Mixing
Dual of a complete topological group
$begingroup$
Let $(G,+)$ be an Hausdorff topological abelian group. Assume that $G$ is complete, i.e. every Cauchy sequence in $G$ converges in $G$.
Let $widehat G:=operatorname{Hom}_{text{cont}}(G, S^{1})$ be the dual group of $G$ and put on $widehat G$ the compact-open topology.
Here my question:
Is also $widehat{G}$ a complete group?
Thanks in advance
general-topology group-theory topological-groups duality-theorems
asked Feb 2 at 17:24
notsurenotsure
297
$endgroup$
add a comment |
$begingroup$
Let $(G,+)$ be an Hausdorff topological abelian group. Assume that $G$ is complete, i.e. every Cauchy sequence in $G$ converges in $G$.
Let $widehat G:=operatorname{Hom}_{text{cont}}(G, S^{1})$ be the dual group of $G$ and put on $widehat G$ the compact-open topology.
Here my question:
Is also $widehat{G}$ a complete group?
Thanks in advance
general-topology group-theory topological-groups duality-theorems
asked Feb 2 at 17:24
notsurenotsure
297
$endgroup$
$begingroup$
This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness
$endgroup$
– Max
Feb 2 at 18:43
add a comment |
$begingroup$
Let $(G,+)$ be an Hausdorff topological abelian group. Assume that $G$ is complete, i.e. every Cauchy sequence in $G$ converges in $G$.
Let $widehat G:=operatorname{Hom}_{text{cont}}(G, S^{1})$ be the dual group of $G$ and put on $widehat G$ the compact-open topology.
Here my question:
Is also $widehat{G}$ a complete group?
Thanks in advance
general-topology group-theory topological-groups duality-theorems
asked Feb 2 at 17:24
notsurenotsure
297
$endgroup$
Let $(G,+)$ be an Hausdorff topological abelian group. Assume that $G$ is complete, i.e. every Cauchy sequence in $G$ converges in $G$.
Let $widehat G:=operatorname{Hom}_{text{cont}}(G, S^{1})$ be the dual group of $G$ and put on $widehat G$ the compact-open topology.
Here my question:
Is also $widehat{G}$ a complete group?
Thanks in advance
general-topology group-theory topological-groups duality-theorems
general-topology group-theory topological-groups duality-theorems
asked Feb 2 at 17:24
notsurenotsure
297
asked Feb 2 at 17:24
notsurenotsure
297
edited Feb 2 at 18:05
notsure
asked Feb 2 at 17:24
notsurenotsure
297
asked Feb 2 at 17:24
notsurenotsure
297
asked Feb 2 at 17:24
notsurenotsure
297
297
$begingroup$
This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness
$endgroup$
– Max
Feb 2 at 18:43
add a comment |
$begingroup$
This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness
$endgroup$
– Max
Feb 2 at 18:43
$begingroup$
This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness
$endgroup$
– Max
Feb 2 at 18:43
$begingroup$
This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness
$endgroup$
– Max
Feb 2 at 18:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.
To prove $widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $xin G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_nto f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $finwidehat{G}$, and $f_nto f$ in $widehat{G}$ since the convergence is uniform on compact sets.
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
add a comment |
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1 Answer
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$begingroup$
I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.
To prove $widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $xin G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_nto f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $finwidehat{G}$, and $f_nto f$ in $widehat{G}$ since the convergence is uniform on compact sets.
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
add a comment |
$begingroup$
I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.
To prove $widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $xin G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_nto f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $finwidehat{G}$, and $f_nto f$ in $widehat{G}$ since the convergence is uniform on compact sets.
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
add a comment |
$begingroup$
I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.
To prove $widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $xin G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_nto f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $finwidehat{G}$, and $f_nto f$ in $widehat{G}$ since the convergence is uniform on compact sets.
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
$endgroup$
I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.
To prove $widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $xin G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_nto f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $finwidehat{G}$, and $f_nto f$ in $widehat{G}$ since the convergence is uniform on compact sets.
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
answered Feb 2 at 22:35
Eric WofseyEric Wofsey
193k14221352
193k14221352
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
add a comment |
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Your answer can be adapted for nets anyway, it seems to me.
$endgroup$
– Max
Feb 2 at 22:41
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
Sure, but that doesn't eliminate the need for $G$ to be compactly generated.
$endgroup$
– Eric Wofsey
Feb 2 at 22:48
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ?
$endgroup$
– Max
Feb 2 at 22:51
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them.
$endgroup$
– Eric Wofsey
Feb 2 at 22:53
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
$begingroup$
Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too)
$endgroup$
– Max
Feb 2 at 22:55
add a comment |
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$begingroup$
This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness
$endgroup$
– Max
Feb 2 at 18:43