Convergent Sequence Analysis
$begingroup$
Suppose that $(a_n)to a$ with $a in mathbb R$
If $A leq a_n leq B$ then $A leq a leq B$
Proof:
Consider the sequences $(b_n)=A$ and $(c_n) = B$
Thus $(b_n)to A$ & $(c_n)to B$
Noting $(b_n) leq(a_n)leq(c_n) $ $forall n in mathbb N$
Thus $A leq a leq B$
$blacksquare$
is this a valid proof? thanks.
sequences-and-series convergence
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
$endgroup$
add a comment |
$begingroup$
Suppose that $(a_n)to a$ with $a in mathbb R$
If $A leq a_n leq B$ then $A leq a leq B$
Proof:
Consider the sequences $(b_n)=A$ and $(c_n) = B$
Thus $(b_n)to A$ & $(c_n)to B$
Noting $(b_n) leq(a_n)leq(c_n) $ $forall n in mathbb N$
Thus $A leq a leq B$
$blacksquare$
is this a valid proof? thanks.
sequences-and-series convergence
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
$endgroup$
$begingroup$
And your question?
$endgroup$
– Martín Vacas Vignolo
Dec 19 '18 at 22:42
$begingroup$
The proof is ok if you really can use the fact that if $a_nto a$, $b_nto b$ and $a_nleq b_n$ for all $n$ then $aleq b$.
$endgroup$
– Mark
Dec 19 '18 at 23:01
add a comment |
$begingroup$
Suppose that $(a_n)to a$ with $a in mathbb R$
If $A leq a_n leq B$ then $A leq a leq B$
Proof:
Consider the sequences $(b_n)=A$ and $(c_n) = B$
Thus $(b_n)to A$ & $(c_n)to B$
Noting $(b_n) leq(a_n)leq(c_n) $ $forall n in mathbb N$
Thus $A leq a leq B$
$blacksquare$
is this a valid proof? thanks.
sequences-and-series convergence
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
$endgroup$
Suppose that $(a_n)to a$ with $a in mathbb R$
If $A leq a_n leq B$ then $A leq a leq B$
Proof:
Consider the sequences $(b_n)=A$ and $(c_n) = B$
Thus $(b_n)to A$ & $(c_n)to B$
Noting $(b_n) leq(a_n)leq(c_n) $ $forall n in mathbb N$
Thus $A leq a leq B$
$blacksquare$
is this a valid proof? thanks.
sequences-and-series convergence
sequences-and-series convergence
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
edited Dec 19 '18 at 22:43
PolynomialC
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
asked Dec 19 '18 at 22:39
PolynomialCPolynomialC
826
826
$begingroup$
And your question?
$endgroup$
– Martín Vacas Vignolo
Dec 19 '18 at 22:42
$begingroup$
The proof is ok if you really can use the fact that if $a_nto a$, $b_nto b$ and $a_nleq b_n$ for all $n$ then $aleq b$.
$endgroup$
– Mark
Dec 19 '18 at 23:01
add a comment |
$begingroup$
And your question?
$endgroup$
– Martín Vacas Vignolo
Dec 19 '18 at 22:42
$begingroup$
The proof is ok if you really can use the fact that if $a_nto a$, $b_nto b$ and $a_nleq b_n$ for all $n$ then $aleq b$.
$endgroup$
– Mark
Dec 19 '18 at 23:01
$begingroup$
And your question?
$endgroup$
– Martín Vacas Vignolo
Dec 19 '18 at 22:42
$begingroup$
And your question?
$endgroup$
– Martín Vacas Vignolo
Dec 19 '18 at 22:42
$begingroup$
The proof is ok if you really can use the fact that if $a_nto a$, $b_nto b$ and $a_nleq b_n$ for all $n$ then $aleq b$.
$endgroup$
– Mark
Dec 19 '18 at 23:01
$begingroup$
The proof is ok if you really can use the fact that if $a_nto a$, $b_nto b$ and $a_nleq b_n$ for all $n$ then $aleq b$.
$endgroup$
– Mark
Dec 19 '18 at 23:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not a valid proof. You proved your statement only for two specific sequences, not in the general case.
Suppose that $a<A$. Take $varepsilon=A-a$. There is a natural $N$ such that$$ngeqslant Nimplieslvert a_n-arvert<varepsilon=A-a.$$But then$$a_N-aleqslantlvert a_N-arvert<A-A$$and therefore $a_N<A$, which is a contradiction.
By a similar argument, you can't have $a>B$.
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not a valid proof. You proved your statement only for two specific sequences, not in the general case.
Suppose that $a<A$. Take $varepsilon=A-a$. There is a natural $N$ such that$$ngeqslant Nimplieslvert a_n-arvert<varepsilon=A-a.$$But then$$a_N-aleqslantlvert a_N-arvert<A-A$$and therefore $a_N<A$, which is a contradiction.
By a similar argument, you can't have $a>B$.
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
add a comment |
$begingroup$
This is not a valid proof. You proved your statement only for two specific sequences, not in the general case.
Suppose that $a<A$. Take $varepsilon=A-a$. There is a natural $N$ such that$$ngeqslant Nimplieslvert a_n-arvert<varepsilon=A-a.$$But then$$a_N-aleqslantlvert a_N-arvert<A-A$$and therefore $a_N<A$, which is a contradiction.
By a similar argument, you can't have $a>B$.
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
add a comment |
$begingroup$
This is not a valid proof. You proved your statement only for two specific sequences, not in the general case.
Suppose that $a<A$. Take $varepsilon=A-a$. There is a natural $N$ such that$$ngeqslant Nimplieslvert a_n-arvert<varepsilon=A-a.$$But then$$a_N-aleqslantlvert a_N-arvert<A-A$$and therefore $a_N<A$, which is a contradiction.
By a similar argument, you can't have $a>B$.
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
This is not a valid proof. You proved your statement only for two specific sequences, not in the general case.
Suppose that $a<A$. Take $varepsilon=A-a$. There is a natural $N$ such that$$ngeqslant Nimplieslvert a_n-arvert<varepsilon=A-a.$$But then$$a_N-aleqslantlvert a_N-arvert<A-A$$and therefore $a_N<A$, which is a contradiction.
By a similar argument, you can't have $a>B$.
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 19 '18 at 23:56
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
add a comment |
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
Thanks very much.
$endgroup$
– PolynomialC
Dec 19 '18 at 23:57
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
I'm failing to see your contradiction, you're saying that $a_N <A$ is a contradiction yet it agrees with your supposition that $a < A$
$endgroup$
– PolynomialC
Dec 20 '18 at 0:05
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
It contradicts the hypothesis that$$(forall ninmathbb{N}):Aleqslant a_nleqslant B.$$
$endgroup$
– José Carlos Santos
Dec 20 '18 at 0:08
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
$begingroup$
Ah that makes sense, great!
$endgroup$
– PolynomialC
Dec 20 '18 at 0:09
add a comment |
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$begingroup$
And your question?
$endgroup$
– Martín Vacas Vignolo
Dec 19 '18 at 22:42
$begingroup$
The proof is ok if you really can use the fact that if $a_nto a$, $b_nto b$ and $a_nleq b_n$ for all $n$ then $aleq b$.
$endgroup$
– Mark
Dec 19 '18 at 23:01