Talk:Indecomposable continuum

A precise defintion, please?

I can't figure out the definition here. All I have is this, from the lead

an indecomposable continuum is a continuum that is not the union of any two of its proper subcontinua.

what does that mean? If I take the unit interval, and then consider the right 2/3rds of it, and the left 2/3rds of it, then their union forms the whole interval. So is the unit interval decomposable?

If I take the buckethandle, and consider left 4/7ths of it, and the right 4/7ths of it, wouldn't the whole bucket-handle be the union of these two pieces? What am I missing? 67.198.37.16 (talk) 17:37, 27 November 2017 (UTC)[reply]

OK to partly answer my own question: a subcontinuum is, by definition, connected and compact. So the 4/7th's construction above will not qualify for that. This paper: "Bucket handles and Solenoids", Notes by Carl Eberhart, March 2004 suggests that the buckethandle can be constructed as an infinite union of semi-circular arcs; so is the issue that the buckethandle cannot be decomposed into just two arcs?

That's not obvious either. There is a unique label for each arc: the Cantor set can be thought of as the "endpoints" of an infinite binary tree. Any/every branch of that tree can be uniquely labeled by a countably-infinite string of left and right moves. Any irrational number can serve as such a label: any irrational real has a unique representation as an infinite string of ones and zeros. So why can't I just pick such a number, and say "ah ha, pick the corresponding arcs, remove a short segment from that/those arcs, and, bingo, this falls apart into two pieces?"

Perhaps the argument is that these two pieces I built above are not disjoint in some Hausdorf sense? i.e. they are not T_2 or T_1 or even T_0 seperable? I agree that they're not seperable in this way, but seperability does not seem to enter into the definition of decomposability. So I'm still confused. 67.198.37.16 (talk) 18:01, 30 November 2017 (UTC)[reply]

I see now. Something is connected when its not disconnected. Two sets are disconnected when they are seperable. So in fact, seperability does enter into the definition, in this indirect way. The construction I give above removes a chunk of the buckethandle, but it does not separate it, and therefore, the buckethandle remains connected despite this removal. I believe that it is worth articulating all this with examples and a discussion. Its not "obvious" without carefully examining all the definitions of all the terminology. 67.198.37.16 (talk) 21:20, 30 November 2017 (UTC)[reply]

This can't be right

The section History contains this sentence:

"In 1910 L. E. J. Brouwer described an indecomposable continuum which disproved a conjecture made by Arthur Moritz Schoenflies that the joint boundary of two open, connected, disjoint sets in $\mathbb {R} ^{2}$ was the union of two closed, connected proper subsets."

It is extremely easy to find two disjoint connected open sets in the plane, the intersection of whose boundaries is not the union of two closed connected proper subsets. (Imagine a thickened capital E next to its mirror image, so that their common boundary is the union of three disjoint closed intervals.)

I will hazard a guess that Schoenflies's (eventually disproven) conjecture was this:

"If the intersection of the boundaries of two disjoint open sets in the plane is a continuum, then it must be the union of two closed. connected proper subsets." 2601:200:C000:1A0:B4E6:9BA4:980:C6B4 (talk) 20:09, 2 August 2021 (UTC)[reply]

Yes, it's unclear - "common boundary" means

\partial X_{1}=\partial X_{2}

. I'll rephrase. NPalgan2 (talk) 21:44, 2 August 2021 (UTC)[reply]

Lead needs improvement

"... an indecomposable continuum is a continuum that is indecomposable, i.e. ..."

This statement is a vacuous tautology. Instead the sentence should proceed directly to the material which follows the "i.e.", which is the actual definition:

"In point-set topology, an indecomposable continuum is a continuum that cannot be expressed as the union of any two of its proper subcontinua." 71.188.95.60 (talk) 17:53, 18 January 2024 (UTC)[reply]

Also, "proper" in "proper subcontinua" wikilinks to "Subset#Definitions", but it is not intuitively obvious that a subcontinuum is a kind of subset. 71.188.95.60 (talk) 18:10, 18 January 2024 (UTC)[reply]