Question: Let $X =$ {0, 1, 2, ...} and $T$ = { $\emptyset$, $X$, {0}, {1}, {0,2}, {0,1,3}, {0,1,2,4}, {0,1,2,3,5}, ... } $\cup$ {{0,1,2}, {0,1,2,3}, {0,1,2,3,4}, ... }. It is easily verified that $T$ forms a topology on $X$. Burdick has shown (Amer. Math. Monthly January 2006 p. 83) that the singleton {0} generates infinitely many distinct sets in this space under the operations of closure, complement and union. Q: Is this the only countably infinite topological space (up to homeomorphism) in which a finite seed set generates an infinite family under those three operations?

The answer seems to be yes. Burdick's and my solutions to the above Monthly problem both used this space. Even before seeing his solution I was led to think it's the only one that works.

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(edited April 9 to correct the topology description)

Answer: 
[corrections applied, per Ilya and Anton]

Consider the formal series
$$
f(x) = \sum\_{k=0}^\infty (\sum\_A A^k) x^k
$$
where $A$ runs through $p \times p$ matrices. It is equal to $\sum\_A (I - Ax)^{-1}$, a rational function with values in scalar matrices. Thus, for some $d$, if the first $d$ coefficients vanish all coefficients must vanish.

To determine $d$, we need to bound the degree of the numerator and denominator of $f(x)$. Note by Cramer's rule, each $(I - Ax)^{-1}$ is a matrix of degree $(p-1)$ polynomials divided by the degree $p$ determinant of $(I - Ax)$.

There are $p^p$ different denominators, i.e. polynomials of the form 
$$
\det(I - Ax) = c\_p x^p + c\_{p-1} x^{p-1} + \cdots + c\_1 x + 1
$$
The $c\_i$ are unconstrained, they appear at least once in case $A$ is the companion matrix with $-c\_p,-c\_{p-1},\ldots$ along the last column. So we can write the sum as
$$
\sum\_{c\_p,\ldots,c\_1} \frac{\text{something of degree $p-1$}}{c\_p x^p + \cdots + c\_1 x + 1} 
$$
There are $(p-1)$ linear denominators, $p (p-1)$ quadratic denominators, $p^2 (p-1)$ cubic denominators and so on, so the product of all the $c\_p x^p + \cdots + 1$ has degree
$$
d = (p-1) + 2p(p-1) + 3p^2(p-1) + \cdots + p p^{p-1}(p-1)
$$
i.e. $d = p^{p+1} - \frac{p^p - 1}{p - 1}$. So $f(x)$ is a polynomial of degree less than $d$ divided by a polynomial of degree $d$.

Ilya's answer shows that the first $p^{p+1} - p$ Taylor coefficients of $f(x)$ vanish, and when $p \geq 3$ this is larger than $d$ so all the remaining Taylor coefficients are vanish as well. When $p = 2$ Will points out the sums are not always zero, in fact for $p=2$
$$
f(x) = \frac{x^5}{(x+1)^2 (x^2 + x + 1)} I
$$
My answer was nonsense, sorry.