Does Heine-Borel Theorem hold in $prod_alpha in A mathbbC$, where $A$ is any index set? [closed]
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Does Heine-Borel Theorem hold in $prod_alpha in A mathbbC$, where $A$ is any set? That is is, every closed and bounded set compact in $prod_alpha in A mathbbC$?
In finite case this is true. I am not sure about infinite product.
real-analysis complex-analysis functional-analysis
asked Sep 10 at 18:44
Arindam
303213
closed as unclear what you're asking by Nate Eldredge, user99914, Xander Henderson, Deepesh Meena, Lord Shark the Unknown Sep 11 at 4:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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down vote
favorite
Does Heine-Borel Theorem hold in $prod_alpha in A mathbbC$, where $A$ is any set? That is is, every closed and bounded set compact in $prod_alpha in A mathbbC$?
In finite case this is true. I am not sure about infinite product.
real-analysis complex-analysis functional-analysis
asked Sep 10 at 18:44
Arindam
303213
closed as unclear what you're asking by Nate Eldredge, user99914, Xander Henderson, Deepesh Meena, Lord Shark the Unknown Sep 11 at 4:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
6
What is bounded in your product set?
– Lord Shark the Unknown
Sep 10 at 19:09
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does Heine-Borel Theorem hold in $prod_alpha in A mathbbC$, where $A$ is any set? That is is, every closed and bounded set compact in $prod_alpha in A mathbbC$?
In finite case this is true. I am not sure about infinite product.
real-analysis complex-analysis functional-analysis
asked Sep 10 at 18:44
Arindam
303213
Does Heine-Borel Theorem hold in $prod_alpha in A mathbbC$, where $A$ is any set? That is is, every closed and bounded set compact in $prod_alpha in A mathbbC$?
In finite case this is true. I am not sure about infinite product.
real-analysis complex-analysis functional-analysis
real-analysis complex-analysis functional-analysis
asked Sep 10 at 18:44
Arindam
303213
asked Sep 10 at 18:44
Arindam
303213
edited Sep 10 at 23:23
asked Sep 10 at 18:44
Arindam
303213
asked Sep 10 at 18:44
Arindam
303213
asked Sep 10 at 18:44
Arindam
303213
303213
closed as unclear what you're asking by Nate Eldredge, user99914, Xander Henderson, Deepesh Meena, Lord Shark the Unknown Sep 11 at 4:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Nate Eldredge, user99914, Xander Henderson, Deepesh Meena, Lord Shark the Unknown Sep 11 at 4:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
6
What is bounded in your product set?
– Lord Shark the Unknown
Sep 10 at 19:09
add a comment |Â
6
What is bounded in your product set?
– Lord Shark the Unknown
Sep 10 at 19:09
6
6
What is bounded in your product set?
– Lord Shark the Unknown
Sep 10 at 19:09
What is bounded in your product set?
– Lord Shark the Unknown
Sep 10 at 19:09
add a comment |Â
1 Answer
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There is no standard notion of "bounded" for a subset of $mathbbC^A$ in general. When $A$ is finite then all norms on $mathbbC^A$ are equivalent and give a canonical notion of boundedness, but this all breaks down if $A$ is infinite. Indeed, if $A$ is uncountable, then $mathbbC^A$ is not even metrizable (in the product topology).
However, if you take "bounded" to mean "bounded on each coordinate separately", then the answer is yes. That is, suppose $SsubseteqmathbbC^A$ is a closed set such that for each $ain A$, the set $f(a):fin SsubseteqmathbbC$ is bounded. Then $S$ is compact. This follows from Tychonoff's theorem: any product of compact spaces is compact. So, letting $B_a$ be a closed ball containing $f(a):fin S$, the product $prod_ain AB_a$ is compact and $S$ is a closed subset of it, and so $S$ is also compact.
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
1
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
There is no standard notion of "bounded" for a subset of $mathbbC^A$ in general. When $A$ is finite then all norms on $mathbbC^A$ are equivalent and give a canonical notion of boundedness, but this all breaks down if $A$ is infinite. Indeed, if $A$ is uncountable, then $mathbbC^A$ is not even metrizable (in the product topology).
However, if you take "bounded" to mean "bounded on each coordinate separately", then the answer is yes. That is, suppose $SsubseteqmathbbC^A$ is a closed set such that for each $ain A$, the set $f(a):fin SsubseteqmathbbC$ is bounded. Then $S$ is compact. This follows from Tychonoff's theorem: any product of compact spaces is compact. So, letting $B_a$ be a closed ball containing $f(a):fin S$, the product $prod_ain AB_a$ is compact and $S$ is a closed subset of it, and so $S$ is also compact.
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
1
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
add a comment |Â
up vote
5
down vote
There is no standard notion of "bounded" for a subset of $mathbbC^A$ in general. When $A$ is finite then all norms on $mathbbC^A$ are equivalent and give a canonical notion of boundedness, but this all breaks down if $A$ is infinite. Indeed, if $A$ is uncountable, then $mathbbC^A$ is not even metrizable (in the product topology).
However, if you take "bounded" to mean "bounded on each coordinate separately", then the answer is yes. That is, suppose $SsubseteqmathbbC^A$ is a closed set such that for each $ain A$, the set $f(a):fin SsubseteqmathbbC$ is bounded. Then $S$ is compact. This follows from Tychonoff's theorem: any product of compact spaces is compact. So, letting $B_a$ be a closed ball containing $f(a):fin S$, the product $prod_ain AB_a$ is compact and $S$ is a closed subset of it, and so $S$ is also compact.
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
1
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
add a comment |Â
up vote
5
down vote
up vote
5
down vote
There is no standard notion of "bounded" for a subset of $mathbbC^A$ in general. When $A$ is finite then all norms on $mathbbC^A$ are equivalent and give a canonical notion of boundedness, but this all breaks down if $A$ is infinite. Indeed, if $A$ is uncountable, then $mathbbC^A$ is not even metrizable (in the product topology).
However, if you take "bounded" to mean "bounded on each coordinate separately", then the answer is yes. That is, suppose $SsubseteqmathbbC^A$ is a closed set such that for each $ain A$, the set $f(a):fin SsubseteqmathbbC$ is bounded. Then $S$ is compact. This follows from Tychonoff's theorem: any product of compact spaces is compact. So, letting $B_a$ be a closed ball containing $f(a):fin S$, the product $prod_ain AB_a$ is compact and $S$ is a closed subset of it, and so $S$ is also compact.
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
There is no standard notion of "bounded" for a subset of $mathbbC^A$ in general. When $A$ is finite then all norms on $mathbbC^A$ are equivalent and give a canonical notion of boundedness, but this all breaks down if $A$ is infinite. Indeed, if $A$ is uncountable, then $mathbbC^A$ is not even metrizable (in the product topology).
However, if you take "bounded" to mean "bounded on each coordinate separately", then the answer is yes. That is, suppose $SsubseteqmathbbC^A$ is a closed set such that for each $ain A$, the set $f(a):fin SsubseteqmathbbC$ is bounded. Then $S$ is compact. This follows from Tychonoff's theorem: any product of compact spaces is compact. So, letting $B_a$ be a closed ball containing $f(a):fin S$, the product $prod_ain AB_a$ is compact and $S$ is a closed subset of it, and so $S$ is also compact.
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
answered Sep 10 at 23:36
Eric Wofsey
167k12196312
167k12196312
1
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
add a comment |Â
1
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
1
1
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
To the proposer: Tychonoff's theorem (a.k.a. the Tychonoff Theorem) is not something I would call obvious.
– DanielWainfleet
Sep 11 at 0:34
add a comment |Â
6
What is bounded in your product set?
– Lord Shark the Unknown
Sep 10 at 19:09