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It is known that there are only 14 reasonable tensor norms in $Ban$. On the other hand it is well known fact for topologists that one can obtain only 14 different sets from a given set applying closure and complementation operations.

It seems like a pure coincidence but... both constructions starts with the single object (a set/ the injective tensor norm), then apply only predescribed operations (closure and complement/ transpose, dual and right projective associate) and get only 14 different objects in the end.

Is this a simple coincidence or there is something deep behind the scenes? Maybe someone knows other examples?

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  • $\begingroup$ Maybe, it's just the idempotence of certain compositions? $\endgroup$
    – Alex Degtyarev
    Commented Jan 28, 2014 at 11:00
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    $\begingroup$ Take a look at the relations satisfied by the operators. I don't know about the Grothendieck example, but the topology operators satisfy clcl=cl, ccc=c, cccl=cl and some other equational relationships which I may bother to work out later. Since composition is (usually) associative, in each case you are looking at a (something resembling a) two-generated semigroup which satisfies some relations. If the relations are the same, maybe you have a connection; if they aren't maybe you don't. Gerhard "Or Maybe It Lies Deeper" Paseman, 2014.01.28 $\endgroup$
    – Gerhard Paseman
    Commented Jan 28, 2014 at 17:30
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    $\begingroup$ In particular, see if you can find an isomorphism between the two semigroups of operators, even if you can't tell from the relations of the generators. That isomorphism (if it exists) would give you something. Another possibility is that both of them are non-isomorphic subsemigroups of a larger semigroup, which if you found it might reveal more structure of larger classes of objects. Regarding your example, it suggests to me the poset (size 19?) formed by class operators H, S, and P for universal algebraic classes. Gerhard "Promoting The General Algebraic Approach" Paseman, 2014.01.28 $\endgroup$
    – Gerhard Paseman
    Commented Jan 28, 2014 at 17:37
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    $\begingroup$ If you do decide to pursue this and write it up, you might find David Sherman's article on variations of the Kuratowski 14-set problem useful. He references a recent paper of Gardner and Jackson which gives other variations and mentions the H,S,P, P_s example of Pigozzi that I mentioned above. Without knowing anything technical about tensor norms, my feeling is that you are dealing with a different semigroup of operators, which might tie in with one of the (partially ordered) semigroups studied in these papers. Gerhard "Ask Me About System Design" Paseman, 2014.01.28 $\endgroup$
    – Gerhard Paseman
    Commented Jan 28, 2014 at 18:42
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    $\begingroup$ And Bhargava showed that there were a total of 14 composition laws Gauss could have chosen for binary quadratic forms. $\endgroup$
    – Will Jagy
    Commented Jan 29, 2014 at 6:08

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I asked David Sherman this question and this was his response:

In the Kuratowski setup, if you omit complementation, you get 7 elements. Adding complementation gives you a disjoint upside-down copy.

In the Grothendieck setup, if you omit duality, you get 8 elements (two five-element chains with common inf and sup, like an O). Adding duality gives you an upside-down copy, but it is not disjoint -- two elements agree between the copies. ''

So, it seems that the answer is NO.

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