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Why we define the completeness of a space by the converge of a Cauchy sequence rather than a normal sequence?
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The intuition of the completeness to me is that the limit of any sequence converges to the point inside the set itself.
But why we define a set to be complete as any Cauchy sequence converge into the set itself? It seems a more complex definition than a simple convergent sequence. Why we use Cauchy sequence rather than a simple sequence?
For the simple sequence, can we use $lim_nrightarrow infty x_n=x$?
real-analysis functional-analysis convergence cauchy-sequences
asked Aug 10 at 3:51
maple
8332922
 |Â
show 2 more comments
up vote
2
down vote
favorite
The intuition of the completeness to me is that the limit of any sequence converges to the point inside the set itself.
But why we define a set to be complete as any Cauchy sequence converge into the set itself? It seems a more complex definition than a simple convergent sequence. Why we use Cauchy sequence rather than a simple sequence?
For the simple sequence, can we use $lim_nrightarrow infty x_n=x$?
real-analysis functional-analysis convergence cauchy-sequences
asked Aug 10 at 3:51
maple
8332922
2
What do you mean by normal sequence or simple sequence?
– Aweygan
Aug 10 at 3:54
Because, Cauchy sequences always converge; in the case the space is not complete, they just converge to something that is outside of the space.
– onurcanbektas
Aug 10 at 3:54
@Aweygan, can we define it as the edited question? Or do you mean that convergent is defined by Cauchy sequence?
– maple
Aug 10 at 3:57
1
Your intuition is off. In any metric space with more than one point, there are always sequences that don't converge. In particular, a sequence that is not a Cauchy sequence can't converge. For example, a sequence that alternates between two different points.
– Robert Israel
Aug 10 at 3:58
@RobertIsrael Can I just require that the sequence is convergent?
– maple
Aug 10 at 4:00
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The intuition of the completeness to me is that the limit of any sequence converges to the point inside the set itself.
But why we define a set to be complete as any Cauchy sequence converge into the set itself? It seems a more complex definition than a simple convergent sequence. Why we use Cauchy sequence rather than a simple sequence?
For the simple sequence, can we use $lim_nrightarrow infty x_n=x$?
real-analysis functional-analysis convergence cauchy-sequences
asked Aug 10 at 3:51
maple
8332922
The intuition of the completeness to me is that the limit of any sequence converges to the point inside the set itself.
But why we define a set to be complete as any Cauchy sequence converge into the set itself? It seems a more complex definition than a simple convergent sequence. Why we use Cauchy sequence rather than a simple sequence?
For the simple sequence, can we use $lim_nrightarrow infty x_n=x$?
real-analysis functional-analysis convergence cauchy-sequences
asked Aug 10 at 3:51
maple
8332922
edited Aug 10 at 3:56
asked Aug 10 at 3:51
maple
8332922
asked Aug 10 at 3:51
maple
8332922
asked Aug 10 at 3:51
maple
8332922
8332922
2
What do you mean by normal sequence or simple sequence?
– Aweygan
Aug 10 at 3:54
Because, Cauchy sequences always converge; in the case the space is not complete, they just converge to something that is outside of the space.
– onurcanbektas
Aug 10 at 3:54
@Aweygan, can we define it as the edited question? Or do you mean that convergent is defined by Cauchy sequence?
– maple
Aug 10 at 3:57
1
Your intuition is off. In any metric space with more than one point, there are always sequences that don't converge. In particular, a sequence that is not a Cauchy sequence can't converge. For example, a sequence that alternates between two different points.
– Robert Israel
Aug 10 at 3:58
@RobertIsrael Can I just require that the sequence is convergent?
– maple
Aug 10 at 4:00
 |Â
show 2 more comments
2
What do you mean by normal sequence or simple sequence?
– Aweygan
Aug 10 at 3:54
Because, Cauchy sequences always converge; in the case the space is not complete, they just converge to something that is outside of the space.
– onurcanbektas
Aug 10 at 3:54
@Aweygan, can we define it as the edited question? Or do you mean that convergent is defined by Cauchy sequence?
– maple
Aug 10 at 3:57
1
Your intuition is off. In any metric space with more than one point, there are always sequences that don't converge. In particular, a sequence that is not a Cauchy sequence can't converge. For example, a sequence that alternates between two different points.
– Robert Israel
Aug 10 at 3:58
@RobertIsrael Can I just require that the sequence is convergent?
– maple
Aug 10 at 4:00
2
2
What do you mean by normal sequence or simple sequence?
– Aweygan
Aug 10 at 3:54
What do you mean by normal sequence or simple sequence?
– Aweygan
Aug 10 at 3:54
Because, Cauchy sequences always converge; in the case the space is not complete, they just converge to something that is outside of the space.
– onurcanbektas
Aug 10 at 3:54
Because, Cauchy sequences always converge; in the case the space is not complete, they just converge to something that is outside of the space.
– onurcanbektas
Aug 10 at 3:54
@Aweygan, can we define it as the edited question? Or do you mean that convergent is defined by Cauchy sequence?
– maple
Aug 10 at 3:57
@Aweygan, can we define it as the edited question? Or do you mean that convergent is defined by Cauchy sequence?
– maple
Aug 10 at 3:57
1
1
Your intuition is off. In any metric space with more than one point, there are always sequences that don't converge. In particular, a sequence that is not a Cauchy sequence can't converge. For example, a sequence that alternates between two different points.
– Robert Israel
Aug 10 at 3:58
Your intuition is off. In any metric space with more than one point, there are always sequences that don't converge. In particular, a sequence that is not a Cauchy sequence can't converge. For example, a sequence that alternates between two different points.
– Robert Israel
Aug 10 at 3:58
@RobertIsrael Can I just require that the sequence is convergent?
– maple
Aug 10 at 4:00
@RobertIsrael Can I just require that the sequence is convergent?
– maple
Aug 10 at 4:00
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
13
down vote
accepted
If I understand correctly, what you're doing is positing an existing "ambient complete space" $Y$ within which the space $X$ you care about lives. That is, you want to say
$(*)quad$ $X$ is complete if every $colorredmboxconvergent$ sequence of elements of $X$ converges to an element of $X$
However, this has a serious problem: what does the red "convergent" mean? If we take that to mean convergent in the sense of $X$, then every space is complete in this sense.
To make $(*)$ work as a notion of completeness, we need to have an ambient space $Y$ which we're using as our guide to what sequences "really converge." In many cases it's clear what $Y$ should be - e.g. (with the usual metric) for $X=mathbbQ$ we clearly want $Y=mathbbR$ - but in general this starts us down a dangerous road: for a really weird metric space $X$, how would you go about finding the right $Y$?
Instead, we want to define completeness in a "self-contained" way: the statement "$X$ is complete" should only make reference to $X$ itself, not any presupposed ambient space. This is where Cauchy sequences come in: to tell if a sequence of elements of $X$ is Cauchy, we don't need any ambient space to live in - Cauchyness is determined fully within $X$. Intuitively, a sequence is Cauchy if it "ought to" converge, and that's where we get the right definition of completeness from:
$(**)quad$ $X$ is complete if every Cauchy sequence in $X$ converges in $X$.
Incidentally, with this definition in hand we can appropriately make $(*)$ a theorem, as follows:
First, we show that every metric space $X$ has a completion $hatX$. Roughly speaking, points in $hatX$ are "named" by Cauchy sequences from $X$, and for each point $x$ in $X$ the constant sequence $(x,x,x,x,...)$ "names" $x$ in $hatX$ so that we can think of $X$ as literally being a subset of $hatX$. The details are more complicated - for one thing, multiple Cauchy sequences might name the same point! - but this is the basic idea.
This is the ambient space we wanted in $(*)$! We can now prove that $X$ is complete iff every sequence of elements of $X$ which is convergent in the sense of $hatX$, converges to something in $X$ (in the sense of either $X$ or $hatX$; they'll agree on this).
Remark: This is an example of a more general phenomenon: that in mathematics, we frequently want to consider objects "on their own" as opposed to embedded in some larger "background object." This often makes things harder to visualize, but the payoff is huge. For one thing, it broadens the range of objects we can talk about (e.g. in this case, $(**)$ lets us talk about the (in)completeness of metric spaces without any obvious "background"). For another, it can free us from ultimately misleading intuitions. A good example of this is the idea of intrinsic dimension: if we insist on thinking of surfaces as embedded in an ambient space it's natural to say that the hollow sphere is three-dimensional while the Klein bottle is four-dimensional, but the right way to think about things turns out to be that they're each two-dimensional.
answered Aug 10 at 4:05
Noah Schweber
111k9141265
Very nicely explained!
– amsmath
Aug 10 at 4:09
2
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
1
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
1
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
If I understand correctly, what you're doing is positing an existing "ambient complete space" $Y$ within which the space $X$ you care about lives. That is, you want to say
$(*)quad$ $X$ is complete if every $colorredmboxconvergent$ sequence of elements of $X$ converges to an element of $X$
However, this has a serious problem: what does the red "convergent" mean? If we take that to mean convergent in the sense of $X$, then every space is complete in this sense.
To make $(*)$ work as a notion of completeness, we need to have an ambient space $Y$ which we're using as our guide to what sequences "really converge." In many cases it's clear what $Y$ should be - e.g. (with the usual metric) for $X=mathbbQ$ we clearly want $Y=mathbbR$ - but in general this starts us down a dangerous road: for a really weird metric space $X$, how would you go about finding the right $Y$?
Instead, we want to define completeness in a "self-contained" way: the statement "$X$ is complete" should only make reference to $X$ itself, not any presupposed ambient space. This is where Cauchy sequences come in: to tell if a sequence of elements of $X$ is Cauchy, we don't need any ambient space to live in - Cauchyness is determined fully within $X$. Intuitively, a sequence is Cauchy if it "ought to" converge, and that's where we get the right definition of completeness from:
$(**)quad$ $X$ is complete if every Cauchy sequence in $X$ converges in $X$.
Incidentally, with this definition in hand we can appropriately make $(*)$ a theorem, as follows:
First, we show that every metric space $X$ has a completion $hatX$. Roughly speaking, points in $hatX$ are "named" by Cauchy sequences from $X$, and for each point $x$ in $X$ the constant sequence $(x,x,x,x,...)$ "names" $x$ in $hatX$ so that we can think of $X$ as literally being a subset of $hatX$. The details are more complicated - for one thing, multiple Cauchy sequences might name the same point! - but this is the basic idea.
This is the ambient space we wanted in $(*)$! We can now prove that $X$ is complete iff every sequence of elements of $X$ which is convergent in the sense of $hatX$, converges to something in $X$ (in the sense of either $X$ or $hatX$; they'll agree on this).
Remark: This is an example of a more general phenomenon: that in mathematics, we frequently want to consider objects "on their own" as opposed to embedded in some larger "background object." This often makes things harder to visualize, but the payoff is huge. For one thing, it broadens the range of objects we can talk about (e.g. in this case, $(**)$ lets us talk about the (in)completeness of metric spaces without any obvious "background"). For another, it can free us from ultimately misleading intuitions. A good example of this is the idea of intrinsic dimension: if we insist on thinking of surfaces as embedded in an ambient space it's natural to say that the hollow sphere is three-dimensional while the Klein bottle is four-dimensional, but the right way to think about things turns out to be that they're each two-dimensional.
answered Aug 10 at 4:05
Noah Schweber
111k9141265
Very nicely explained!
– amsmath
Aug 10 at 4:09
2
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
1
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
1
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
add a comment |Â
up vote
13
down vote
accepted
If I understand correctly, what you're doing is positing an existing "ambient complete space" $Y$ within which the space $X$ you care about lives. That is, you want to say
$(*)quad$ $X$ is complete if every $colorredmboxconvergent$ sequence of elements of $X$ converges to an element of $X$
However, this has a serious problem: what does the red "convergent" mean? If we take that to mean convergent in the sense of $X$, then every space is complete in this sense.
To make $(*)$ work as a notion of completeness, we need to have an ambient space $Y$ which we're using as our guide to what sequences "really converge." In many cases it's clear what $Y$ should be - e.g. (with the usual metric) for $X=mathbbQ$ we clearly want $Y=mathbbR$ - but in general this starts us down a dangerous road: for a really weird metric space $X$, how would you go about finding the right $Y$?
Instead, we want to define completeness in a "self-contained" way: the statement "$X$ is complete" should only make reference to $X$ itself, not any presupposed ambient space. This is where Cauchy sequences come in: to tell if a sequence of elements of $X$ is Cauchy, we don't need any ambient space to live in - Cauchyness is determined fully within $X$. Intuitively, a sequence is Cauchy if it "ought to" converge, and that's where we get the right definition of completeness from:
$(**)quad$ $X$ is complete if every Cauchy sequence in $X$ converges in $X$.
Incidentally, with this definition in hand we can appropriately make $(*)$ a theorem, as follows:
First, we show that every metric space $X$ has a completion $hatX$. Roughly speaking, points in $hatX$ are "named" by Cauchy sequences from $X$, and for each point $x$ in $X$ the constant sequence $(x,x,x,x,...)$ "names" $x$ in $hatX$ so that we can think of $X$ as literally being a subset of $hatX$. The details are more complicated - for one thing, multiple Cauchy sequences might name the same point! - but this is the basic idea.
This is the ambient space we wanted in $(*)$! We can now prove that $X$ is complete iff every sequence of elements of $X$ which is convergent in the sense of $hatX$, converges to something in $X$ (in the sense of either $X$ or $hatX$; they'll agree on this).
Remark: This is an example of a more general phenomenon: that in mathematics, we frequently want to consider objects "on their own" as opposed to embedded in some larger "background object." This often makes things harder to visualize, but the payoff is huge. For one thing, it broadens the range of objects we can talk about (e.g. in this case, $(**)$ lets us talk about the (in)completeness of metric spaces without any obvious "background"). For another, it can free us from ultimately misleading intuitions. A good example of this is the idea of intrinsic dimension: if we insist on thinking of surfaces as embedded in an ambient space it's natural to say that the hollow sphere is three-dimensional while the Klein bottle is four-dimensional, but the right way to think about things turns out to be that they're each two-dimensional.
answered Aug 10 at 4:05
Noah Schweber
111k9141265
Very nicely explained!
– amsmath
Aug 10 at 4:09
2
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
1
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
1
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
add a comment |Â
up vote
13
down vote
accepted
up vote
13
down vote
accepted
If I understand correctly, what you're doing is positing an existing "ambient complete space" $Y$ within which the space $X$ you care about lives. That is, you want to say
$(*)quad$ $X$ is complete if every $colorredmboxconvergent$ sequence of elements of $X$ converges to an element of $X$
However, this has a serious problem: what does the red "convergent" mean? If we take that to mean convergent in the sense of $X$, then every space is complete in this sense.
To make $(*)$ work as a notion of completeness, we need to have an ambient space $Y$ which we're using as our guide to what sequences "really converge." In many cases it's clear what $Y$ should be - e.g. (with the usual metric) for $X=mathbbQ$ we clearly want $Y=mathbbR$ - but in general this starts us down a dangerous road: for a really weird metric space $X$, how would you go about finding the right $Y$?
Instead, we want to define completeness in a "self-contained" way: the statement "$X$ is complete" should only make reference to $X$ itself, not any presupposed ambient space. This is where Cauchy sequences come in: to tell if a sequence of elements of $X$ is Cauchy, we don't need any ambient space to live in - Cauchyness is determined fully within $X$. Intuitively, a sequence is Cauchy if it "ought to" converge, and that's where we get the right definition of completeness from:
$(**)quad$ $X$ is complete if every Cauchy sequence in $X$ converges in $X$.
Incidentally, with this definition in hand we can appropriately make $(*)$ a theorem, as follows:
First, we show that every metric space $X$ has a completion $hatX$. Roughly speaking, points in $hatX$ are "named" by Cauchy sequences from $X$, and for each point $x$ in $X$ the constant sequence $(x,x,x,x,...)$ "names" $x$ in $hatX$ so that we can think of $X$ as literally being a subset of $hatX$. The details are more complicated - for one thing, multiple Cauchy sequences might name the same point! - but this is the basic idea.
This is the ambient space we wanted in $(*)$! We can now prove that $X$ is complete iff every sequence of elements of $X$ which is convergent in the sense of $hatX$, converges to something in $X$ (in the sense of either $X$ or $hatX$; they'll agree on this).
Remark: This is an example of a more general phenomenon: that in mathematics, we frequently want to consider objects "on their own" as opposed to embedded in some larger "background object." This often makes things harder to visualize, but the payoff is huge. For one thing, it broadens the range of objects we can talk about (e.g. in this case, $(**)$ lets us talk about the (in)completeness of metric spaces without any obvious "background"). For another, it can free us from ultimately misleading intuitions. A good example of this is the idea of intrinsic dimension: if we insist on thinking of surfaces as embedded in an ambient space it's natural to say that the hollow sphere is three-dimensional while the Klein bottle is four-dimensional, but the right way to think about things turns out to be that they're each two-dimensional.
answered Aug 10 at 4:05
Noah Schweber
111k9141265
If I understand correctly, what you're doing is positing an existing "ambient complete space" $Y$ within which the space $X$ you care about lives. That is, you want to say
$(*)quad$ $X$ is complete if every $colorredmboxconvergent$ sequence of elements of $X$ converges to an element of $X$
However, this has a serious problem: what does the red "convergent" mean? If we take that to mean convergent in the sense of $X$, then every space is complete in this sense.
To make $(*)$ work as a notion of completeness, we need to have an ambient space $Y$ which we're using as our guide to what sequences "really converge." In many cases it's clear what $Y$ should be - e.g. (with the usual metric) for $X=mathbbQ$ we clearly want $Y=mathbbR$ - but in general this starts us down a dangerous road: for a really weird metric space $X$, how would you go about finding the right $Y$?
Instead, we want to define completeness in a "self-contained" way: the statement "$X$ is complete" should only make reference to $X$ itself, not any presupposed ambient space. This is where Cauchy sequences come in: to tell if a sequence of elements of $X$ is Cauchy, we don't need any ambient space to live in - Cauchyness is determined fully within $X$. Intuitively, a sequence is Cauchy if it "ought to" converge, and that's where we get the right definition of completeness from:
$(**)quad$ $X$ is complete if every Cauchy sequence in $X$ converges in $X$.
Incidentally, with this definition in hand we can appropriately make $(*)$ a theorem, as follows:
First, we show that every metric space $X$ has a completion $hatX$. Roughly speaking, points in $hatX$ are "named" by Cauchy sequences from $X$, and for each point $x$ in $X$ the constant sequence $(x,x,x,x,...)$ "names" $x$ in $hatX$ so that we can think of $X$ as literally being a subset of $hatX$. The details are more complicated - for one thing, multiple Cauchy sequences might name the same point! - but this is the basic idea.
This is the ambient space we wanted in $(*)$! We can now prove that $X$ is complete iff every sequence of elements of $X$ which is convergent in the sense of $hatX$, converges to something in $X$ (in the sense of either $X$ or $hatX$; they'll agree on this).
Remark: This is an example of a more general phenomenon: that in mathematics, we frequently want to consider objects "on their own" as opposed to embedded in some larger "background object." This often makes things harder to visualize, but the payoff is huge. For one thing, it broadens the range of objects we can talk about (e.g. in this case, $(**)$ lets us talk about the (in)completeness of metric spaces without any obvious "background"). For another, it can free us from ultimately misleading intuitions. A good example of this is the idea of intrinsic dimension: if we insist on thinking of surfaces as embedded in an ambient space it's natural to say that the hollow sphere is three-dimensional while the Klein bottle is four-dimensional, but the right way to think about things turns out to be that they're each two-dimensional.
answered Aug 10 at 4:05
Noah Schweber
111k9141265
edited Aug 10 at 4:19
answered Aug 10 at 4:05
Noah Schweber
111k9141265
answered Aug 10 at 4:05
Noah Schweber
111k9141265
answered Aug 10 at 4:05
Noah Schweber
111k9141265
111k9141265
Very nicely explained!
– amsmath
Aug 10 at 4:09
2
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
1
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
1
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
add a comment |Â
Very nicely explained!
– amsmath
Aug 10 at 4:09
2
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
1
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
1
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
Very nicely explained!
– amsmath
Aug 10 at 4:09
Very nicely explained!
– amsmath
Aug 10 at 4:09
2
2
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
Oh wow! They enabled scrolling for equations. The web designers of this website are going for a clunky look.
– user583012
Aug 10 at 4:09
1
1
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
@fatherBrown Did that improve the formatting?
– Noah Schweber
Aug 10 at 4:13
1
1
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
@NoahSchweber Yes, but the complain wasn't about your choices. The web designers are the ones that should have not made those scroll bars appear where are not needed, specially the vertical one.
– user583012
Aug 10 at 4:16
add a comment |Â
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2
What do you mean by normal sequence or simple sequence?
– Aweygan
Aug 10 at 3:54
Because, Cauchy sequences always converge; in the case the space is not complete, they just converge to something that is outside of the space.
– onurcanbektas
Aug 10 at 3:54
@Aweygan, can we define it as the edited question? Or do you mean that convergent is defined by Cauchy sequence?
– maple
Aug 10 at 3:57
1
Your intuition is off. In any metric space with more than one point, there are always sequences that don't converge. In particular, a sequence that is not a Cauchy sequence can't converge. For example, a sequence that alternates between two different points.
– Robert Israel
Aug 10 at 3:58
@RobertIsrael Can I just require that the sequence is convergent?
– maple
Aug 10 at 4:00