Re: -- sequences of integers, closed under some operations

On Wed, 7 Jan 2009, quasi wrote:

Fix a positive integer n.

Let S be the set of all sequences of n integers.

Call a subset A of S "closed" if

(1) if a is in A, then p(a) is in A, where p(a) is any permutation of a.

(2) if a is in A, then for any f in Z[x], the sequence

f(a[1]), f(a[2]), ..., f(a[n])

is in A.

How am I to understand S? The best possible interpetation that seems to fit the context is S = Z^n. That you are considering finite sequences.

Remark: It's easily seen that the closed sets, as defined above, yield
a topology on S.

An Alexadroff space, that is space in which infininte intersections of open sets are open. Thus your spaces they are either discrete or not T1.

Do you intende the closed sets to be the topology, ie the open sets of the space, or do you want them to be the collection of closed sets of the space? I'm naturally assuming the former.

Question: Does there exist a finite subset of S whose closure is S?

The subspace of constant sequences is discrete.
If A is a subset of S and k a constant sequence of S,
then k in A iff k in permutation closure of A.

Since there are infinitely many constant sequences, the answer is no.

.

Relevant Pages

  • Some Miscellaneous Topology Problems
    ... REALS ARE A QUOTIENT SPACE OF CONVERGENT SEQUENCES ... RADIALLY OPEN TOPOLOGY ... REGULARLY OPEN SETS ... Ryszard Engelking, GENERAL TOPOLOGY, Monografie ...
    (sci.math)
  • Re: -- sequences of integers, closed under some operations
    ... Let S be the set of all sequences of n integers. ... Thus your spaces they are either discrete or not T1. ... The subspace of constant sequences is discrete. ... then k in A iff k in permutation closure of A. ...
    (sci.math)