Mixing
How many non-homeomorphic topological structures are there on a finite set of order $n$?
$begingroup$
Given a finite set of order $n$, how many different (that is, non-homeomorphic) topological structures are there on this set?
It is a question about topology but my feeling is that it is essentially a question in combinatorics.
combinatorics general-topology
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
$endgroup$
add a comment |
$begingroup$
Given a finite set of order $n$, how many different (that is, non-homeomorphic) topological structures are there on this set?
It is a question about topology but my feeling is that it is essentially a question in combinatorics.
combinatorics general-topology
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
$endgroup$
3
$begingroup$
This is a very hard problem. Some efforts are here: math.uchicago.edu/~may/FINITE/REUNotes2010/FiniteSpaces.pdf
$endgroup$
– Randall
Jan 28 at 1:50
1
$begingroup$
For low values of n: oeis.org/A000798
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– AHusain
Jan 28 at 1:58
$begingroup$
See the references in the OEIS sequence A001930 which is the number of non-homeomorphic (or unlabeled) topologies. A000798 is the total number of different topologies.
$endgroup$
– Somos
Jan 28 at 2:42
$begingroup$
This may be a duplicate. Also see the discussion and links available at math.stackexchange.com/questions/1582662/…
$endgroup$
– Ben
Jan 28 at 3:15
$begingroup$
You will find information about the combinatorial-language version of the problem at that question - e.g. finite spaces are equivalent to a certain class of posets.
$endgroup$
– Ben
Jan 28 at 3:18
add a comment |
$begingroup$
Given a finite set of order $n$, how many different (that is, non-homeomorphic) topological structures are there on this set?
It is a question about topology but my feeling is that it is essentially a question in combinatorics.
combinatorics general-topology
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
$endgroup$
Given a finite set of order $n$, how many different (that is, non-homeomorphic) topological structures are there on this set?
It is a question about topology but my feeling is that it is essentially a question in combinatorics.
combinatorics general-topology
combinatorics general-topology
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
edited Jan 28 at 1:51
Zuriel
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
asked Jan 28 at 1:43
ZurielZuriel
1,8881228
1,8881228
3
$begingroup$
This is a very hard problem. Some efforts are here: math.uchicago.edu/~may/FINITE/REUNotes2010/FiniteSpaces.pdf
$endgroup$
– Randall
Jan 28 at 1:50
1
$begingroup$
For low values of n: oeis.org/A000798
$endgroup$
– AHusain
Jan 28 at 1:58
$begingroup$
See the references in the OEIS sequence A001930 which is the number of non-homeomorphic (or unlabeled) topologies. A000798 is the total number of different topologies.
$endgroup$
– Somos
Jan 28 at 2:42
$begingroup$
This may be a duplicate. Also see the discussion and links available at math.stackexchange.com/questions/1582662/…
$endgroup$
– Ben
Jan 28 at 3:15
$begingroup$
You will find information about the combinatorial-language version of the problem at that question - e.g. finite spaces are equivalent to a certain class of posets.
$endgroup$
– Ben
Jan 28 at 3:18
add a comment |
3
$begingroup$
This is a very hard problem. Some efforts are here: math.uchicago.edu/~may/FINITE/REUNotes2010/FiniteSpaces.pdf
$endgroup$
– Randall
Jan 28 at 1:50
1
$begingroup$
For low values of n: oeis.org/A000798
$endgroup$
– AHusain
Jan 28 at 1:58
$begingroup$
See the references in the OEIS sequence A001930 which is the number of non-homeomorphic (or unlabeled) topologies. A000798 is the total number of different topologies.
$endgroup$
– Somos
Jan 28 at 2:42
$begingroup$
This may be a duplicate. Also see the discussion and links available at math.stackexchange.com/questions/1582662/…
$endgroup$
– Ben
Jan 28 at 3:15
$begingroup$
You will find information about the combinatorial-language version of the problem at that question - e.g. finite spaces are equivalent to a certain class of posets.
$endgroup$
– Ben
Jan 28 at 3:18
3
3
$begingroup$
This is a very hard problem. Some efforts are here: math.uchicago.edu/~may/FINITE/REUNotes2010/FiniteSpaces.pdf
$endgroup$
– Randall
Jan 28 at 1:50
$begingroup$
This is a very hard problem. Some efforts are here: math.uchicago.edu/~may/FINITE/REUNotes2010/FiniteSpaces.pdf
$endgroup$
– Randall
Jan 28 at 1:50
1
1
$begingroup$
For low values of n: oeis.org/A000798
$endgroup$
– AHusain
Jan 28 at 1:58
$begingroup$
For low values of n: oeis.org/A000798
$endgroup$
– AHusain
Jan 28 at 1:58
$begingroup$
See the references in the OEIS sequence A001930 which is the number of non-homeomorphic (or unlabeled) topologies. A000798 is the total number of different topologies.
$endgroup$
– Somos
Jan 28 at 2:42
$begingroup$
See the references in the OEIS sequence A001930 which is the number of non-homeomorphic (or unlabeled) topologies. A000798 is the total number of different topologies.
$endgroup$
– Somos
Jan 28 at 2:42
$begingroup$
This may be a duplicate. Also see the discussion and links available at math.stackexchange.com/questions/1582662/…
$endgroup$
– Ben
Jan 28 at 3:15
$begingroup$
This may be a duplicate. Also see the discussion and links available at math.stackexchange.com/questions/1582662/…
$endgroup$
– Ben
Jan 28 at 3:15
$begingroup$
You will find information about the combinatorial-language version of the problem at that question - e.g. finite spaces are equivalent to a certain class of posets.
$endgroup$
– Ben
Jan 28 at 3:18
$begingroup$
You will find information about the combinatorial-language version of the problem at that question - e.g. finite spaces are equivalent to a certain class of posets.
$endgroup$
– Ben
Jan 28 at 3:18
add a comment |
1 Answer
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A topology $tau$ on the set $X$ is (generated by) a set of subsets of $X$. So $tauin P(P(X))$.
Let $mid Xmid=n$. Then $mid P(P(X))mid =2^{{2^n}}$. This gives (at least) an upper bound on the number.
Since the subset must contain the $emptyset$ and the whole set X, the inequality is strict.
Because of this, we should be able to (marginally, to use @Arturo Magidin's description ) improve it to $2^{2^n-2}=frac{2^{2^n}}4$.
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
2
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
add a comment |
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$begingroup$
A topology $tau$ on the set $X$ is (generated by) a set of subsets of $X$. So $tauin P(P(X))$.
Let $mid Xmid=n$. Then $mid P(P(X))mid =2^{{2^n}}$. This gives (at least) an upper bound on the number.
Since the subset must contain the $emptyset$ and the whole set X, the inequality is strict.
Because of this, we should be able to (marginally, to use @Arturo Magidin's description ) improve it to $2^{2^n-2}=frac{2^{2^n}}4$.
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
2
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
add a comment |
$begingroup$
A topology $tau$ on the set $X$ is (generated by) a set of subsets of $X$. So $tauin P(P(X))$.
Let $mid Xmid=n$. Then $mid P(P(X))mid =2^{{2^n}}$. This gives (at least) an upper bound on the number.
Since the subset must contain the $emptyset$ and the whole set X, the inequality is strict.
Because of this, we should be able to (marginally, to use @Arturo Magidin's description ) improve it to $2^{2^n-2}=frac{2^{2^n}}4$.
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
2
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
add a comment |
$begingroup$
A topology $tau$ on the set $X$ is (generated by) a set of subsets of $X$. So $tauin P(P(X))$.
Let $mid Xmid=n$. Then $mid P(P(X))mid =2^{{2^n}}$. This gives (at least) an upper bound on the number.
Since the subset must contain the $emptyset$ and the whole set X, the inequality is strict.
Because of this, we should be able to (marginally, to use @Arturo Magidin's description ) improve it to $2^{2^n-2}=frac{2^{2^n}}4$.
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
$endgroup$
A topology $tau$ on the set $X$ is (generated by) a set of subsets of $X$. So $tauin P(P(X))$.
Let $mid Xmid=n$. Then $mid P(P(X))mid =2^{{2^n}}$. This gives (at least) an upper bound on the number.
Since the subset must contain the $emptyset$ and the whole set X, the inequality is strict.
Because of this, we should be able to (marginally, to use @Arturo Magidin's description ) improve it to $2^{2^n-2}=frac{2^{2^n}}4$.
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
edited Jan 28 at 2:36
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
answered Jan 28 at 2:05
Chris CusterChris Custer
14.2k3827
14.2k3827
2
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
add a comment |
2
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
2
2
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
This is rather naive... you can improve it marginally without too much work by noting that whatever $tau$ is, it must include both $varnothing$ and the whole set.
$endgroup$
– Arturo Magidin
Jan 28 at 2:07
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@ArturoMagidin not to mention the action of the group of self-bijections of $n$, since we're counting homeomorphism classes. Well, on second thought, I guess the orbits of this action are probably too complicated to easily lower the bound.
$endgroup$
– Kevin Carlson
Jan 28 at 2:15
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
$begingroup$
@KevinCarlson That strikes me as a little harder (and not nearly as mindless as mine), since the orbits are not all the same size...
$endgroup$
– Arturo Magidin
Jan 28 at 2:17
add a comment |
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3
$begingroup$
This is a very hard problem. Some efforts are here: math.uchicago.edu/~may/FINITE/REUNotes2010/FiniteSpaces.pdf
$endgroup$
– Randall
Jan 28 at 1:50
1
$begingroup$
For low values of n: oeis.org/A000798
$endgroup$
– AHusain
Jan 28 at 1:58
$begingroup$
See the references in the OEIS sequence A001930 which is the number of non-homeomorphic (or unlabeled) topologies. A000798 is the total number of different topologies.
$endgroup$
– Somos
Jan 28 at 2:42
$begingroup$
This may be a duplicate. Also see the discussion and links available at math.stackexchange.com/questions/1582662/…
$endgroup$
– Ben
Jan 28 at 3:15
$begingroup$
You will find information about the combinatorial-language version of the problem at that question - e.g. finite spaces are equivalent to a certain class of posets.
$endgroup$
– Ben
Jan 28 at 3:18