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In continuum theory we frequently use the fact that two points in a continuum are contained in an irreducible subcontinuum.

A continuum $X$ is a compact connected metric space. A subcontinuum $K\subset X$ is irreducible between $x,y\in X$ provided that no proper subcontinuum of $K$ contains both $x$ and $y$.

Theorem. If $X$ is a continuum and $x,y\in X$, then there is an irreducible continuum $K\subset X$ with $x,y\in K$.

This is an easy consequence of Zorn's lemma, since the intersection of any decreasing chain of continua is a continuum.

My question is about what choice principle is actually needed here. Can the Theorem be proved using only Dependent Choice, or Countable Choice? My reason for asking is that I tend to "believe in" Dependent Choice, while I'm not so sure about Zorn's lemma.

Jeremy Brazas pointed out in the comments that the Theorem follows from Brouwer's Reduction Theorem. Here is a proof of that: enter image description here

Is Dependent Choice not used in this proof?

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    $\begingroup$ Would that be a consequence of Uryshon's lemma? $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 17, 2022 at 22:08
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    $\begingroup$ It is a consequence of the Brouwer Reduction Theorem, which I don’t believe requires even countable choice. I think this application of it is in Whyburn’s “Analytic Topology.” $\endgroup$
    – Jeremy Brazas
    Commented Dec 17, 2022 at 22:16
  • $\begingroup$ @JeremyBrazas Thank you, I have updated my question to include a proof of the Brouwer theorem. But I cannot tell whether DC is used. $\endgroup$
    – D.S. Lipham
    Commented Dec 17, 2022 at 22:38
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    $\begingroup$ Maybe, a Whitney size function can be used. See Nadler: Introduction to continuum theory, Exercises 4.33 (for construction) and 4.34, 4.35 (how to avoid Zorns Lemma and Brouwer reduction theorem when searching for irreducible continua). $\endgroup$
    – Benjamin Vejnar
    Commented Dec 19, 2022 at 7:43
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    $\begingroup$ @BenjaminVejnar That reduces it to the theorem that continuous functions on compact spaces assume maximum and minimum values and that uses the axiom of choice for countable families (which follows from DC) $\endgroup$
    – KP Hart
    Commented Dec 21, 2022 at 19:44

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