$N$ should not be complete!!
$begingroup$
The sequence ($frac{1}{n}$) is a Cauchy sequence in $N$ but does not converge in $N$ as it converges to $0$.Then $N$ should not be complete. But $N$ is the closed subset of $R$ which is a complete metric space .Hence, $N$ should be complete and from here also link text Please clear my doubt.
And thanks for help in advance.
general-topology metric-spaces
asked Dec 1 '18 at 19:53
SejySejy
445
$endgroup$
add a comment |
$begingroup$
The sequence ($frac{1}{n}$) is a Cauchy sequence in $N$ but does not converge in $N$ as it converges to $0$.Then $N$ should not be complete. But $N$ is the closed subset of $R$ which is a complete metric space .Hence, $N$ should be complete and from here also link text Please clear my doubt.
And thanks for help in advance.
general-topology metric-spaces
asked Dec 1 '18 at 19:53
SejySejy
445
$endgroup$
2
$begingroup$
$1/n$ is not a sequence in $mathbb{N}$, it is a sequence in, say, $mathbb{Q}$ or $mathbb{R}$.
$endgroup$
– Ian
Dec 1 '18 at 20:01
1
$begingroup$
"The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $frac 1n$ is not a natural number. So ${frac 1n} not subset mathbb N$ and it is not a "sequence in $mathbb N$".
$endgroup$
– fleablood
Dec 1 '18 at 20:07
add a comment |
$begingroup$
The sequence ($frac{1}{n}$) is a Cauchy sequence in $N$ but does not converge in $N$ as it converges to $0$.Then $N$ should not be complete. But $N$ is the closed subset of $R$ which is a complete metric space .Hence, $N$ should be complete and from here also link text Please clear my doubt.
And thanks for help in advance.
general-topology metric-spaces
asked Dec 1 '18 at 19:53
SejySejy
445
$endgroup$
The sequence ($frac{1}{n}$) is a Cauchy sequence in $N$ but does not converge in $N$ as it converges to $0$.Then $N$ should not be complete. But $N$ is the closed subset of $R$ which is a complete metric space .Hence, $N$ should be complete and from here also link text Please clear my doubt.
And thanks for help in advance.
general-topology metric-spaces
general-topology metric-spaces
asked Dec 1 '18 at 19:53
SejySejy
445
asked Dec 1 '18 at 19:53
SejySejy
445
asked Dec 1 '18 at 19:53
SejySejy
445
asked Dec 1 '18 at 19:53
SejySejy
445
asked Dec 1 '18 at 19:53
SejySejy
445
445
2
$begingroup$
$1/n$ is not a sequence in $mathbb{N}$, it is a sequence in, say, $mathbb{Q}$ or $mathbb{R}$.
$endgroup$
– Ian
Dec 1 '18 at 20:01
1
$begingroup$
"The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $frac 1n$ is not a natural number. So ${frac 1n} not subset mathbb N$ and it is not a "sequence in $mathbb N$".
$endgroup$
– fleablood
Dec 1 '18 at 20:07
add a comment |
2
$begingroup$
$1/n$ is not a sequence in $mathbb{N}$, it is a sequence in, say, $mathbb{Q}$ or $mathbb{R}$.
$endgroup$
– Ian
Dec 1 '18 at 20:01
1
$begingroup$
"The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $frac 1n$ is not a natural number. So ${frac 1n} not subset mathbb N$ and it is not a "sequence in $mathbb N$".
$endgroup$
– fleablood
Dec 1 '18 at 20:07
2
2
$begingroup$
$1/n$ is not a sequence in $mathbb{N}$, it is a sequence in, say, $mathbb{Q}$ or $mathbb{R}$.
$endgroup$
– Ian
Dec 1 '18 at 20:01
$begingroup$
$1/n$ is not a sequence in $mathbb{N}$, it is a sequence in, say, $mathbb{Q}$ or $mathbb{R}$.
$endgroup$
– Ian
Dec 1 '18 at 20:01
1
1
$begingroup$
"The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $frac 1n$ is not a natural number. So ${frac 1n} not subset mathbb N$ and it is not a "sequence in $mathbb N$".
$endgroup$
– fleablood
Dec 1 '18 at 20:07
$begingroup$
"The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $frac 1n$ is not a natural number. So ${frac 1n} not subset mathbb N$ and it is not a "sequence in $mathbb N$".
$endgroup$
– fleablood
Dec 1 '18 at 20:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$frac 1n$ is not a Cauchy sequence in $mathbb N$ as $frac 1n not in mathbb N$ for any $n ne 1$.
To be a cauchy sequence OF NATURAL NUMBERS there must come a point where all the terms are less than $1$ apart. As all the terms ARE natural numbers[$*$] that means there comes a point where all the terms are equal. In other words the sequence is "constant for all but finite terms". And such sequences do converge to the constant.
More formally. If ${m_i} subset mathbb N$ is causchy so that for any $epsilon > 0$ then there is an $M$ so that $n,p > M implies |m_n - m_p| < epsilon$. Then if $epsilon < 1$ then there is some $M$ so that $n,p > M implies |m_n - m_p | < 1$ which means $m_n = m_p$ for ALL $n,p> M$. And $m_n = m_p = c$ for some $c in mathbb N$.
SO $m_ito c$. Yep. It's complete.
[$*$] That's what "being a sequence in $X$" means. It means all the terms are elements of $X$. So ${frac 1n}$ must certainly is NOT a sequence "in $mathbb N$".
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
$endgroup$
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
add a comment |
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1 Answer
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$begingroup$
$frac 1n$ is not a Cauchy sequence in $mathbb N$ as $frac 1n not in mathbb N$ for any $n ne 1$.
To be a cauchy sequence OF NATURAL NUMBERS there must come a point where all the terms are less than $1$ apart. As all the terms ARE natural numbers[$*$] that means there comes a point where all the terms are equal. In other words the sequence is "constant for all but finite terms". And such sequences do converge to the constant.
More formally. If ${m_i} subset mathbb N$ is causchy so that for any $epsilon > 0$ then there is an $M$ so that $n,p > M implies |m_n - m_p| < epsilon$. Then if $epsilon < 1$ then there is some $M$ so that $n,p > M implies |m_n - m_p | < 1$ which means $m_n = m_p$ for ALL $n,p> M$. And $m_n = m_p = c$ for some $c in mathbb N$.
SO $m_ito c$. Yep. It's complete.
[$*$] That's what "being a sequence in $X$" means. It means all the terms are elements of $X$. So ${frac 1n}$ must certainly is NOT a sequence "in $mathbb N$".
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
$endgroup$
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
add a comment |
$begingroup$
$frac 1n$ is not a Cauchy sequence in $mathbb N$ as $frac 1n not in mathbb N$ for any $n ne 1$.
To be a cauchy sequence OF NATURAL NUMBERS there must come a point where all the terms are less than $1$ apart. As all the terms ARE natural numbers[$*$] that means there comes a point where all the terms are equal. In other words the sequence is "constant for all but finite terms". And such sequences do converge to the constant.
More formally. If ${m_i} subset mathbb N$ is causchy so that for any $epsilon > 0$ then there is an $M$ so that $n,p > M implies |m_n - m_p| < epsilon$. Then if $epsilon < 1$ then there is some $M$ so that $n,p > M implies |m_n - m_p | < 1$ which means $m_n = m_p$ for ALL $n,p> M$. And $m_n = m_p = c$ for some $c in mathbb N$.
SO $m_ito c$. Yep. It's complete.
[$*$] That's what "being a sequence in $X$" means. It means all the terms are elements of $X$. So ${frac 1n}$ must certainly is NOT a sequence "in $mathbb N$".
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
$endgroup$
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
add a comment |
$begingroup$
$frac 1n$ is not a Cauchy sequence in $mathbb N$ as $frac 1n not in mathbb N$ for any $n ne 1$.
To be a cauchy sequence OF NATURAL NUMBERS there must come a point where all the terms are less than $1$ apart. As all the terms ARE natural numbers[$*$] that means there comes a point where all the terms are equal. In other words the sequence is "constant for all but finite terms". And such sequences do converge to the constant.
More formally. If ${m_i} subset mathbb N$ is causchy so that for any $epsilon > 0$ then there is an $M$ so that $n,p > M implies |m_n - m_p| < epsilon$. Then if $epsilon < 1$ then there is some $M$ so that $n,p > M implies |m_n - m_p | < 1$ which means $m_n = m_p$ for ALL $n,p> M$. And $m_n = m_p = c$ for some $c in mathbb N$.
SO $m_ito c$. Yep. It's complete.
[$*$] That's what "being a sequence in $X$" means. It means all the terms are elements of $X$. So ${frac 1n}$ must certainly is NOT a sequence "in $mathbb N$".
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
$endgroup$
$frac 1n$ is not a Cauchy sequence in $mathbb N$ as $frac 1n not in mathbb N$ for any $n ne 1$.
To be a cauchy sequence OF NATURAL NUMBERS there must come a point where all the terms are less than $1$ apart. As all the terms ARE natural numbers[$*$] that means there comes a point where all the terms are equal. In other words the sequence is "constant for all but finite terms". And such sequences do converge to the constant.
More formally. If ${m_i} subset mathbb N$ is causchy so that for any $epsilon > 0$ then there is an $M$ so that $n,p > M implies |m_n - m_p| < epsilon$. Then if $epsilon < 1$ then there is some $M$ so that $n,p > M implies |m_n - m_p | < 1$ which means $m_n = m_p$ for ALL $n,p> M$. And $m_n = m_p = c$ for some $c in mathbb N$.
SO $m_ito c$. Yep. It's complete.
[$*$] That's what "being a sequence in $X$" means. It means all the terms are elements of $X$. So ${frac 1n}$ must certainly is NOT a sequence "in $mathbb N$".
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
answered Dec 1 '18 at 20:05
fleabloodfleablood
70.8k22686
70.8k22686
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
add a comment |
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
$begingroup$
Thank u so much all you .I was not able to find where I was going wrong !!
$endgroup$
– Sejy
Dec 1 '18 at 21:15
add a comment |
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$begingroup$
$1/n$ is not a sequence in $mathbb{N}$, it is a sequence in, say, $mathbb{Q}$ or $mathbb{R}$.
$endgroup$
– Ian
Dec 1 '18 at 20:01
1
$begingroup$
"The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $frac 1n$ is not a natural number. So ${frac 1n} not subset mathbb N$ and it is not a "sequence in $mathbb N$".
$endgroup$
– fleablood
Dec 1 '18 at 20:07