I am interested in the following result:

Theorem. A locally compact Hausdorff topological space $X$ without isolated points has at least cardinality $\mathfrak{c}$.

To prove it, one can find two closed sets $P_0$ and $P_1$ without isolated point and such that $P_0 \cap P_1= \emptyset$; then one find $P_{00} \subset P_0$ and $P_{01} \subset P_{0}$ (resp. $P_{10} \subset P_1$ and $P_{11} \subset P_1$) with the same properties, and so on.

For every sequence $a=(a_n) \in \{0,1\}^{\mathbb{N}}$, there exists $f(a) \in \bigcap\limits_{k \geq 1} P_{a_0...a_k}$ (thanks to compactness). Thus, $f$ defines a one-to-one function from $\{0,1\}^{\mathbb{N}}$ to $X$, hence $|X| \geq \mathfrak{c}$.

But to chose an element in $\bigcap\limits_{k \geq 1} P_{a_0...a_k}$, we need the axiom of choice (not needed in the metric case beacuse the intersection is reduced to a singleton).

Is there another argument to avoid it? Is there any other known proof of this result?

  • 1
    $\begingroup$ Your construction already used dependent choice in the construction of the sets $P_{a_0\dots a_k}$ (a good indicator of the use of dependent choice is the appearance of "and so on" at some point in the argument). So even if you should manage to avoid choosing an element in each $\bigcap_{k \geq 1} P_{a_0\dots a_k}$ by some clever trick, your proof will not be without any use of (some form of) the axiom of choice. $\endgroup$ – Martin Jan 18 '13 at 11:58
  • $\begingroup$ How do you define locally compact? $\endgroup$ – Asaf Karagila Jan 18 '13 at 21:22
  • $\begingroup$ @AsafKaragila: For me, a space is locally compact if each point has a fundamental system of compact neighborhoods. $\endgroup$ – Seirios Jan 18 '13 at 21:39

After a lot of work, and practically writing most of the proofs and constructions myself for the past week, I figured "Hey, that can't be new", and like many other good things related to the axiom of choice, it turned out that the answer was known in the 1960's.

In 1962 Läuchli [4] constructed a topological space which is Hausdorff compact and strongly connected (every real valued function is constant). This space was used as an example for a failure of Urysohn's lemma without the axiom of choice. The construction appears in [3, Ch. 10] as exercise 6.

Of course it still remains to show it has no isolated points, but if $x$ was an isolated point then $X\setminus\{x\}$ was clopen, and $\{x\}$ was clopen. The characteristics function of $\{x\}$ is a continuous real valued function, and this is a contradiction to the strong connectedness.

Two decades later Brunner extended this work [1] and showed that in fact in Mostowski's ordered model [3, Ch. 4] every closed intervals of atoms makes a Läuchli continuum. The atoms in that model are Dedekind-finite, and so every closed interval make a Dedekind-finite Läuchli continuum, which is what we wanted. Brunner also shows that if a continuum has a Dedekind-finite power set then it is a Läuchli continuum.

Another twenty years or so pass, and Creed & Truss [2] approach a closely related question via model theoretical goggles as well, as the atoms in the Mostowski model are an $o$-amorphous set.


  1. Brunner, Norbert Geordnete Läuchli Kontinuen. (German) [Ordered Läuchli continua] Fund. Math. 117 (1983), no. 1, 67–73. (English summary: MR0712214)

  2. Creed, P.; Truss, J. K. On $o$-amorphous sets. Ann. Pure Appl. Logic 101 (2000), no. 2-3, 185–226.

  3. Jech, Thomas J. The axiom of choice. Studies in Logic and the Foundations of Mathematics, Vol. 75. North-Holland Publishing Co., Amsterdam-London; Amercan Elsevier Publishing Co., Inc., New York, 1973. xi+202 pp.

  4. Läuchli, H. Auswahlaxiom in der Algebra. (German) Comment. Math. Helv. 37 1962/1963 1–18. (English summary: MR143705)


I'd like to thank Jonas Teuwen for helping me in translating a part of Brunner's paper.

  • $\begingroup$ It took me good ten days to figure out exactly what is the construction in order to realize what papers I need to search for. I will also admit a total lack of surprise when I found out Läuchli did it, and that Truss had some involvement in the topic. $\endgroup$ – Asaf Karagila Jan 29 '13 at 21:09

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