Mixing
Why don't we use closed covers to define compactness of metric space?
$begingroup$
I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?
real-analysis general-topology metric-spaces compactness
edited Feb 2 at 23:16
peterh
2,20851731
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
$endgroup$
|
show 12 more comments
$begingroup$
I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?
real-analysis general-topology metric-spaces compactness
edited Feb 2 at 23:16
peterh
2,20851731
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
$endgroup$
8
$begingroup$
No infinite $T_1$ space would be compact: cover it by singletons
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:19
4
$begingroup$
Because then the only "compact" metric spaces are the finite ones. No new info.
$endgroup$
– Henno Brandsma
Feb 2 at 15:23
3
$begingroup$
Because definitions are supposed to be useful and a definition of compact in which almost no space is compact wouldn't be. The definition by open sets happens to be the correct one, but I don't know if there is any better explanation than "it works"
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:24
3
$begingroup$
I think the question is quite interesting. Topology is always based on open sets, but an equivalent definition of topology can be stated using closed sets. And it seems here we can do too. To every open covering one can associated a closed covering just by taking complements. And if the space is compact, there exists a finite open subcovering and thus a finite closed covering. So, in my opinion, the question is not as easy to answer as it may suggest in some comments. On the other hand, answering a question simply by saying "it works" doesn't seem to be very clever.
$endgroup$
– Dog_69
Feb 2 at 17:19
3
$begingroup$
@Dog_69 I mean responses, usually in comnents, that give a trivial or non-explanatory answer to a good question—like "that's just how it is" or whatever—instead of helping the questioner understand anything. "Why" questions seem especially prone to this, because there's usually a trivial reason as well as the deeper one that's being asked for.
$endgroup$
– timtfj
Feb 2 at 21:34
|
show 12 more comments
$begingroup$
I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?
real-analysis general-topology metric-spaces compactness
edited Feb 2 at 23:16
peterh
2,20851731
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
$endgroup$
I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?
real-analysis general-topology metric-spaces compactness
real-analysis general-topology metric-spaces compactness
edited Feb 2 at 23:16
peterh
2,20851731
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
edited Feb 2 at 23:16
peterh
2,20851731
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
edited Feb 2 at 23:16
peterh
2,20851731
edited Feb 2 at 23:16
peterh
2,20851731
edited Feb 2 at 23:16
peterh
2,20851731
2,20851731
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
asked Feb 2 at 15:16
Arjun BanerjeeArjun Banerjee
568110
568110
8
$begingroup$
No infinite $T_1$ space would be compact: cover it by singletons
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:19
4
$begingroup$
Because then the only "compact" metric spaces are the finite ones. No new info.
$endgroup$
– Henno Brandsma
Feb 2 at 15:23
3
$begingroup$
Because definitions are supposed to be useful and a definition of compact in which almost no space is compact wouldn't be. The definition by open sets happens to be the correct one, but I don't know if there is any better explanation than "it works"
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:24
3
$begingroup$
I think the question is quite interesting. Topology is always based on open sets, but an equivalent definition of topology can be stated using closed sets. And it seems here we can do too. To every open covering one can associated a closed covering just by taking complements. And if the space is compact, there exists a finite open subcovering and thus a finite closed covering. So, in my opinion, the question is not as easy to answer as it may suggest in some comments. On the other hand, answering a question simply by saying "it works" doesn't seem to be very clever.
$endgroup$
– Dog_69
Feb 2 at 17:19
3
$begingroup$
@Dog_69 I mean responses, usually in comnents, that give a trivial or non-explanatory answer to a good question—like "that's just how it is" or whatever—instead of helping the questioner understand anything. "Why" questions seem especially prone to this, because there's usually a trivial reason as well as the deeper one that's being asked for.
$endgroup$
– timtfj
Feb 2 at 21:34
|
show 12 more comments
8
$begingroup$
No infinite $T_1$ space would be compact: cover it by singletons
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:19
4
$begingroup$
Because then the only "compact" metric spaces are the finite ones. No new info.
$endgroup$
– Henno Brandsma
Feb 2 at 15:23
3
$begingroup$
Because definitions are supposed to be useful and a definition of compact in which almost no space is compact wouldn't be. The definition by open sets happens to be the correct one, but I don't know if there is any better explanation than "it works"
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:24
3
$begingroup$
I think the question is quite interesting. Topology is always based on open sets, but an equivalent definition of topology can be stated using closed sets. And it seems here we can do too. To every open covering one can associated a closed covering just by taking complements. And if the space is compact, there exists a finite open subcovering and thus a finite closed covering. So, in my opinion, the question is not as easy to answer as it may suggest in some comments. On the other hand, answering a question simply by saying "it works" doesn't seem to be very clever.
$endgroup$
– Dog_69
Feb 2 at 17:19
3
$begingroup$
@Dog_69 I mean responses, usually in comnents, that give a trivial or non-explanatory answer to a good question—like "that's just how it is" or whatever—instead of helping the questioner understand anything. "Why" questions seem especially prone to this, because there's usually a trivial reason as well as the deeper one that's being asked for.
$endgroup$
– timtfj
Feb 2 at 21:34
8
8
$begingroup$
No infinite $T_1$ space would be compact: cover it by singletons
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:19
$begingroup$
No infinite $T_1$ space would be compact: cover it by singletons
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:19
4
4
$begingroup$
Because then the only "compact" metric spaces are the finite ones. No new info.
$endgroup$
– Henno Brandsma
Feb 2 at 15:23
$begingroup$
Because then the only "compact" metric spaces are the finite ones. No new info.
$endgroup$
– Henno Brandsma
Feb 2 at 15:23
3
3
$begingroup$
Because definitions are supposed to be useful and a definition of compact in which almost no space is compact wouldn't be. The definition by open sets happens to be the correct one, but I don't know if there is any better explanation than "it works"
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:24
$begingroup$
Because definitions are supposed to be useful and a definition of compact in which almost no space is compact wouldn't be. The definition by open sets happens to be the correct one, but I don't know if there is any better explanation than "it works"
$endgroup$
– Alessandro Codenotti
Feb 2 at 15:24
3
3
$begingroup$
I think the question is quite interesting. Topology is always based on open sets, but an equivalent definition of topology can be stated using closed sets. And it seems here we can do too. To every open covering one can associated a closed covering just by taking complements. And if the space is compact, there exists a finite open subcovering and thus a finite closed covering. So, in my opinion, the question is not as easy to answer as it may suggest in some comments. On the other hand, answering a question simply by saying "it works" doesn't seem to be very clever.
$endgroup$
– Dog_69
Feb 2 at 17:19
$begingroup$
I think the question is quite interesting. Topology is always based on open sets, but an equivalent definition of topology can be stated using closed sets. And it seems here we can do too. To every open covering one can associated a closed covering just by taking complements. And if the space is compact, there exists a finite open subcovering and thus a finite closed covering. So, in my opinion, the question is not as easy to answer as it may suggest in some comments. On the other hand, answering a question simply by saying "it works" doesn't seem to be very clever.
$endgroup$
– Dog_69
Feb 2 at 17:19
3
3
$begingroup$
@Dog_69 I mean responses, usually in comnents, that give a trivial or non-explanatory answer to a good question—like "that's just how it is" or whatever—instead of helping the questioner understand anything. "Why" questions seem especially prone to this, because there's usually a trivial reason as well as the deeper one that's being asked for.
$endgroup$
– timtfj
Feb 2 at 21:34
$begingroup$
@Dog_69 I mean responses, usually in comnents, that give a trivial or non-explanatory answer to a good question—like "that's just how it is" or whatever—instead of helping the questioner understand anything. "Why" questions seem especially prone to this, because there's usually a trivial reason as well as the deeper one that's being asked for.
$endgroup$
– timtfj
Feb 2 at 21:34
|
show 12 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?
(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)
This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:
- A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
- A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
- If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
- Every sequence of points from a subset of $Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way
and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.
In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.
Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?
And formulating compactness correctly helps us better understand the original space, $Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.
Keep asking these questions. Every definition is made for a reason.
$endgroup$
3
$begingroup$
To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
$endgroup$
– Hagen von Eitzen
Feb 2 at 18:44
1
$begingroup$
As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
$endgroup$
– user159517
Feb 2 at 19:41
$begingroup$
Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
$endgroup$
– MJD
Feb 2 at 21:06
1
$begingroup$
@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
$endgroup$
– Owen
Feb 2 at 21:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?
(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)
This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:
- A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
- A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
- If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
- Every sequence of points from a subset of $Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way
and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.
In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.
Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?
And formulating compactness correctly helps us better understand the original space, $Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.
Keep asking these questions. Every definition is made for a reason.
$endgroup$
3
$begingroup$
To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
$endgroup$
– Hagen von Eitzen
Feb 2 at 18:44
1
$begingroup$
As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
$endgroup$
– user159517
Feb 2 at 19:41
$begingroup$
Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
$endgroup$
– MJD
Feb 2 at 21:06
1
$begingroup$
@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
$endgroup$
– Owen
Feb 2 at 21:46
add a comment |
$begingroup$
It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?
(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)
This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:
- A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
- A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
- If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
- Every sequence of points from a subset of $Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way
and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.
In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.
Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?
And formulating compactness correctly helps us better understand the original space, $Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.
Keep asking these questions. Every definition is made for a reason.
$endgroup$
3
$begingroup$
To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
$endgroup$
– Hagen von Eitzen
Feb 2 at 18:44
1
$begingroup$
As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
$endgroup$
– user159517
Feb 2 at 19:41
$begingroup$
Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
$endgroup$
– MJD
Feb 2 at 21:06
1
$begingroup$
@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
$endgroup$
– Owen
Feb 2 at 21:46
add a comment |
$begingroup$
It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?
(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)
This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:
- A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
- A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
- If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
- Every sequence of points from a subset of $Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way
and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.
In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.
Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?
And formulating compactness correctly helps us better understand the original space, $Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.
Keep asking these questions. Every definition is made for a reason.
$endgroup$
It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?
(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)
This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:
- A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
- A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
- If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
- Every sequence of points from a subset of $Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way
and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.
In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.
Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?
And formulating compactness correctly helps us better understand the original space, $Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.
Keep asking these questions. Every definition is made for a reason.
edited Feb 2 at 21:05
community wiki
4 revs
MJD
3
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To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
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– Hagen von Eitzen
Feb 2 at 18:44
1
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As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
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– user159517
Feb 2 at 19:41
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Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
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– MJD
Feb 2 at 21:06
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@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
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– Owen
Feb 2 at 21:46
add a comment |
3
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To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
$endgroup$
– Hagen von Eitzen
Feb 2 at 18:44
1
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As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
$endgroup$
– user159517
Feb 2 at 19:41
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Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
$endgroup$
– MJD
Feb 2 at 21:06
1
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@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
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– Owen
Feb 2 at 21:46
3
3
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To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
$endgroup$
– Hagen von Eitzen
Feb 2 at 18:44
$begingroup$
To add to the last sentence: And it is always a worthwhile exercise to learn about that reason by playing with definitions as the OP did here (replace open with closed cover) or e.g. swapping the quantors in the $epsilondelta$ definition of continuity or or or ... and see how that breaks the nice properties the concept given by original definition has
$endgroup$
– Hagen von Eitzen
Feb 2 at 18:44
1
1
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As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
$endgroup$
– user159517
Feb 2 at 19:41
$begingroup$
As I read it, the "only if" in 3 is not true - any constant function $f$ is a counterexample.
$endgroup$
– user159517
Feb 2 at 19:41
$begingroup$
Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
$endgroup$
– MJD
Feb 2 at 21:06
$begingroup$
Thanks. I have corrected the statement of (3) to make the quantifiers clearer.
$endgroup$
– MJD
Feb 2 at 21:06
1
1
$begingroup$
@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
$endgroup$
– Owen
Feb 2 at 21:46
$begingroup$
@MJD the same issue of the quantifiers would apply to (1) and (2), I believe (though I think it's clear what you mean).
$endgroup$
– Owen
Feb 2 at 21:46
add a comment |
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No infinite $T_1$ space would be compact: cover it by singletons
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– Alessandro Codenotti
Feb 2 at 15:19
4
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Because then the only "compact" metric spaces are the finite ones. No new info.
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– Henno Brandsma
Feb 2 at 15:23
3
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Because definitions are supposed to be useful and a definition of compact in which almost no space is compact wouldn't be. The definition by open sets happens to be the correct one, but I don't know if there is any better explanation than "it works"
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– Alessandro Codenotti
Feb 2 at 15:24
3
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I think the question is quite interesting. Topology is always based on open sets, but an equivalent definition of topology can be stated using closed sets. And it seems here we can do too. To every open covering one can associated a closed covering just by taking complements. And if the space is compact, there exists a finite open subcovering and thus a finite closed covering. So, in my opinion, the question is not as easy to answer as it may suggest in some comments. On the other hand, answering a question simply by saying "it works" doesn't seem to be very clever.
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– Dog_69
Feb 2 at 17:19
3
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@Dog_69 I mean responses, usually in comnents, that give a trivial or non-explanatory answer to a good question—like "that's just how it is" or whatever—instead of helping the questioner understand anything. "Why" questions seem especially prone to this, because there's usually a trivial reason as well as the deeper one that's being asked for.
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– timtfj
Feb 2 at 21:34