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Moore in his classic book "Foundations of Point Set Theory" defines a "continuum" to be a set of points that is both closed and connected (but not necessarily compact). He also calls "non-degenerate" any set of points containing more than one point. Now every example in the book of a non-degenerate continuum contains at least one proper non-degenerate sub-continuum. But nowhere in the book is there a theorem stating that if C is a non-degenerate continuum, then C always contains a proper non-degenerate sub-continuum (unless further restrictions are placed on C-such as being locally compact). Does such an "unrestricted theorem" actually not follow from the Moore space axioms.?

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You could use a title that asks the whole question: Does every non-degenerate continuum contain a proper non-degenerate sub-continuum? –  Joel David Hamkins Jul 28 '10 at 19:03
    
Since "closed" is a relative notion, it sounds like Moore's continua are subsets of some ambient topological space. Maybe Euclidean $n$-space for some $n$? Please clarify. –  Pete L. Clark Jul 28 '10 at 19:38
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The title may be a bit confusing: what is usually called a Moore space is a CW complex whose reduced homology in given degree is isomorphic to a given abelian group and is zero in other degrees. –  algori Jul 28 '10 at 21:29
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The title should be changed; "Moore space" has a specific meaning different from what you intend... –  Romeo Nov 4 '10 at 6:54

1 Answer 1

Here is a partial answer.

There is a space (Knaster-Kuratowski fan) which is a subset of RxR, and which is connected but becomes totally disconnected upon removal of one point. There may be extensions of this example in which removal of any point causes the extension to be disconnected. I do not know of (but can posit the existence of) a space in which removal of any point causes the subspace to be totally disconnected. This might then be a candidate for a Moore space with no proper nondegenerate Moore spaces.

Gerhard "Ask Me About System Design" Paseman, 2010.07.28

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Such a space cannot be compact, of course, as every compact connected Hausdorff space has at least two non-cutpoints. –  Henno Brandsma Jul 28 '10 at 20:19

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