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Any Cauchy net in metric space is convergent?
Any Cauchy net in a metric space is convergent?
I know that if (X,d) is metric space COMPLETE then cauchy net in X is convergent in X.
But, only (X,d) is metric space (not complete) It is true?
pd:
As additional doubt. In general, any succession of Cauchy in a set X is always converges in (probability another) some set in somewhere?
functional-analysis metric-spaces cauchy-sequences complete-spaces nets
asked Nov 21 '18 at 0:39
eraldcoil
363111
add a comment |
Any Cauchy net in a metric space is convergent?
I know that if (X,d) is metric space COMPLETE then cauchy net in X is convergent in X.
But, only (X,d) is metric space (not complete) It is true?
pd:
As additional doubt. In general, any succession of Cauchy in a set X is always converges in (probability another) some set in somewhere?
functional-analysis metric-spaces cauchy-sequences complete-spaces nets
asked Nov 21 '18 at 0:39
eraldcoil
363111
The definition of a complete metric space is precisely that every Cauchy net converges in the metric space, so this fails every time a metric space is not complete. user348279's answer gives an example of a non-complete metric space.
– Monstrous Moonshiner
Nov 21 '18 at 1:02
add a comment |
Any Cauchy net in a metric space is convergent?
I know that if (X,d) is metric space COMPLETE then cauchy net in X is convergent in X.
But, only (X,d) is metric space (not complete) It is true?
pd:
As additional doubt. In general, any succession of Cauchy in a set X is always converges in (probability another) some set in somewhere?
functional-analysis metric-spaces cauchy-sequences complete-spaces nets
asked Nov 21 '18 at 0:39
eraldcoil
363111
Any Cauchy net in a metric space is convergent?
I know that if (X,d) is metric space COMPLETE then cauchy net in X is convergent in X.
But, only (X,d) is metric space (not complete) It is true?
pd:
As additional doubt. In general, any succession of Cauchy in a set X is always converges in (probability another) some set in somewhere?
functional-analysis metric-spaces cauchy-sequences complete-spaces nets
functional-analysis metric-spaces cauchy-sequences complete-spaces nets
asked Nov 21 '18 at 0:39
eraldcoil
363111
asked Nov 21 '18 at 0:39
eraldcoil
363111
edited Nov 21 '18 at 0:45
asked Nov 21 '18 at 0:39
eraldcoil
363111
asked Nov 21 '18 at 0:39
eraldcoil
363111
asked Nov 21 '18 at 0:39
eraldcoil
363111
363111
The definition of a complete metric space is precisely that every Cauchy net converges in the metric space, so this fails every time a metric space is not complete. user348279's answer gives an example of a non-complete metric space.
– Monstrous Moonshiner
Nov 21 '18 at 1:02
add a comment |
The definition of a complete metric space is precisely that every Cauchy net converges in the metric space, so this fails every time a metric space is not complete. user348279's answer gives an example of a non-complete metric space.
– Monstrous Moonshiner
Nov 21 '18 at 1:02
The definition of a complete metric space is precisely that every Cauchy net converges in the metric space, so this fails every time a metric space is not complete. user348279's answer gives an example of a non-complete metric space.
– Monstrous Moonshiner
Nov 21 '18 at 1:02
The definition of a complete metric space is precisely that every Cauchy net converges in the metric space, so this fails every time a metric space is not complete. user348279's answer gives an example of a non-complete metric space.
– Monstrous Moonshiner
Nov 21 '18 at 1:02
add a comment |
1 Answer
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No: in particular, the statement is false for sequences, and all sequences are nets (they're those nets for which the directed set is the natural numbers with the obvious order relation). For an explicit example, take any Cauchy sequence of rationals that converges in the reals to $pi$. In the metric space $mathbb{Q}$, this sequence does not converge.
I'm not at all sure what you mean by your edit, but I think that you're asking if, for every metric space $X$, there is some complete metric space that includes both $X$ and all limits of $X$. If that's correct, then the answer is yes: you simply take all of those things and throw them together, and you're done: this is called the completion of $X$, and is, for example, one of the standard definitions/constructions of $mathbb{R}$: it is precisely the completion of the rationals.
answered Nov 21 '18 at 0:44
user3482749
2,687414
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
No: in particular, the statement is false for sequences, and all sequences are nets (they're those nets for which the directed set is the natural numbers with the obvious order relation). For an explicit example, take any Cauchy sequence of rationals that converges in the reals to $pi$. In the metric space $mathbb{Q}$, this sequence does not converge.
I'm not at all sure what you mean by your edit, but I think that you're asking if, for every metric space $X$, there is some complete metric space that includes both $X$ and all limits of $X$. If that's correct, then the answer is yes: you simply take all of those things and throw them together, and you're done: this is called the completion of $X$, and is, for example, one of the standard definitions/constructions of $mathbb{R}$: it is precisely the completion of the rationals.
answered Nov 21 '18 at 0:44
user3482749
2,687414
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
add a comment |
No: in particular, the statement is false for sequences, and all sequences are nets (they're those nets for which the directed set is the natural numbers with the obvious order relation). For an explicit example, take any Cauchy sequence of rationals that converges in the reals to $pi$. In the metric space $mathbb{Q}$, this sequence does not converge.
I'm not at all sure what you mean by your edit, but I think that you're asking if, for every metric space $X$, there is some complete metric space that includes both $X$ and all limits of $X$. If that's correct, then the answer is yes: you simply take all of those things and throw them together, and you're done: this is called the completion of $X$, and is, for example, one of the standard definitions/constructions of $mathbb{R}$: it is precisely the completion of the rationals.
answered Nov 21 '18 at 0:44
user3482749
2,687414
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
add a comment |
No: in particular, the statement is false for sequences, and all sequences are nets (they're those nets for which the directed set is the natural numbers with the obvious order relation). For an explicit example, take any Cauchy sequence of rationals that converges in the reals to $pi$. In the metric space $mathbb{Q}$, this sequence does not converge.
I'm not at all sure what you mean by your edit, but I think that you're asking if, for every metric space $X$, there is some complete metric space that includes both $X$ and all limits of $X$. If that's correct, then the answer is yes: you simply take all of those things and throw them together, and you're done: this is called the completion of $X$, and is, for example, one of the standard definitions/constructions of $mathbb{R}$: it is precisely the completion of the rationals.
answered Nov 21 '18 at 0:44
user3482749
2,687414
No: in particular, the statement is false for sequences, and all sequences are nets (they're those nets for which the directed set is the natural numbers with the obvious order relation). For an explicit example, take any Cauchy sequence of rationals that converges in the reals to $pi$. In the metric space $mathbb{Q}$, this sequence does not converge.
I'm not at all sure what you mean by your edit, but I think that you're asking if, for every metric space $X$, there is some complete metric space that includes both $X$ and all limits of $X$. If that's correct, then the answer is yes: you simply take all of those things and throw them together, and you're done: this is called the completion of $X$, and is, for example, one of the standard definitions/constructions of $mathbb{R}$: it is precisely the completion of the rationals.
answered Nov 21 '18 at 0:44
user3482749
2,687414
edited Nov 21 '18 at 0:52
answered Nov 21 '18 at 0:44
user3482749
2,687414
answered Nov 21 '18 at 0:44
user3482749
2,687414
answered Nov 21 '18 at 0:44
user3482749
2,687414
2,687414
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
add a comment |
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
I do indeed, oops.
– user3482749
Nov 21 '18 at 0:52
add a comment |
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The definition of a complete metric space is precisely that every Cauchy net converges in the metric space, so this fails every time a metric space is not complete. user348279's answer gives an example of a non-complete metric space.
– Monstrous Moonshiner
Nov 21 '18 at 1:02