If a sequence converges, do its tail terms equal its limit?
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A sequence $(x_n)$ in a metric space $X$ is said to converge if there is a point $x in X$ such that for every $epsilon > 0$ there is an integer $N$ such that $n geq N$ implies that $d(x_n, x) < epsilon$.
I've seen proofs where having found that $d(A,B) < epsilon$, we state that the choice of $epsilon$ was arbitrary and thus that $A = B$. I was wondering if we could do the same here and conclude that for $n geq N$, $x_n = x$. I'd also appreciate an explanation of when we can and cannot make such a conclusion; sometimes it seems like we're just letting $epsilon$ be $0$ when it suits us.
real-analysis analysis
asked Nov 14 at 4:45
slothropp
62
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A sequence $(x_n)$ in a metric space $X$ is said to converge if there is a point $x in X$ such that for every $epsilon > 0$ there is an integer $N$ such that $n geq N$ implies that $d(x_n, x) < epsilon$.
I've seen proofs where having found that $d(A,B) < epsilon$, we state that the choice of $epsilon$ was arbitrary and thus that $A = B$. I was wondering if we could do the same here and conclude that for $n geq N$, $x_n = x$. I'd also appreciate an explanation of when we can and cannot make such a conclusion; sometimes it seems like we're just letting $epsilon$ be $0$ when it suits us.
real-analysis analysis
asked Nov 14 at 4:45
slothropp
62
3
No, the $N$ depends on your choice of $epsilon$. Take $x_{n}=1/n$ in $mathbb{R}$ for example. It converges to $0$ and the tails are never constant.
– weirdo
Nov 14 at 4:49
1
You're never letting $epsilon = 0$ when it suits you. It's only that eventually all terms will be arbitrarily close to your limit $x$. Think of what happens when $x_n = 1/n$ for example.
– JavaMan
Nov 14 at 4:49
That would imply that the sequence is eventually the constant sequence which isn’t always the case. Consider the case where $X$ is the real line and $d$ is the Euclidean metric and the sequence $left(frac{1}{k}right)$. Is it true that there is some integer $N$ such that for every $n geq N$, we have $0 = frac{1}{n}$?
– user328442
Nov 14 at 4:50
If $d(A,B)<e$ for every POSITIVE $e$ then $d(A,B)=0$ and therefore $A=B.$
– DanielWainfleet
Nov 14 at 11:23
add a comment |
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up vote
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down vote
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A sequence $(x_n)$ in a metric space $X$ is said to converge if there is a point $x in X$ such that for every $epsilon > 0$ there is an integer $N$ such that $n geq N$ implies that $d(x_n, x) < epsilon$.
I've seen proofs where having found that $d(A,B) < epsilon$, we state that the choice of $epsilon$ was arbitrary and thus that $A = B$. I was wondering if we could do the same here and conclude that for $n geq N$, $x_n = x$. I'd also appreciate an explanation of when we can and cannot make such a conclusion; sometimes it seems like we're just letting $epsilon$ be $0$ when it suits us.
real-analysis analysis
asked Nov 14 at 4:45
slothropp
62
A sequence $(x_n)$ in a metric space $X$ is said to converge if there is a point $x in X$ such that for every $epsilon > 0$ there is an integer $N$ such that $n geq N$ implies that $d(x_n, x) < epsilon$.
I've seen proofs where having found that $d(A,B) < epsilon$, we state that the choice of $epsilon$ was arbitrary and thus that $A = B$. I was wondering if we could do the same here and conclude that for $n geq N$, $x_n = x$. I'd also appreciate an explanation of when we can and cannot make such a conclusion; sometimes it seems like we're just letting $epsilon$ be $0$ when it suits us.
real-analysis analysis
real-analysis analysis
asked Nov 14 at 4:45
slothropp
62
asked Nov 14 at 4:45
slothropp
62
asked Nov 14 at 4:45
slothropp
62
asked Nov 14 at 4:45
slothropp
62
asked Nov 14 at 4:45
slothropp
62
62
3
No, the $N$ depends on your choice of $epsilon$. Take $x_{n}=1/n$ in $mathbb{R}$ for example. It converges to $0$ and the tails are never constant.
– weirdo
Nov 14 at 4:49
1
You're never letting $epsilon = 0$ when it suits you. It's only that eventually all terms will be arbitrarily close to your limit $x$. Think of what happens when $x_n = 1/n$ for example.
– JavaMan
Nov 14 at 4:49
That would imply that the sequence is eventually the constant sequence which isn’t always the case. Consider the case where $X$ is the real line and $d$ is the Euclidean metric and the sequence $left(frac{1}{k}right)$. Is it true that there is some integer $N$ such that for every $n geq N$, we have $0 = frac{1}{n}$?
– user328442
Nov 14 at 4:50
If $d(A,B)<e$ for every POSITIVE $e$ then $d(A,B)=0$ and therefore $A=B.$
– DanielWainfleet
Nov 14 at 11:23
add a comment |
3
No, the $N$ depends on your choice of $epsilon$. Take $x_{n}=1/n$ in $mathbb{R}$ for example. It converges to $0$ and the tails are never constant.
– weirdo
Nov 14 at 4:49
1
You're never letting $epsilon = 0$ when it suits you. It's only that eventually all terms will be arbitrarily close to your limit $x$. Think of what happens when $x_n = 1/n$ for example.
– JavaMan
Nov 14 at 4:49
That would imply that the sequence is eventually the constant sequence which isn’t always the case. Consider the case where $X$ is the real line and $d$ is the Euclidean metric and the sequence $left(frac{1}{k}right)$. Is it true that there is some integer $N$ such that for every $n geq N$, we have $0 = frac{1}{n}$?
– user328442
Nov 14 at 4:50
If $d(A,B)<e$ for every POSITIVE $e$ then $d(A,B)=0$ and therefore $A=B.$
– DanielWainfleet
Nov 14 at 11:23
3
3
No, the $N$ depends on your choice of $epsilon$. Take $x_{n}=1/n$ in $mathbb{R}$ for example. It converges to $0$ and the tails are never constant.
– weirdo
Nov 14 at 4:49
No, the $N$ depends on your choice of $epsilon$. Take $x_{n}=1/n$ in $mathbb{R}$ for example. It converges to $0$ and the tails are never constant.
– weirdo
Nov 14 at 4:49
1
1
You're never letting $epsilon = 0$ when it suits you. It's only that eventually all terms will be arbitrarily close to your limit $x$. Think of what happens when $x_n = 1/n$ for example.
– JavaMan
Nov 14 at 4:49
You're never letting $epsilon = 0$ when it suits you. It's only that eventually all terms will be arbitrarily close to your limit $x$. Think of what happens when $x_n = 1/n$ for example.
– JavaMan
Nov 14 at 4:49
That would imply that the sequence is eventually the constant sequence which isn’t always the case. Consider the case where $X$ is the real line and $d$ is the Euclidean metric and the sequence $left(frac{1}{k}right)$. Is it true that there is some integer $N$ such that for every $n geq N$, we have $0 = frac{1}{n}$?
– user328442
Nov 14 at 4:50
That would imply that the sequence is eventually the constant sequence which isn’t always the case. Consider the case where $X$ is the real line and $d$ is the Euclidean metric and the sequence $left(frac{1}{k}right)$. Is it true that there is some integer $N$ such that for every $n geq N$, we have $0 = frac{1}{n}$?
– user328442
Nov 14 at 4:50
If $d(A,B)<e$ for every POSITIVE $e$ then $d(A,B)=0$ and therefore $A=B.$
– DanielWainfleet
Nov 14 at 11:23
If $d(A,B)<e$ for every POSITIVE $e$ then $d(A,B)=0$ and therefore $A=B.$
– DanielWainfleet
Nov 14 at 11:23
add a comment |
1 Answer
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Note that $left{frac{1}{n} : n in mathbb{Z}right}$ is a sequence of non-zero terms which converges to $0$.
When we have something like $d(x,y) < epsilon$ for all $epsilon > 0$, we never "let $epsilon = 0$". Rather we are able to deduce that $x = y$. For if $x ne y$, then $d = d(x,y) > 0$, which would contradict our assumption (i.e. $epsilon = d$ would be a positive number for which $d(x,y) < epsilon$ fails to hold). Therefore, it must be the case that $x = y$.
answered Nov 14 at 4:51
suchan
18910
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Note that $left{frac{1}{n} : n in mathbb{Z}right}$ is a sequence of non-zero terms which converges to $0$.
When we have something like $d(x,y) < epsilon$ for all $epsilon > 0$, we never "let $epsilon = 0$". Rather we are able to deduce that $x = y$. For if $x ne y$, then $d = d(x,y) > 0$, which would contradict our assumption (i.e. $epsilon = d$ would be a positive number for which $d(x,y) < epsilon$ fails to hold). Therefore, it must be the case that $x = y$.
answered Nov 14 at 4:51
suchan
18910
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
add a comment |
up vote
0
down vote
Note that $left{frac{1}{n} : n in mathbb{Z}right}$ is a sequence of non-zero terms which converges to $0$.
When we have something like $d(x,y) < epsilon$ for all $epsilon > 0$, we never "let $epsilon = 0$". Rather we are able to deduce that $x = y$. For if $x ne y$, then $d = d(x,y) > 0$, which would contradict our assumption (i.e. $epsilon = d$ would be a positive number for which $d(x,y) < epsilon$ fails to hold). Therefore, it must be the case that $x = y$.
answered Nov 14 at 4:51
suchan
18910
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that $left{frac{1}{n} : n in mathbb{Z}right}$ is a sequence of non-zero terms which converges to $0$.
When we have something like $d(x,y) < epsilon$ for all $epsilon > 0$, we never "let $epsilon = 0$". Rather we are able to deduce that $x = y$. For if $x ne y$, then $d = d(x,y) > 0$, which would contradict our assumption (i.e. $epsilon = d$ would be a positive number for which $d(x,y) < epsilon$ fails to hold). Therefore, it must be the case that $x = y$.
answered Nov 14 at 4:51
suchan
18910
Note that $left{frac{1}{n} : n in mathbb{Z}right}$ is a sequence of non-zero terms which converges to $0$.
When we have something like $d(x,y) < epsilon$ for all $epsilon > 0$, we never "let $epsilon = 0$". Rather we are able to deduce that $x = y$. For if $x ne y$, then $d = d(x,y) > 0$, which would contradict our assumption (i.e. $epsilon = d$ would be a positive number for which $d(x,y) < epsilon$ fails to hold). Therefore, it must be the case that $x = y$.
answered Nov 14 at 4:51
suchan
18910
edited Nov 14 at 4:56
answered Nov 14 at 4:51
suchan
18910
answered Nov 14 at 4:51
suchan
18910
answered Nov 14 at 4:51
suchan
18910
18910
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
add a comment |
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
Thanks for your answer! Is my intuition correct in that if $x neq y$, there would be a lower bound for $epsilon$?
– slothropp
Nov 14 at 5:29
add a comment |
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No, the $N$ depends on your choice of $epsilon$. Take $x_{n}=1/n$ in $mathbb{R}$ for example. It converges to $0$ and the tails are never constant.
– weirdo
Nov 14 at 4:49
1
You're never letting $epsilon = 0$ when it suits you. It's only that eventually all terms will be arbitrarily close to your limit $x$. Think of what happens when $x_n = 1/n$ for example.
– JavaMan
Nov 14 at 4:49
That would imply that the sequence is eventually the constant sequence which isn’t always the case. Consider the case where $X$ is the real line and $d$ is the Euclidean metric and the sequence $left(frac{1}{k}right)$. Is it true that there is some integer $N$ such that for every $n geq N$, we have $0 = frac{1}{n}$?
– user328442
Nov 14 at 4:50
If $d(A,B)<e$ for every POSITIVE $e$ then $d(A,B)=0$ and therefore $A=B.$
– DanielWainfleet
Nov 14 at 11:23