# When are compact sets NOT Borel?

The theory of Radon measures relies a lot on the hypothesis that compact subsets of a topological space are Borel (i.e., in the $$\sigma$$-algebra generated by the open sets). This is an okay assumption in Hausdorff spaces (where the bulk of the introductory theory takes place) because all compact subsets are closed and hence Borel. However, this answer remarks that there ARE topological spaces containing non-Borel compact sets.

To flesh out the linked answer, let $$X$$ be any set containing more than one point. Then let $$\tau \subseteq \mathscr{P}(X)$$ be the trivial topology (e.g., $$\tau = \{\varnothing, X\}$$). Then the Borel $$\sigma$$-algebra is then $$\mathcal{B}_X = \tau = \{\varnothing,X\}$$. Any singleton $$\{x\} \subseteq X$$ is then clearly not Borel, yet it is compact since all singletons are so (in fact, all subsets in the trivial topology are trivially compact).

My question is this: What conditions must be placed on the topology to ensure compact subsets are or aren't Borel? We have the obvious extremes above. However, I am aware that "Hausdorff" is not a necessary condition. For an example from algebraic geometry, the Zariski topology on the prime spectrum of a ring is $$T_0$$ yet has the peculiar property that the basic open sets are compact. To be more specific, every subset of $$\text{Spec}(\mathbb{Z})$$ is Borel (every nonzero point is closed, and the whole space is countable, so zero is Borel too).

This gives us a few avenues of attack:

(1). Are all compact subsets of a $$T_0$$ topological space Borel? Or,

(2). Is there an example of a $$T_0$$ space having a non-Borel compact subset? (Note the trivial topology is $$T_0$$ iff it's also discrete.)

(3). Same questions, but for $$T_1$$ spaces instead.

Any insight is appreciated!

• Possibly relevant: Harley/McNulty, When is a point Borel?, Pacific Journal of Mathematics 80 #1 (January 1979), 151-157 (freely available on the internet). Nov 7, 2022 at 7:22
• For (3): if you look at $\Bbb A^n_{\Bbb C}$ or $\Bbb P^n_{\Bbb C}$ with the Zariski topology (in the classical sense), then these spaces are Noetherian and T1, but not all subsets are Borel. This is particularly easy for $n=1$, because the Borel $\sigma$-algebra of the cofinite topology is the algebra of countable-or-cocountable subsets, but it should be true more generally that if a subset $S$ is Borel in the Zariski topology, then there is an Euclidean meagre set $X$ with such that $S\subseteq X$ or $S^\complement\subseteq X$. Nov 7, 2022 at 8:46
• I believe we don't actually care about all compact sets here, but considering only the saturated compact sets should be enough. A set is saturated, iff it is equal to the intersection of all open sets containing it.
– Arno
Nov 7, 2022 at 13:10
• @Arno I think you are right, at least in the lens of "useful" Radon measure theory. But the point remains that historically, inner regularity requires Borel and compact, yet the Borel $\sigma$-algebra as it's normally defined need not include compact sets. Besides, I'm not sure if even saturated compact sets need be Borel--can't the intersections be uncountable? Nov 8, 2022 at 2:49
• @NickF Saturated sets are a notion that is only relevant in non-T1 spaces, since T1 spaces are exactly the topological spaces where all the subsets are saturated. The intersections can be uncountable, and in fact there is another term for subsets that are intersection of countable families of open sets: the $G_\delta$ sets, which are obviously Borel. Nov 8, 2022 at 15:38

The first thing that occurred to me is to change the topology (and hence the meaning of compact) without changing the Borel $$\sigma$$-algebra. This leads to an example showing that $$T_0$$ is not enough.

Consider the real line with the non-trivial open subsets of the form $$(-\infty,a)$$ (= the left order topology). This generates the same Borel $$\sigma$$-algebra as the usual topology, because we can write the intervals $$(-\infty,a]$$ as countable intersections of open sets etc.

Let $$S$$ be a non Lebesgue measurable subset of $$[0,1]$$ (a Vitali set). Then $$S\cup\{1\}$$ is still non-measurable, and hence non-Borel. But it becomes compact in the left order topology.

The construction of Vitali sets relies (as far as I know) on the axiom of choice. I don't know what happens if a different set theory is used.

• I wonder if this would work to get a $T_1$ counterexample--what would happen if you threw the sets $\mathbb{R} \setminus \{x\}$ into the left order topology to make it $T_1$? Nov 8, 2022 at 2:41
• Also, I think that the overall answer to this question is fundamentally tied to the axiom of choice anyway, because in fact full choice is needed to prove "compact in Hausdorff implies closed" (or alternatively we make the space second-countable Hausdorff) Nov 8, 2022 at 2:42
• @NickF ZF does prove that compact in Hausdorff is closed. Let $K$ be compact and let $x\notin K$. Assign to each $y\in K$ the set $\mathcal V_y=\{A\in \tau\,:\, y\in A\land \exists B\in \tau, x\in B\land A\cap B=\varnothing\}$. Then, consider $\mathcal V=\bigcup_{y\in K}\mathcal V_y$. This is a covering of $K$, because by T2 $y\in\bigcup\mathcal V_y$. So there is a finite subcovering $\{A_1,\cdots, A_m\}\subseteq\mathcal V$. By construction, for each $A_i$ there is some $B\in\tau$ such that $A_i\cap B=\varnothing$ and $x\in B$. Choose one $B_i$ for each ($m$ choices) and intersect. Nov 8, 2022 at 16:50
• @SassatelliGiulio Thanks for that! Maybe I was misremembering the proof. Nov 8, 2022 at 17:00