Is there a set that is both a sigma algebra, $\Sigma$, and a topology, $\tau$, but not a powerset, $\mathcal{P}$?

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    $\begingroup$ What do you think? $\endgroup$ – Asaf Karagila Sep 24 '12 at 19:01
  • $\begingroup$ I think it is if and only if. i.e., powerset $\iff$ the set is both a sigma algebra and a topology. So the answer to my question is no. $\endgroup$ – torrho Sep 24 '12 at 19:43

This can be easily done on large enough infinite sets, for example consider $$\mathcal A=\{A\subseteq\mathbb R\mid A\subseteq\mathbb N\text{ or }\mathbb R\setminus A\subseteq\mathbb N\}.$$

Let us see that this is a topology and a $\sigma$-algebra:

  1. Clearly $\mathbb R$ and $\varnothing$ are in this collection.
  2. It is closed under countable intersections, suppose $A_n$ is a collection of sets in $\cal A$. If the intersection is empty then we are done, otherwise
    • if there is at least one set which is a subset of $\mathbb N$ then obviously their intersection is a subset of $\mathbb N$ and therefore is in $\cal A$; otherwise
    • all sets come are subsets of $\mathbb R\setminus B$, where $B\subseteq\mathbb N$, then by DeMorgan this is again true.
  3. Closed under any sort of unions, take $\{A_i\mid i\in I\}$ any collection of sets from $\cal A$, and by similar (but dual) arguments to before the union is in $\cal A$.
  4. Closure under complement is trivial.

It is quite obvious that $\cal A\neq P(\Bbb R)$ since the rationals (for example) are not in $\cal A$.

(Note that this is essentially equivalent to two copies of $\cal P(\Bbb N)$ though, but it's not quite that.)

  • 1
    $\begingroup$ Isn't this simply $\mathcal A=\mathcal P(\mathbb N)\subset \mathcal P(\mathbb R)$? And how is $\mathbb R\in\mathcal A$? Or maybe I don't understand your definition right. But for me $\exists B\subset \mathbb N\colon A\cup B=\mathbb N\land A\cap B=\emptyset \lor A\subseteq \mathbb N$ is simply equivalent to $A\subseteq \mathbb N$. $\endgroup$ – Hagen von Eitzen Sep 24 '12 at 20:13
  • $\begingroup$ @Hagen $\varnothing$? $\endgroup$ – Asaf Karagila Sep 24 '12 at 20:28
  • $\begingroup$ Is this supposed to mean that the choice $B=\emptyset$ shows that $A=\mathbb R$ is an element of $\mathcal A$? $\emptyset\subseteq \mathbb N$ - check. $\mathbb R\cup \emptyset=\mathbb N$ - fail. $\endgroup$ – Hagen von Eitzen Sep 24 '12 at 20:31
  • $\begingroup$ @Hagen: I found the definition of $\mathcal{A}$ a bit hard to parse sensibly, but Asaf appears to mean the set of all $A\subseteq\Bbb R$ such that either $A\subseteq\Bbb N$, or $A\supseteq\Bbb R\setminus\Bbb N$; equivalently, $A\subseteq\Bbb N$ or $\Bbb R\setminus A\subseteq\Bbb N$. $\endgroup$ – Brian M. Scott Sep 24 '12 at 21:08
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    $\begingroup$ @Clive: There was a typo, it is now corrected. $\endgroup$ – Asaf Karagila Sep 25 '12 at 6:01

Yes. If $X$ is any set containing at least two elements then the collection $\{ X, \varnothing \}$ is both a $\sigma$-algebra and a topology on $X$ but is not the powerset.


If you add the Hausdorff condition, then you can achieve the full power set result, as I explain below. Meanwhile, in general, here is a sense in which every such algebra/topology is in fact a power set.

Theorem. Every family $\cal{A}$ of sets that is both a $\sigma$-algebra and a topology on a set $X$ (or even just an algebra and a topology) is isomorphic, as a partial order under $\subseteq$, to a power set order. Specifically, $\cal{A}$ is isomorphic to the power set of the collection of minimal nonempty elements of $\cal{A}$, which form a partition of $X$.

Proof. Suppose that $\cal{A}$ is closed under complement and is a topology on $X$. Thus, $\cal{A}$ is closed under complement and arbitrary unions, and hence by de Morgan it is closed under arbitrary intersection. For each $x\in X$, let $[x]$ be the intersection of all $A\in\cal{A}$ with $x\in A$. Thus, $[x]\in \cal{A}$, and these are simply the minimal nonempty elements of $\cal{A}$. These partition $X$, and form the equivalence classes of the relation $x\sim y$ iff $x$ and $y$ are in all the same elements of $\cal{A}$. Every element of $\cal{A}$ is a union of the various classes $[x]$, and furthermore, every union of classes $[x]$ is in $\cal{A}$. So the partial order $\langle \cal{A},{\subseteq}\rangle$ is isomorphic to the power set of the classes $\langle P(X/\sim),\subseteq\rangle$, since every element $A\in \cal{A}$ breaks into the union of pieces of the partition $A=\bigcup_{a\in A}[a]$. QED

Let us now add a separation requirement to the topology to achieve the full power set result.

Theorem. Every family $\cal{A}$ of sets that is both an algebra and a Hausdorff topology on $X$ is the full power set of $X$.

Proof. If the topology is Hausdorff, or merely even $T_1$, then the relation $x\sim y$ I mention above is just $x=y$, and so $[x]=\{x\}$. So all the singletons are in $\cal{A}$, which is closed under arbitrary unions. So ${\cal A}=P(X)$. QED


Hint: a topology on $X$ that contains all singletons $\{x\}$ for $x \in X$ is the power set. Try a very simple $\sigma$-algebra on a finite set.


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