Prove uniqueness of limit of convergent sequence in $mathbbR^n$ [duplicate]
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Every convergent sequence in a metric space has a unique limit.
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Using the definition, that for every $epsilon>0$, there exists a $NinmathbbN$, so that $d(v_i,L)$ for all $igeq N$, where $v_i$ is the $i$th vector in the sequence. And $L$ is the limit.
I assumed that let there be two limits, $L$ and $M$.
Now I choose my epsilon to be $frac13d(L,M)$, but after that, how do I proceed? Plain algebra doesn't help.
Note that $d(a,b)$ is the distance between vectors $a$ and $b$ in $mathbbR^n$.
sequences-and-series multivariable-calculus proof-writing
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
marked as duplicate by Cameron Buie, Community♦ Sep 5 at 12:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
0
down vote
favorite
This question already has an answer here:
Every convergent sequence in a metric space has a unique limit.
3 answers
Using the definition, that for every $epsilon>0$, there exists a $NinmathbbN$, so that $d(v_i,L)$ for all $igeq N$, where $v_i$ is the $i$th vector in the sequence. And $L$ is the limit.
I assumed that let there be two limits, $L$ and $M$.
Now I choose my epsilon to be $frac13d(L,M)$, but after that, how do I proceed? Plain algebra doesn't help.
Note that $d(a,b)$ is the distance between vectors $a$ and $b$ in $mathbbR^n$.
sequences-and-series multivariable-calculus proof-writing
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
marked as duplicate by Cameron Buie, Community♦ Sep 5 at 12:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
You might find the triangle inequality useful.
– Paul
Sep 5 at 11:40
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Every convergent sequence in a metric space has a unique limit.
3 answers
Using the definition, that for every $epsilon>0$, there exists a $NinmathbbN$, so that $d(v_i,L)$ for all $igeq N$, where $v_i$ is the $i$th vector in the sequence. And $L$ is the limit.
I assumed that let there be two limits, $L$ and $M$.
Now I choose my epsilon to be $frac13d(L,M)$, but after that, how do I proceed? Plain algebra doesn't help.
Note that $d(a,b)$ is the distance between vectors $a$ and $b$ in $mathbbR^n$.
sequences-and-series multivariable-calculus proof-writing
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
This question already has an answer here:
Every convergent sequence in a metric space has a unique limit.
3 answers
Using the definition, that for every $epsilon>0$, there exists a $NinmathbbN$, so that $d(v_i,L)$ for all $igeq N$, where $v_i$ is the $i$th vector in the sequence. And $L$ is the limit.
I assumed that let there be two limits, $L$ and $M$.
Now I choose my epsilon to be $frac13d(L,M)$, but after that, how do I proceed? Plain algebra doesn't help.
Note that $d(a,b)$ is the distance between vectors $a$ and $b$ in $mathbbR^n$.
This question already has an answer here:
Every convergent sequence in a metric space has a unique limit.
3 answers
sequences-and-series multivariable-calculus proof-writing
sequences-and-series multivariable-calculus proof-writing
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
edited Sep 5 at 11:29
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
asked Sep 5 at 11:27
Aditya Agarwal
2,86911536
2,86911536
marked as duplicate by Cameron Buie, Community♦ Sep 5 at 12:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Cameron Buie, Community♦ Sep 5 at 12:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
You might find the triangle inequality useful.
– Paul
Sep 5 at 11:40
add a comment |Â
1
You might find the triangle inequality useful.
– Paul
Sep 5 at 11:40
1
1
You might find the triangle inequality useful.
– Paul
Sep 5 at 11:40
You might find the triangle inequality useful.
– Paul
Sep 5 at 11:40
add a comment |Â
1 Answer
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I like to think about it geometrically first. Your sequence points are getting really close to $L$ and to $M$ at the same time, but these two points are a positive distance away. Eventually, they're going to have to choose a point to get closer to.
In fact, if you choose $varepsilon = d(L, M)/2$, then the open ball centred at $L$ with radius $varepsilon$ doesn't even intersect the open ball centred at $M$ with radius $varepsilon$. This fact comes courtesy of the triangle inequality. See if you can prove this by contradiction. Then, see if you can apply the limit definition to show why this a problem for this sequence.
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
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1 Answer
1
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1 Answer
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active
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active
oldest
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active
oldest
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up vote
0
down vote
I like to think about it geometrically first. Your sequence points are getting really close to $L$ and to $M$ at the same time, but these two points are a positive distance away. Eventually, they're going to have to choose a point to get closer to.
In fact, if you choose $varepsilon = d(L, M)/2$, then the open ball centred at $L$ with radius $varepsilon$ doesn't even intersect the open ball centred at $M$ with radius $varepsilon$. This fact comes courtesy of the triangle inequality. See if you can prove this by contradiction. Then, see if you can apply the limit definition to show why this a problem for this sequence.
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
add a comment |Â
up vote
0
down vote
I like to think about it geometrically first. Your sequence points are getting really close to $L$ and to $M$ at the same time, but these two points are a positive distance away. Eventually, they're going to have to choose a point to get closer to.
In fact, if you choose $varepsilon = d(L, M)/2$, then the open ball centred at $L$ with radius $varepsilon$ doesn't even intersect the open ball centred at $M$ with radius $varepsilon$. This fact comes courtesy of the triangle inequality. See if you can prove this by contradiction. Then, see if you can apply the limit definition to show why this a problem for this sequence.
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I like to think about it geometrically first. Your sequence points are getting really close to $L$ and to $M$ at the same time, but these two points are a positive distance away. Eventually, they're going to have to choose a point to get closer to.
In fact, if you choose $varepsilon = d(L, M)/2$, then the open ball centred at $L$ with radius $varepsilon$ doesn't even intersect the open ball centred at $M$ with radius $varepsilon$. This fact comes courtesy of the triangle inequality. See if you can prove this by contradiction. Then, see if you can apply the limit definition to show why this a problem for this sequence.
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
I like to think about it geometrically first. Your sequence points are getting really close to $L$ and to $M$ at the same time, but these two points are a positive distance away. Eventually, they're going to have to choose a point to get closer to.
In fact, if you choose $varepsilon = d(L, M)/2$, then the open ball centred at $L$ with radius $varepsilon$ doesn't even intersect the open ball centred at $M$ with radius $varepsilon$. This fact comes courtesy of the triangle inequality. See if you can prove this by contradiction. Then, see if you can apply the limit definition to show why this a problem for this sequence.
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
answered Sep 5 at 12:26
Theo Bendit
13.5k12045
13.5k12045
add a comment |Â
add a comment |Â
1
You might find the triangle inequality useful.
– Paul
Sep 5 at 11:40