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Let $X$ be finite set, Find all topology on finite set with this condition:

  • For every subset $A \subset X$, either $A$ is open or $A$ is closed.

I find one of them:

$$T=\{ \emptyset, X, \{1\}, \{3\},...,\{n\},\{1,3\},\{1,4\},...\{1,n\},\{1,3,4\},...,\{1,3,4,...,n\}\} = \{A \subset X \bigm| A=X \, \text{or} \, 2 \not\in A \} $$ It is clear that $T$ is topology and $T \neq P(X)$. Note that discrete topology is a trivial answer. So find the number of all topology with above condition.

Lee Mosher
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amir bahadory
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  • Note that every subset of X must be open or close , if we say every subset of X is open and close then answer is only discrete topology. – amir bahadory Dec 29 '15 at 21:42
  • Actually the $T$ you've written is the discrete topology, though written incompletely since you have written all singletons as opens sets. – charlestoncrabb Dec 29 '15 at 21:50
  • No in T ${2}$ is not open – amir bahadory Dec 29 '15 at 22:03
  • @amirbahadory: Your example contains ${3}$ and ${4}$ but not ${3,4}$, so it is not a topology. – Jendrik Stelzner Dec 29 '15 at 22:12
  • No in my example T contain all 2_union of Singleton, and all 3-union of Singleton , and etc, (note that 2-union is all ${a,b}$ that a,b aren't 2, and 3-union is all ${a,b,c}$ that a,b,c aren't 2, and ...} – amir bahadory Dec 29 '15 at 22:29
  • I have edited your question for grammar, punctuation, etc., and to clarify the definition. Let me know if this is not OK. – Lee Mosher Dec 29 '15 at 22:34
  • @LeeMosher You replaced “open or close” with “either […] open or […] closed”, changing the problem itself. In particular the discrete topology, which was mentioned as a trivial answer, is not a solution to the new problem. – Jendrik Stelzner Dec 29 '15 at 22:45
  • @JendrikStelzner: The discrete topology is the one for which every subset $A$ is open, and so certainly it is true that every subset $A$ either $A$ is open or $A$ is closed. – Lee Mosher Dec 29 '15 at 22:51
  • @LeeMosher: Doesn’t “either […] or […]” usually mean exclusive disjunction? If you just mean (inclusive) disjunction why use „either“? – Jendrik Stelzner Dec 29 '15 at 22:58
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    @JendrikStelzner: "either" is used in English to mark the beginning of a disjunction for emphasis. "Either" does not signify exclusive disjunction. – Rob Arthan Dec 29 '15 at 23:07
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    @Jendrik: English either ... or is not quite equivalent to German entweder ... oder; it can (and often does) mean exclusive OR, but it need not. – Brian M. Scott Dec 29 '15 at 23:40

1 Answers1

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A topology on a finite set $X$ is the same as a preorder: given a preorder $\leq$, the collection of all sets $U$ such that $x\in U$ and $y\geq x$ implies $y\in U$ is a topology, and every topology comes from a unique preorder in this way.

The condition that every singleton is either open or closed means that $\leq$ is a partial order (i.e., $x\leq y$ and $y\leq x$ implies $x=y$) and every element of $X$ is either a minimal element or a maximal element. Now suppose $x$ is a maximal element which is not minimal and $y$ is a minimal element that is not maximal. If $\{x,y\}$ is open, then no element other than $x$ can be $>y$, and since $y$ is not maximal, we must have $x>y$. Similarly, if $\{x,y\}$ is closed, then $y<x$ and $y$ is the only element that is $<x$. So we must have $x>y$ and either $x$ is the only element $>y$ or $y$ is the only element $<x$.

It now follows easily that there must either be at most one maximal element that is not minimal or at most one minimal element which is not maximal. Suppose $x$ is the unique maximal element that is not minimal. Then every minimal element that is not maximal must be $<x$, and this completely determines the order. In detail, we can split $X=A\cup B\cup\{x\}$, where $A$ is the set of points which are both maximal and minimal, and $B$ is the set of points which are minimal but not maximal, and then the order relation on $X$ is defined by $s<t$ iff $t=x$ and $s\in B$. Dually, if $X$ has a unique minimal element $y$ that is not maximal, we can split $X=A\cup C\cup\{y\}$, and $X$ is ordered by $s<t$ iff $s=y$ and $t\in C$. Finally, if $X$ has no minimal element that is not maximal or maximal element that is not minimal, then $X$ has the trivial order (i.e., any two distinct elements of $X$ are incomparable).

You can check that in all three cases, every subset of $X$ is either open or closed. In the case where $X=A\cup B\cup\{x\}$, a set is open if it contains $x$, and closed if it does not contain $x$. In the case where $X=A\cup C\cup\{y\}$, a set is closed if it contains $y$, and open if it does not contain $y$. In the case that $X$ has the trivial order, every set is open and closed.

Eric Wofsey
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