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Let $n \in \mathbb N$ be a fixed natural number.

Does there always exists a topological space $(X, \tau)$ such that $\vert \tau \vert=n$ ? I am interested in both cases when the cardinality of $X$ is finite and cardinality of $X$ in infinite?

Its is clear that if $n=2^k$ then we can easily construct required topological spaces in which cardinality of $X$ is $k$ and cardinality of $\tau$ is $2^k$. It is also clear that the above fact is true for numbers other than $2^k$. For example, Sierpinski Space. But I am unable to see the other cases?

P.S: The above question is motivated from this question of Measure Theory about cardinality of sigma algebra.

Community
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Dontknowanything
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2 Answers2

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Let $X = \{1, 2, \dots, n-1\}$ and let,

$$\tau = \{ O_0, O_1, \dots O_{n-1}\}$$

where, $$O_i = \{ 1, 2, \dots i\}.$$

Notice $|\tau| = n$ and it satisfies all the conditions of a topology,

  1. $O_0 = \emptyset$ and $O_{n-1} = X$, thus $\emptyset, X \in \tau$.
  2. If $O_i, O_j \in \tau$ with $i \leq j$, then $O_i \cup O_j = O_j \in \tau$.
  3. If $O_i, O_j \in \tau$ with $i \leq j$, then $O_i \cap O_j = O_i \in \tau$.

EDIT: If $X$ is infinite and if $\{1, 2, \dots, n-2\} \subset X$, then we can pick, $$\tau = \{O_0, O_1, \dots, O_{n-2}, X\}.$$

benguin
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  • I think you can use in a similar way ${1,\ldots,n}, n \in \mathbb{N}$ to define a topology with countable number of sets and $(-\infty,r), r \in \mathbb{R}$ to define a topology with an uncountable number of sets. – miracle173 Aug 25 '16 at 09:25
  • @benguin: Great! Thank you. – Dontknowanything Aug 25 '16 at 09:27
  • @Dontknowanything No problem! These topologies have a nice structure to them. I read a paper awhile back that uses these topologies (which they call chain topologies) to tackle the problem of counting the number of topologies on a finite set (https://cs.uwaterloo.ca/journals/JIS/VOL9/Benoumhani/benoumhani11.pdf). – benguin Aug 25 '16 at 09:33
  • @miracle173 Indeed! – benguin Aug 25 '16 at 09:34
  • @benguin Awesome..I'll surely have a look.Thank you again. – Dontknowanything Aug 25 '16 at 09:35
  • @benguin..Is there any construction if we require the topology to be $T_2$ ? – Dontknowanything Aug 25 '16 at 09:39
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    @Dontknowanything I know in the case that $X$ is finite and Hausdorff, then the only topology on $X$ is the discrete topology which contains $2^{|X|}$ elements so such a construction is not possible at least in the finite case. – benguin Aug 25 '16 at 10:03
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    Looking at some old class notes, the proof of the claim I made in my last comment is proved by using the fact that a finite set has a finite number of open sets so the same argument works for an infinite set and finite number of open sets. So there is no such construction if the topology is Hausdorff, regardless if $X$ is finite or infinite. – benguin Aug 25 '16 at 10:24
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I guess the answer is affirmative for infinite space $X$. Just pick $n-2$ nonempty subsets $\{U_i\}$ of $X$ such that $U_1\subsetneq U_2\subsetneq U_3\subsetneq\cdots\subsetneq U_{n-2}\subsetneq X$, and let $\mathcal{T}=\{\varnothing, U_1, U_2,\ldots,U_{n-2},X\}$. Here is a visualization for $n=5$:

enter image description here

Fei Li
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