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Let $(X,\mathscr{T})$ be a topological space, and define a new topology $\mathscr{T}'$ on $X$, such that the collection of closed sets $\mathscr{C}$ of the topology $\mathscr{T}$ form a sub-basis for $\mathscr{T}'$. Lets refer to $\mathscr{T}'$ as the 'complementary topology' for convenience.

In fact since the intersections of closed sets are again closed and since $X\in \mathscr{C}$, the sub-basis also covers the space, so the sub-basis $\mathscr{C}$ would be a basis for $\mathscr{T}'$.

I was curious as to the properties of this topology. An initial thought was that this may be the discrete topology. Taking for example the standard euclidean topology $(\mathbb{R},\mathscr{T}_{E})$, if we form the 'complementary topology', we have in our basis $\mathscr{C}_E$, that $\{x\}\in\mathscr{C}_E, \, \forall x\in\mathbb{R}$. Since open sets are formed from unions of these sets, any set is open so $\mathscr{T}_{E}'=\mathscr{T}_{D}$, the discrete topology.

This cannot be true in general since taking for example a three point set $\{1,2,3\}$, with the topology $\mathscr{T}=\{\emptyset, \{1\},\{1,2\},\{1,2,3\}\}$, the complementary topology is given by $\mathscr{T}'=\{\emptyset,\{3\},\{2,3\},\{1,2,3\}\}$, which is certainly not the discrete topology.

However in the second example we do have that $(\mathscr{T}')'=\mathscr{T}$.

So I am curious if there are conditions on the topology such that we have either $(\mathscr{T}')'=\mathscr{T}$ or $(\mathscr{T})'=\mathscr{T}_D$? Is there a simple example where neither hold?

Of course if the topology was discrete in the first place both would hold.

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    $\begingroup$ $\mathscr{T}' = \mathscr{T}_D$ if and only if $\mathscr{T}$ is $T_1$. $\endgroup$ – Daniel Fischer Sep 24 '15 at 11:58
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As the family $\mathscr C$ of $\mathscr T$-closed sets is closed under arbitrary intersections, it follows that they form a base for the topology $\mathscr T'$ on $X$.

This in turn means that $\mathscr T'$ consists of the unions of all subfamilies of $\mathscr C$.


As Daniel Fischer mentions in a comment, $\mathscr T' = \mathscr T_\text{D}$ iff $\mathscr T$ is a T1 topology.

  • Proof. Clearly $\mathscr T'$ is discrete iff each singleton $\{ x \}$ is $\mathscr T'$-open. Suppose that $x \in X$ is such that $\{ x \}$ is open. As $\mathscr T'$ is generated by $\mathscr C$ there must be an $F \in \mathscr C$ such that $x \in F \subseteq \{ x \}$, meaning that $\{ x \} \in \mathscr C$. That is, $\{ x \}$ is $\mathscr T$-closed.

To investigate when the equality $(\mathscr T')' = \mathscr T$ holds, note that the inclusion $\mathscr T \subseteq ( \mathscr T')'$ always holds, so we need only worry about the reverse inclusion $\mathscr T \supseteq ( \mathscr T')'$.

Note that if the family $\mathscr C$ of $\mathscr T$-closed sets is closed under arbitrary unions, then $\mathscr T' = \mathscr C$. It follows that in this case we will have $(\mathscr T')' = \mathscr T$.

  • Proof. In this case the family $\mathscr C'$ of $\mathscr T'$-closed sets is exactly $\mathscr T$, which is clearly closed under arbitrary unions, and so — using the above observation — $( \mathscr T')' = \mathscr C' = \mathscr T$.

Suppose now that $\mathscr C$ is not closed under arbitrary unions. Then there is a $\{ F_i : i \in I \} \subseteq \mathscr C$ such that $E = \bigcup_{i \in I} F_i \notin \mathscr C$. However $E$ must be $\mathscr T'$-open, and therefore it is $(\mathscr T')'$-closed. As we have demonstrated a $(\mathscr T')'$-closed set which is not $\mathscr T$-closed, it must be that $(\mathscr T')' \neq \mathscr T$.

So the equality $(\mathscr T')' = \mathscr T$ will hold exactly when $\mathscr C$ is closed under arbitrary intersections. This defines the class of Alexandroff spaces.


So if $\mathscr T$ is neither T1 nore Alexandroff, then $\mathscr T' \neq \mathscr T_\text{D}$ and $(\mathscr T')' \neq \mathscr T$.

A relatively simple example of this is the following. Let $X = \mathbb R \times \{ 0 , 1 \}$, and give it the topology $\mathscr{T}$ where the open sets are of the form $U \times \{ 0 , 1 \}$ where $U \subseteq \mathbb R$ is open in the usual Euclidean topology on $\mathbb R$.

  • As no singleton is closed, clearly $\mathscr T$ is not T1.
  • $\{ [\frac 1n , 1] \times \{0,1\} : n \geq 1 \}$ is a family of closed sets whose union $\bigcup_n ( [ \frac 1n , 1] \times \{0,1\} ) = ( 0 , 1 ] \times \{0,1\}$ is not closed.

We can actually compute the topologies $\mathscr T'$ and $( \mathscr T')'$ relatively easily.

  • $\mathscr T'$ is the family of all $A \times \{ 0,1 \}$ for $A \subseteq \mathbb R$.

  • $(\mathscr T')' = \mathscr T'$. (Note that every $\mathscr T'$-open set is also $\mathscr T'$-closed, and vice versa, and so the family $\mathscr C'$ of $\mathscr T'$-closed sets coincides with $\mathscr T'$. As this is closed under arbitrary unions, it follows from an above observation that $(\mathscr T')' = \mathscr C' = \mathscr T'$.)

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First if $(X,\mathscr{T})$ is a finite topology (i.e. $|\mathscr{T}|<\infty$) then any union of closed sets is a finite union of closed sets so that the set of closed sets is closed by union. Whence the topology $\mathscr{T}'$ you get by taking the topology generated by the closed sets is actually the set of closed sets. Whence if $(X,\mathscr{T})$ is a finite topology you always get $(\mathscr{T}')'=\mathscr{T}$. (this is a particular case of Alexandroff's space, see $major^4$'s answer).

Second, if $(X,\mathscr{T})$ is $T_1$ (by definition : all points are closed) then $\mathscr{T}'$ is $\mathscr{T}_D$ the discrete topology (this is Daniel Fischer's comment).

Hence, looking for a counter example, you need $(X,\mathscr{T})$ an infinite topology, not $T_1$. In particular $X$ should be infinite.

Take $X:=\mathbb{N}$ and define $\mathscr{T}$ by saying that $A$ is open if $A\subseteq \mathcal{P}$ (the prime numbers) and $\mathcal{P}-A$ is finite. This is a topology. Furthermore $2$ is not a closed point.

Then the set of open sets being not stable by infinite intersection (e.g. we can write the set of prime number equal to $1$ mod $4$ as an intersection of open sets) it follows that $\mathscr{T}'$ will contain a set which is not closed for $\mathscr{T}$ so that $(\mathscr{T}')'\neq\mathscr{T}$.

Now we need to show that $\mathscr{T}'$ is not the discrete topology. Since closed sets for $\mathscr{T}$ are included in $\mathcal{P}^c$, it follows that the topology generated by those sets cannot generate a set which contains $2$, in particular $\{2\}\notin \mathscr{T}'$ and the topology cannot be discrete.

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    $\begingroup$ Do you mean $\cal P\subseteq A$, or did you mean $\cal P\setminus A$ is finite? $\endgroup$ – Stefan Hamcke Sep 24 '15 at 12:51
  • $\begingroup$ @StefanHamcke, I meant the complementary inside the prime numbers is finite, thank you. $\endgroup$ – Clément Guérin Sep 24 '15 at 13:10

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