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Question: Is there a topological space with exactly 100 distinct open sets?

My attempt : I took topological space which is countable. Maximum number of open sets in a topological space is $2^n$ which is the set of all possible subsets.

Number of elements in this topological space can't be lesser than 7 since for 6 elements the maximum possible subsets is $2^6< 100$.

First I took 5 elements as singletons and a single 2 element set Y. ($\phi$ is already included in thees 32 subset)These 5 elements form 32 subsets.Hence along with Y and whole set we have 34 sets. On taking union of all possible sets with Y I got 65 sets.

I tried different possibilities like this. But this method leads to nowhere. I couldn't find a clear cut logic though I put efforts on this problem for many days. I searched many websites as well but I didn't get any idea. Seems to be like combinatorics problem but I don't know this subject in detail.

Is there some key concept behin this problem that I am missing. Please help me in proving or disproving this problem.

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    $\begingroup$ This is equivalent to "Is there a distributive lattice with 100 elements." A simple example would be $\{(a,b)\mid 1\leq a,b\leq 10\},$ where $(a,b)\leq(c,d)$ iff $a\leq c$ and $b\leq d.$ $\endgroup$ Nov 4, 2023 at 17:22
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    $\begingroup$ This is a fun question. Let's say the par for an integer N is the minimum number of elements necessary for a topology on those elements to have N open sets. Let's play some topological golf... $\endgroup$ Nov 4, 2023 at 21:26
  • $\begingroup$ As noted in an answer below, the par of 100 is 8. $\endgroup$ Nov 4, 2023 at 23:04
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    $\begingroup$ @ThomasAndrews: Indeed. t's worth noting that there are distributive lattices of every order - and in fact the number of distributive lattices on $n$ elements (up to isomorphism) grows exponentially with $n$. So there are a great many such topological spaces. $\endgroup$
    – Z. A. K.
    Nov 5, 2023 at 9:17
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    $\begingroup$ "Is there some key concept behin this problem that I am missing" Yes, the correspondence between partially ordered sets and Alexandrov spaces. (en.wikipedia.org/wiki/Alexandrov_topology) $\endgroup$
    – Pilcrow
    Nov 5, 2023 at 20:37

3 Answers 3

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Why not the topology $$\mathcal T=\big\{\varnothing ,\{1\},\{1,2\},\{1,2,3\},\{1,2,3,4\},...,\{1,2,...,99\}\big\},$$ on the set $E=\{1,...,99\}$ ?

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    $\begingroup$ Wow. I didn't think about such a possibility $\endgroup$ Nov 4, 2023 at 17:28
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    $\begingroup$ Or $T=\{a_1,\dots\,a_4,b_1,\dots,b_4,c,d\}$, and apply the above idea to the $a_i$, the $b_j$, $c$ and $d$. There remains the task of finding a topology with $p$ open sets, $p$ a prime, with $<p-1$ elements, or proving that $p-1$ is the minimum. $\endgroup$
    – Rosie F
    Nov 5, 2023 at 8:11
  • $\begingroup$ More generally: given a von Neumann ordinal $\alpha$, $(\alpha, \alpha + 1)$ is a topological space with exactly $|\alpha + 1|$ open sets. For $\alpha = 2$, this is the Sierpiński space. A function between such spaces $ f: (\alpha, \alpha+1) \to (\beta, \beta+1)$ is continuous iff it is (weakly) monotone. $\endgroup$ Nov 6, 2023 at 13:55
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Consider a set $X$ with $n$ elements, $x_1$ through $x_n$. Let $\tau$ be the topology consisting of the empty set, and all sets of the form $\{x_1,\ldots,x_k\}$, with $k=1,\ldots, n$.

This is a topology: it contains the empty set, the whole set, the intersection of any two open sets is open (hence so is any finite intersection), and the union of any family of open sets is open. The topology has exactly $n+1$ open sets.

You can generalize this by replacing each $x_i$ with subsets of your space, provided the sets are either pairwise disjoint, or form a chain under inclusion.

You can get topologies with any desired cardinal of open sets (except $0$, and $1$ if you want to omit the empty space), by taking an ordinal $\alpha$ and letting the open sets be all initial segments of $\alpha$.

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An example of a topological space with $8$ points and $100$ open sets.

You can easily verify that the topology on the set $X=\{1,2,3,4,5,6,7,8\}$ generated by the sets $$\{1\},\{2\},\{1,2,3\},\ \{4\},\{5\},\{4,5,6\},\ \{7\},\ \{8\}$$ has exactly $100$ open sets. Namely, the open sets are the sets of the form $U_3\cup U_6\cup U_7\cup U_8$ where $$U_3\in\{\varnothing,\{1\},\{2\},\{1,2\},\{1,2,3\}\},$$ $$U_6\in\{\varnothing,\{4\},\{5\},\{4,5\},\{4,5,6\}\},$$ $$U_7\in\{\varnothing,\{7\}\},$$ $$U_8\in\{\varnothing,\{8\}\}.$$

P.S. As pointed out in a comment by @MW, there is no example with $7$ points. A non-discrete space with $n$ points has at least $2^{n-2}$ non-open sets since, if $\{p\}$ is not open, then for each partition $X\setminus\{p\}=S_1\cup S_2$ ($S_1\cap S_2=\varnothing$), at least one of the sets $S_i\cup\{p\}$ is not open. Thus a non-discrete space with $7$ points has at most $2^7-2^5=96$ open sets.

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    $\begingroup$ Comment on the original example: Note that $100=(4+1)(4+1)(1+1)(1+1)$ and this set is $\{p^1,...,p^4,q^1,...,q^4,r^1,s^1\}$ for primes $p,q,r,s$. $\endgroup$ Nov 4, 2023 at 21:23
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    $\begingroup$ @StevenClontz In other words, the poset was the disjoint union of $4$ chains of respective lengths $4,4,1,1$. I figured it would take less writing to describe it in terms of divisibility. $\endgroup$
    – bof
    Nov 4, 2023 at 21:26
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    $\begingroup$ Worth noting this is the best you can do, since if there are only $7$ points, and some singleton (say $\{1\}$) is not open, then there are at least $32$ non-open sets (since for $S\subseteq \{2,3,4,5,6,7\}:=T$ either $S\cup\{1\}$ or $T\backslash S \cup \{1\}$ must be nonopen). $\endgroup$
    – M W
    Nov 4, 2023 at 21:40
  • $\begingroup$ @MW Thanks! Silly me, I've been trying to find an example with $7$ points. I will edit your comment into my answer. $\endgroup$
    – bof
    Nov 4, 2023 at 22:15

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