exercise on convergent sequences The Next CEO of Stack OverflowCompleteness and Cauchy SequencesBounded sequence and every convergent subsequence converges to LProve that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.Prove that if every PROPER sub sequence of a sequence is convergent , then the sequence is convergentSequence of elements having a convergent subsequence -NBHM $2014$Convergent sequences out of bounded sequencesBolzano-Weierstrass for metric spacesA map between metric spaces preserving convergent sequences is continuousConvergent sequences in a metric spaceConvergent and Cauchy sequences in metric spaces
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exercise on convergent sequences
The Next CEO of Stack OverflowCompleteness and Cauchy SequencesBounded sequence and every convergent subsequence converges to LProve that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.Prove that if every PROPER sub sequence of a sequence is convergent , then the sequence is convergentSequence of elements having a convergent subsequence -NBHM $2014$Convergent sequences out of bounded sequencesBolzano-Weierstrass for metric spacesA map between metric spaces preserving convergent sequences is continuousConvergent sequences in a metric spaceConvergent and Cauchy sequences in metric spaces
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How can I prove that a sequence in a metric space converges to an element iff every sub-sequence has a sub-sub-sequence convergent to that element.
Thank for the help!!
real-analysis analysis convergence metric-spaces
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
$endgroup$
add a comment |
$begingroup$
How can I prove that a sequence in a metric space converges to an element iff every sub-sequence has a sub-sub-sequence convergent to that element.
Thank for the help!!
real-analysis analysis convergence metric-spaces
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
$endgroup$
$begingroup$
I think "succession" should be "sequence".
$endgroup$
– zhw.
Mar 18 at 18:42
add a comment |
$begingroup$
How can I prove that a sequence in a metric space converges to an element iff every sub-sequence has a sub-sub-sequence convergent to that element.
Thank for the help!!
real-analysis analysis convergence metric-spaces
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
$endgroup$
How can I prove that a sequence in a metric space converges to an element iff every sub-sequence has a sub-sub-sequence convergent to that element.
Thank for the help!!
real-analysis analysis convergence metric-spaces
real-analysis analysis convergence metric-spaces
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
edited Mar 19 at 10:12
Alessandro Pecile
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
asked Mar 18 at 16:05
Alessandro PecileAlessandro Pecile
755
755
$begingroup$
I think "succession" should be "sequence".
$endgroup$
– zhw.
Mar 18 at 18:42
add a comment |
$begingroup$
I think "succession" should be "sequence".
$endgroup$
– zhw.
Mar 18 at 18:42
$begingroup$
I think "succession" should be "sequence".
$endgroup$
– zhw.
Mar 18 at 18:42
$begingroup$
I think "succession" should be "sequence".
$endgroup$
– zhw.
Mar 18 at 18:42
add a comment |
1 Answer
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Hint: If $x_n$ does not converge to $L,$ then for some $epsilon>0,$ there is no $N$ such that $nge N$ implies $d(x_n,L) <epsilon.$ Argue that this implies there is a subsequence $x_n_k$ such that $d(x_n_k,L) geepsilon$ for all $k.$
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Hint: If $x_n$ does not converge to $L,$ then for some $epsilon>0,$ there is no $N$ such that $nge N$ implies $d(x_n,L) <epsilon.$ Argue that this implies there is a subsequence $x_n_k$ such that $d(x_n_k,L) geepsilon$ for all $k.$
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
$endgroup$
add a comment |
$begingroup$
Hint: If $x_n$ does not converge to $L,$ then for some $epsilon>0,$ there is no $N$ such that $nge N$ implies $d(x_n,L) <epsilon.$ Argue that this implies there is a subsequence $x_n_k$ such that $d(x_n_k,L) geepsilon$ for all $k.$
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
$endgroup$
add a comment |
$begingroup$
Hint: If $x_n$ does not converge to $L,$ then for some $epsilon>0,$ there is no $N$ such that $nge N$ implies $d(x_n,L) <epsilon.$ Argue that this implies there is a subsequence $x_n_k$ such that $d(x_n_k,L) geepsilon$ for all $k.$
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
$endgroup$
Hint: If $x_n$ does not converge to $L,$ then for some $epsilon>0,$ there is no $N$ such that $nge N$ implies $d(x_n,L) <epsilon.$ Argue that this implies there is a subsequence $x_n_k$ such that $d(x_n_k,L) geepsilon$ for all $k.$
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
answered Mar 18 at 18:49
zhw.zhw.
74.8k43175
74.8k43175
add a comment |
add a comment |
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$begingroup$
I think "succession" should be "sequence".
$endgroup$
– zhw.
Mar 18 at 18:42