add a comment |
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
I think that $(p'_n - q'_n)$ is not necessarily a subsequence of $(p-q)$ since the index functions that generate $p', q'$ may be different.
$endgroup$
– Sz. Akos
Dec 18 '18 at 20:35
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
$begingroup$
Note that a subsequence $a_{kn}$ of a sequence $a_n$ contains elements of the sequence $a_n$. Thus, think about $p'_n-q'_n$. In other words, $|p_n - q_n| to 0$ essentialy means that the sequences can be brought infinitely close. The same wouldn't hold for the subsequences, since they are just elements of the initial, but $K$ being compact plays a big role here, explained above.
$endgroup$
– Rebellos
Dec 18 '18 at 20:38
add a comment |
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