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My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other onethis other one. If so, please can you explain why?
PPS: I posted this question also on Math.SE, but no one answered.

My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other one. If so, please can you explain why?
PPS: I posted this question also on Math.SE, but no one answered.

My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other one. If so, please can you explain why?
PPS: I posted this question also on Math.SE, but no one answered.

replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Source Link

My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other one. If so, please can you explain why?
PPS: I posted this question also on Math.SEon Math.SE, but no one answered.

My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other one. If so, please can you explain why?
PPS: I posted this question also on Math.SE, but no one answered.

My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other one. If so, please can you explain why?
PPS: I posted this question also on Math.SE, but no one answered.

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dadexix86
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Simplicial complices on unlabelled vertices

My question is about (abstract) simplicial complices.
In particular, how many are they if I consider $n$ unlabelled vertices?

For example, if $n=4$, the two complices $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\} $$ and $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\} $$ are the same, but not $$ \{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\} $$ (since the last two sides of this one intersect in one vertex).

If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).
They are:

  • dim=2 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
  • dim=1 $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$ $$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
  • dim 0 $$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$

Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.

Thank you in advance, Davide

PS: I am not sure whether or not this question is related to this other one. If so, please can you explain why?
PPS: I posted this question also on Math.SE, but no one answered.