Sequences that converge to the same pointCauchy Sequences and Analysistopology - triangle inequalityConvergence and metric - Proof?Complete metric on the space of sequencesConverge in mean square of sum of random variablesProve that if two sequences converge to the same value but their image sequences converge to different values, then the limit DNEProve that either both sequences converge to the same limit or both diverge.Completion of metric space (via formal limits of Cauchy sequences)Proving completeness of space of converging sequencesFind non convergent Cauchy sequence
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Sequences that converge to the same point
Cauchy Sequences and Analysistopology - triangle inequalityConvergence and metric - Proof?Complete metric on the space of sequencesConverge in mean square of sum of random variablesProve that if two sequences converge to the same value but their image sequences converge to different values, then the limit DNEProve that either both sequences converge to the same limit or both diverge.Completion of metric space (via formal limits of Cauchy sequences)Proving completeness of space of converging sequencesFind non convergent Cauchy sequence
$begingroup$
In $(X,d)$ metric space, If $x_n_ninBbb N$ and $y_n_ninBbb N$ are two sequences which converge to the same point, then prove that
$$lim_ntoinfty d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
$endgroup$
add a comment |
$begingroup$
In $(X,d)$ metric space, If $x_n_ninBbb N$ and $y_n_ninBbb N$ are two sequences which converge to the same point, then prove that
$$lim_ntoinfty d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
$endgroup$
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
add a comment |
$begingroup$
In $(X,d)$ metric space, If $x_n_ninBbb N$ and $y_n_ninBbb N$ are two sequences which converge to the same point, then prove that
$$lim_ntoinfty d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
$endgroup$
In $(X,d)$ metric space, If $x_n_ninBbb N$ and $y_n_ninBbb N$ are two sequences which converge to the same point, then prove that
$$lim_ntoinfty d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
edited May 11 '17 at 13:25
user228113
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
82
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
add a comment |
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_ntoinfty d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_n), (y_n)$ by definition Note that $d(x_n,y_n) leq d(x_n,l) + d(l, y_n)$ for all $n$. If $varepsilon > 0$, then there are $N_1,N_2$, by assumption, giving $d(x_n,l), d(l,y_n) < varepsilon/2$ for all $n geq N_1$ and $geq N_2$. So $d(x_n,l) + d(l,y_n) < varepsilon$ for all $n geq max N_1,N_2 $; so $d(x_n,y_n) < varepsilon$ for all $n geq max N_1,N_2 $.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
add a comment |
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
add a comment |
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
answered May 11 '17 at 13:20
user379195
add a comment |
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_ntoinfty d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_ntoinfty d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_ntoinfty d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_ntoinfty d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
91.4k394161
add a comment |
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_n), (y_n)$ by definition Note that $d(x_n,y_n) leq d(x_n,l) + d(l, y_n)$ for all $n$. If $varepsilon > 0$, then there are $N_1,N_2$, by assumption, giving $d(x_n,l), d(l,y_n) < varepsilon/2$ for all $n geq N_1$ and $geq N_2$. So $d(x_n,l) + d(l,y_n) < varepsilon$ for all $n geq max N_1,N_2 $; so $d(x_n,y_n) < varepsilon$ for all $n geq max N_1,N_2 $.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_n), (y_n)$ by definition Note that $d(x_n,y_n) leq d(x_n,l) + d(l, y_n)$ for all $n$. If $varepsilon > 0$, then there are $N_1,N_2$, by assumption, giving $d(x_n,l), d(l,y_n) < varepsilon/2$ for all $n geq N_1$ and $geq N_2$. So $d(x_n,l) + d(l,y_n) < varepsilon$ for all $n geq max N_1,N_2 $; so $d(x_n,y_n) < varepsilon$ for all $n geq max N_1,N_2 $.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_n), (y_n)$ by definition Note that $d(x_n,y_n) leq d(x_n,l) + d(l, y_n)$ for all $n$. If $varepsilon > 0$, then there are $N_1,N_2$, by assumption, giving $d(x_n,l), d(l,y_n) < varepsilon/2$ for all $n geq N_1$ and $geq N_2$. So $d(x_n,l) + d(l,y_n) < varepsilon$ for all $n geq max N_1,N_2 $; so $d(x_n,y_n) < varepsilon$ for all $n geq max N_1,N_2 $.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
Let $l$ be the limit of the sequences $(x_n), (y_n)$ by definition Note that $d(x_n,y_n) leq d(x_n,l) + d(l, y_n)$ for all $n$. If $varepsilon > 0$, then there are $N_1,N_2$, by assumption, giving $d(x_n,l), d(l,y_n) < varepsilon/2$ for all $n geq N_1$ and $geq N_2$. So $d(x_n,l) + d(l,y_n) < varepsilon$ for all $n geq max N_1,N_2 $; so $d(x_n,y_n) < varepsilon$ for all $n geq max N_1,N_2 $.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
17.3k21546
add a comment |
add a comment |
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$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27