The following is a well-known theorem:

Let $X$ be a compact Hausdorff space. Then $x$ and $y$ belong to the same quasicomponent if and only if they belong to the same component of $X$.

In Munkre's Topology Exercise 37.4, he presents an argument using the Zorn's Lemma:

(a) Let $\mathscr A$ be the collection of all closed subspaces of $X$ such that $x$ and $y$ lie in the same quasicomponent of $A$. Let $\mathscr B$ be a collection of $\mathscr A$ that is simply ordered by proper inclusion. Show that the intersection of the elements of $\mathscr B$ belongs to $\mathscr A$.

(b) Show $\mathscr A$ has a minimal element $D$. (Zorn's Lemma used here)

(c) Show $D$ is connected.

However, I've found a proof here which apparently doesn't use the Axiom of Choice:

We just need to prove that every quasicomponent $Q$ is connected. Suppose that $Q = X_1 \cup X_2$, where $X_1, X_2$ are two disjoint closed subsets of the space $Q$. Then $X_1$ and $X_2$ are closed in $X$, since $Q$ is closed in $X$. By normality of compact Hausdorff spaces, there exist disjoint open subsets $U, V$ of $X$ containing $X_1, X_2$, respectively. Hence, we have $Q \subseteq U \cup V$ and, by compactness, there exist closed-open sets $F_1, \ldots, F_k$ such that

$$Q \subseteq \bigcap_{i=1}^k F_i \subseteq U \cup V.$$

$F = \bigcap_{i=1}^k F_i$ is clearly closed-open. Since $ \overline{U \cap F} \subseteq \overline{U} \cap F = \overline{U} \cap (U \cup V) \cap F = U \cap F$, the intersection $U \cap F$ is also closed-open. As $x \in U \cap F$, we have $Q \subseteq U \cap F$ and $X_2 \subseteq Q \subseteq U \cap F \subseteq U$. It follows that $X_2 \subseteq U \cap V = \emptyset$, which shows that the set $Q$ is connected.

Question :

Do we really need AC to prove component = quasicomponent in every compact Hausdorff space?

(As Asaf remarks, this reduces to whether or not proving the normality of compact Hausdorff space uses choice)

  • 1
    $\begingroup$ What's a quasicomponent? $\endgroup$
    – Asaf Karagila
    Commented Apr 14, 2019 at 8:15
  • $\begingroup$ @AsafKaragila Given $x,y \in X$, define $x \sim y$ if $X$ cannot be written as a disjoint union of open sets $U$ and $V$ containing $x$ and $y$, respectively. The equivalence classes are called the quasicomponents of $X$. $\endgroup$ Commented Apr 14, 2019 at 8:16
  • $\begingroup$ @AsafKaragila It is easy to see that the quasicomponent of $x$ is the intersection of all closen-and-open subsets of $X$ which contain $x$. $\endgroup$
    – Paul Frost
    Commented Apr 14, 2019 at 9:02
  • $\begingroup$ Okay, having spent more than 10 minutes awake, let me ask a followup. If we are intersecting closed subspaces of a compact space, either finitely many have a non-empty intersection, or the whole thing has a non-empty intersection. As for the second proof, why do you think that compact spaces are normal without choice? $\endgroup$
    – Asaf Karagila
    Commented Apr 14, 2019 at 9:17
  • 2
    $\begingroup$ @user87690 Although AC is used in the usual proof that compact Hausdorff spaces are normal, I don't see that it's really needed. Let me try to do the first part of the proof (the second should be similar), separating (in a Hausdorff space) a compact set $K$ from a point $p\notin K$ by disjoint neighborhoods. Start with the cover of $K$ by all the open sets that are disjoint from neighborhoods of $p$. Get a finite subcover $C$ and, for each $U\in C$ choose an open $V\ni p$ disjoint from $U$ (no AC as $C$ is finite). Then use the union of $C$ and the intersection of the chosen $V$'s. $\endgroup$ Commented Apr 14, 2019 at 23:03

1 Answer 1


Munkres' proof certainly uses AC. The alternative proof doesn't explicitly use AC, but as Asaf Karagila remarks in his comment, it may be hidden in the proof that compact Hausdorff spaces are normal.

Frankly speaking, I believe that most of us are adherents of ZFC, and I personally did not spend much time in questions concerning the use of AC. However, in this case I tried to find a proof without using AC. So let $X$ be a compact Hausdorff space.

1) $X$ is regular.

Let $x \in X$ and $B \subset X$ be closed such that $x \notin B$. For $y \ne x$ let us say that an open neighborhood $U$ of $y$ is of type $H$ (for Hausdorff) if there exists an open neighborhood $V$ of $x$ such that $U \cap V = \emptyset$. Clearly, each $y \ne x$ has such a neighborhood. Let $\mathfrak{U}(y)$ be the set of all open neighborhoods $U$ of $y$ of type $H$ and $\mathfrak{U} = \bigcup_{y \in B} \mathfrak{U}(y)$. This is a cover of $B$ by open sets in $X$. Since $B$ is closed in $X$, it is compact and there exist finitely many $U_i$ in $\mathfrak{U}$ such that $B \subset U^* = \bigcup_{i=1}^n U_i$. Now we can make finitely many choices to get open neigborhoods $V_i$ of $x$ such that $U_i \cap V_i = \emptyset$. Then $V^* = \bigcap_{i=1}^n V_i$ is a an open neighborhood of $x$ such that $U^* \cap V^* = \emptyset$.

As far as I can see this does not use AC. The "standard proof", however, is based on AC by choosing for each $y \ne x$ a pair of open neigborhoods $U_{y}$ of $y$ and $V_{y}$ of $x$ such that $U_{y} \cap V_{y} = \emptyset$.

2) $X$ is normal.

Let $A, B \subset X$ be closed such that $A \cap B = \emptyset$. For $y \notin B$ let us say that an open neighborhood $U$ of $y$ is of type $R$ (for regular) if there exists an open neighborhood $V$ of $B$ such that $U \cap V = \emptyset$. By 1) each $y \notin B$ has such a neighborhood. Adapting the proof of 1), we see that that $A, B$ have disjoint open neighborhoods.

I hope I didn't make a mistake in showing without AC that "compact Hausdorff $\Rightarrow$ normal". But in my opinion the definition of compactness resembles the spirit of AC. It allows to make a choice: For each open cover it assures the existence of a finite subcover, but it is entirely nebulous how this finite subcover can be found. Of course, all finite $X$ are compact, but to prove the compactness of an infinite space $X$ in many cases AC is needed. For example the compactness of infinite products of compact spaces is equivalent to AC.

  • $\begingroup$ From proofwiki, the Heine-Borel Theorem requires countable choice. $\endgroup$ Commented Apr 15, 2019 at 1:10
  • $\begingroup$ Concerning the soft question: I do not see an advantage, but perhaps that is a matter of taste. In my eyes Munkres' proof is less intuitive since it does not directly analyze the quasicomponents. $\endgroup$
    – Paul Frost
    Commented Apr 15, 2019 at 8:46
  • $\begingroup$ I see. That's why I deleted that comment. $\endgroup$ Commented Apr 15, 2019 at 8:46
  • $\begingroup$ Oops, proving the compactness of a closed and bounded set in $\mathbb R^n$ should not require any choice. The generalized Heine-Borel Theorem for metric spaces (A metric space is compact if and only if it is both complete and totally bounded.), however, requires countable choice. $\endgroup$ Commented Apr 15, 2019 at 9:07
  • $\begingroup$ You are certainly right. I deleted the part concerning $[0,1]$ from my answer. I also deleted my previous comment. By the way, here a nice survey concerning AC: Herrlich, Horst. "Choice principles in elementary topology and analysis." Commentationes Mathematicae Universitatis Carolinae 38.3 (1997): 545-552. $\endgroup$
    – Paul Frost
    Commented Apr 15, 2019 at 9:16

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