Mixing
What is a compact element?
$begingroup$
I was reading the definition of an Algebraic Lattice: "An algebraic lattice is a complete lattice L, such that every element x of L is the supremum of the compact elements below x".
Then I looked for the formal definition of a compact element and It apperead really messy to me ( mentioning directed subsets and ideals).
Then I looked, always on wikipedia, for some examples to gain some intuition but all they provided is:
-"The most basic example is obtained by considering the power set of some set, ordered by subset inclusion. Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"."
Which is not helpful at all.
-"The term "compact" is explained by considering the complete lattices of open sets of some topological space, also ordered by subset inclusion. Within this order, the compact elements are just the compact sets. Indeed, the condition for compactness in join-semilattices translates immediately to the corresponding definition."
Which is too complicated for me.
I hope someone can present an informal explanation, or some intuitve examples at least, that may clarify the concept.
lattice-orders
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
$endgroup$
add a comment |
$begingroup$
I was reading the definition of an Algebraic Lattice: "An algebraic lattice is a complete lattice L, such that every element x of L is the supremum of the compact elements below x".
Then I looked for the formal definition of a compact element and It apperead really messy to me ( mentioning directed subsets and ideals).
Then I looked, always on wikipedia, for some examples to gain some intuition but all they provided is:
-"The most basic example is obtained by considering the power set of some set, ordered by subset inclusion. Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"."
Which is not helpful at all.
-"The term "compact" is explained by considering the complete lattices of open sets of some topological space, also ordered by subset inclusion. Within this order, the compact elements are just the compact sets. Indeed, the condition for compactness in join-semilattices translates immediately to the corresponding definition."
Which is too complicated for me.
I hope someone can present an informal explanation, or some intuitve examples at least, that may clarify the concept.
lattice-orders
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
$endgroup$
add a comment |
$begingroup$
I was reading the definition of an Algebraic Lattice: "An algebraic lattice is a complete lattice L, such that every element x of L is the supremum of the compact elements below x".
Then I looked for the formal definition of a compact element and It apperead really messy to me ( mentioning directed subsets and ideals).
Then I looked, always on wikipedia, for some examples to gain some intuition but all they provided is:
-"The most basic example is obtained by considering the power set of some set, ordered by subset inclusion. Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"."
Which is not helpful at all.
-"The term "compact" is explained by considering the complete lattices of open sets of some topological space, also ordered by subset inclusion. Within this order, the compact elements are just the compact sets. Indeed, the condition for compactness in join-semilattices translates immediately to the corresponding definition."
Which is too complicated for me.
I hope someone can present an informal explanation, or some intuitve examples at least, that may clarify the concept.
lattice-orders
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
$endgroup$
I was reading the definition of an Algebraic Lattice: "An algebraic lattice is a complete lattice L, such that every element x of L is the supremum of the compact elements below x".
Then I looked for the formal definition of a compact element and It apperead really messy to me ( mentioning directed subsets and ideals).
Then I looked, always on wikipedia, for some examples to gain some intuition but all they provided is:
-"The most basic example is obtained by considering the power set of some set, ordered by subset inclusion. Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"."
Which is not helpful at all.
-"The term "compact" is explained by considering the complete lattices of open sets of some topological space, also ordered by subset inclusion. Within this order, the compact elements are just the compact sets. Indeed, the condition for compactness in join-semilattices translates immediately to the corresponding definition."
Which is too complicated for me.
I hope someone can present an informal explanation, or some intuitve examples at least, that may clarify the concept.
lattice-orders
lattice-orders
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 31 at 16:38
Gabriele ScarlattiGabriele Scarlatti
380212
380212
add a comment |
add a comment |
1 Answer
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$begingroup$
One of the definition's you're apparently referring to:
(an element $cin P$ is compact if ) for every directed subset $D$ of $P$, if $D$ has a supremum $sup D$ and $c ≤ sup D$ then $c ≤ d$ for some element $d$ of $D$.
This is straightforward enough: if you have a directed subset $D$ which also has a supremum, and $c$ is beneath the supremum, then it's beneath an element of $D$.
That is, $c$ never manages to sit between an entire directed set and its supremum.
In the following, I'm going to try to get some ideas out there to help with the intuition you were having problems with:
Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"." Which is not helpful at all.
"Compactness" as a rule is meant to generalize "smallness", and that's also something "finite" does. That is why they're mean to describe similar types of elements: small/finite/compact.
If you are handy with topological compactness, consider this: Let $(X,mathscr T)$, be a compact topological space. I claim that this is equivalent with $X$ being a compact element of the lattice of open subsets.
Given any open cover $mathscr O$ of $X$, we have a subset of $mathscr T$, one which is not necessarily a directed subset of $mathscr T$, but it certainly has a supremum, namely $cup mathscr O$. It is not hard to see how to make a directed set out of it: you can just take all finite unions of elements of $mathscr O$ and you then have a directed subset $mathscr D$ with the same supremum. Moreover, $Xleq sup(mathscr D)$.
Conversely, every directed subset of open sets whose supremeum contains $X$ is an open cover of $X$.
Given this connection between open covers of $X$ and directed subsets of $mathscr T$ whose supremeum contains $X$, it should be clear that $X$ is topologically compact iff it is a compact element of the lattice of open sets.
Of course, as you know, every finite subset of a topological space is compact. If you were given an open covering, you would just go through each point one by one and grab an open set containing that point, and when you were done you would have a finite collection of subsets of the open cover which covers your finite set. That's an illustration of "compact generalizes finite."
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
$endgroup$
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
add a comment |
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1 Answer
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$begingroup$
One of the definition's you're apparently referring to:
(an element $cin P$ is compact if ) for every directed subset $D$ of $P$, if $D$ has a supremum $sup D$ and $c ≤ sup D$ then $c ≤ d$ for some element $d$ of $D$.
This is straightforward enough: if you have a directed subset $D$ which also has a supremum, and $c$ is beneath the supremum, then it's beneath an element of $D$.
That is, $c$ never manages to sit between an entire directed set and its supremum.
In the following, I'm going to try to get some ideas out there to help with the intuition you were having problems with:
Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"." Which is not helpful at all.
"Compactness" as a rule is meant to generalize "smallness", and that's also something "finite" does. That is why they're mean to describe similar types of elements: small/finite/compact.
If you are handy with topological compactness, consider this: Let $(X,mathscr T)$, be a compact topological space. I claim that this is equivalent with $X$ being a compact element of the lattice of open subsets.
Given any open cover $mathscr O$ of $X$, we have a subset of $mathscr T$, one which is not necessarily a directed subset of $mathscr T$, but it certainly has a supremum, namely $cup mathscr O$. It is not hard to see how to make a directed set out of it: you can just take all finite unions of elements of $mathscr O$ and you then have a directed subset $mathscr D$ with the same supremum. Moreover, $Xleq sup(mathscr D)$.
Conversely, every directed subset of open sets whose supremeum contains $X$ is an open cover of $X$.
Given this connection between open covers of $X$ and directed subsets of $mathscr T$ whose supremeum contains $X$, it should be clear that $X$ is topologically compact iff it is a compact element of the lattice of open sets.
Of course, as you know, every finite subset of a topological space is compact. If you were given an open covering, you would just go through each point one by one and grab an open set containing that point, and when you were done you would have a finite collection of subsets of the open cover which covers your finite set. That's an illustration of "compact generalizes finite."
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
$endgroup$
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
add a comment |
$begingroup$
One of the definition's you're apparently referring to:
(an element $cin P$ is compact if ) for every directed subset $D$ of $P$, if $D$ has a supremum $sup D$ and $c ≤ sup D$ then $c ≤ d$ for some element $d$ of $D$.
This is straightforward enough: if you have a directed subset $D$ which also has a supremum, and $c$ is beneath the supremum, then it's beneath an element of $D$.
That is, $c$ never manages to sit between an entire directed set and its supremum.
In the following, I'm going to try to get some ideas out there to help with the intuition you were having problems with:
Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"." Which is not helpful at all.
"Compactness" as a rule is meant to generalize "smallness", and that's also something "finite" does. That is why they're mean to describe similar types of elements: small/finite/compact.
If you are handy with topological compactness, consider this: Let $(X,mathscr T)$, be a compact topological space. I claim that this is equivalent with $X$ being a compact element of the lattice of open subsets.
Given any open cover $mathscr O$ of $X$, we have a subset of $mathscr T$, one which is not necessarily a directed subset of $mathscr T$, but it certainly has a supremum, namely $cup mathscr O$. It is not hard to see how to make a directed set out of it: you can just take all finite unions of elements of $mathscr O$ and you then have a directed subset $mathscr D$ with the same supremum. Moreover, $Xleq sup(mathscr D)$.
Conversely, every directed subset of open sets whose supremeum contains $X$ is an open cover of $X$.
Given this connection between open covers of $X$ and directed subsets of $mathscr T$ whose supremeum contains $X$, it should be clear that $X$ is topologically compact iff it is a compact element of the lattice of open sets.
Of course, as you know, every finite subset of a topological space is compact. If you were given an open covering, you would just go through each point one by one and grab an open set containing that point, and when you were done you would have a finite collection of subsets of the open cover which covers your finite set. That's an illustration of "compact generalizes finite."
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
$endgroup$
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
add a comment |
$begingroup$
One of the definition's you're apparently referring to:
(an element $cin P$ is compact if ) for every directed subset $D$ of $P$, if $D$ has a supremum $sup D$ and $c ≤ sup D$ then $c ≤ d$ for some element $d$ of $D$.
This is straightforward enough: if you have a directed subset $D$ which also has a supremum, and $c$ is beneath the supremum, then it's beneath an element of $D$.
That is, $c$ never manages to sit between an entire directed set and its supremum.
In the following, I'm going to try to get some ideas out there to help with the intuition you were having problems with:
Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"." Which is not helpful at all.
"Compactness" as a rule is meant to generalize "smallness", and that's also something "finite" does. That is why they're mean to describe similar types of elements: small/finite/compact.
If you are handy with topological compactness, consider this: Let $(X,mathscr T)$, be a compact topological space. I claim that this is equivalent with $X$ being a compact element of the lattice of open subsets.
Given any open cover $mathscr O$ of $X$, we have a subset of $mathscr T$, one which is not necessarily a directed subset of $mathscr T$, but it certainly has a supremum, namely $cup mathscr O$. It is not hard to see how to make a directed set out of it: you can just take all finite unions of elements of $mathscr O$ and you then have a directed subset $mathscr D$ with the same supremum. Moreover, $Xleq sup(mathscr D)$.
Conversely, every directed subset of open sets whose supremeum contains $X$ is an open cover of $X$.
Given this connection between open covers of $X$ and directed subsets of $mathscr T$ whose supremeum contains $X$, it should be clear that $X$ is topologically compact iff it is a compact element of the lattice of open sets.
Of course, as you know, every finite subset of a topological space is compact. If you were given an open covering, you would just go through each point one by one and grab an open set containing that point, and when you were done you would have a finite collection of subsets of the open cover which covers your finite set. That's an illustration of "compact generalizes finite."
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
$endgroup$
One of the definition's you're apparently referring to:
(an element $cin P$ is compact if ) for every directed subset $D$ of $P$, if $D$ has a supremum $sup D$ and $c ≤ sup D$ then $c ≤ d$ for some element $d$ of $D$.
This is straightforward enough: if you have a directed subset $D$ which also has a supremum, and $c$ is beneath the supremum, then it's beneath an element of $D$.
That is, $c$ never manages to sit between an entire directed set and its supremum.
In the following, I'm going to try to get some ideas out there to help with the intuition you were having problems with:
Within this complete lattice, the compact elements are exactly the finite sets. This justifies the name "finite element"." Which is not helpful at all.
"Compactness" as a rule is meant to generalize "smallness", and that's also something "finite" does. That is why they're mean to describe similar types of elements: small/finite/compact.
If you are handy with topological compactness, consider this: Let $(X,mathscr T)$, be a compact topological space. I claim that this is equivalent with $X$ being a compact element of the lattice of open subsets.
Given any open cover $mathscr O$ of $X$, we have a subset of $mathscr T$, one which is not necessarily a directed subset of $mathscr T$, but it certainly has a supremum, namely $cup mathscr O$. It is not hard to see how to make a directed set out of it: you can just take all finite unions of elements of $mathscr O$ and you then have a directed subset $mathscr D$ with the same supremum. Moreover, $Xleq sup(mathscr D)$.
Conversely, every directed subset of open sets whose supremeum contains $X$ is an open cover of $X$.
Given this connection between open covers of $X$ and directed subsets of $mathscr T$ whose supremeum contains $X$, it should be clear that $X$ is topologically compact iff it is a compact element of the lattice of open sets.
Of course, as you know, every finite subset of a topological space is compact. If you were given an open covering, you would just go through each point one by one and grab an open set containing that point, and when you were done you would have a finite collection of subsets of the open cover which covers your finite set. That's an illustration of "compact generalizes finite."
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
edited Feb 3 at 18:02
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
answered Jan 31 at 17:05
rschwiebrschwieb
108k12103253
108k12103253
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
add a comment |
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
""Compactness" as a rule is meant to generalize "smallness"" I disagree with this to a certain extent: most obviously, subsets of compact sets need not be compact.
$endgroup$
– Noah Schweber
Jan 31 at 17:07
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
$begingroup$
@NoahSchweber It's not a perfect analogy, I agree. It is not completely monotonic. But the sentiment was good enough for Arhangelski'i to use in his courses.
$endgroup$
– rschwieb
Jan 31 at 17:09
add a comment |
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