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March 3

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Human-made mazes (complex branching passages)

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What are some of the most complex mazes (complex branching passages) made by humans? They can be in buildings or tunnels or any other structures made by humans. (In planning this question about three months ago, I decided that it would be best if I first asked a more basic question. That question is archived at Wikipedia:Reference desk/Archives/Mathematics/2012 December 7#Mazes (complex branching passages).)
Wavelength (talk) 03:30, 3 March 2013 (UTC)[reply]

I'm not sure I understand what you would classify as a "maze." Does something like the National Highway System (United States) count? —Bkell (talk) 04:07, 3 March 2013 (UTC)[reply]
Yes, it does. I am looking for examples in various categories: (1) buildings, (2) tunnels, (3) urban streets, (4) highways, and (5) other structures.
Wavelength (talk) 15:36, 3 March 2013 (UTC)[reply]
How about the electrical wiring of a Boeing 747, or the world's telecommunication network, or the UK's electricity supply system from generation to domestic consumer (much wider than the UK, as there is an interconnector with France, who may well have physical links with other countries)?←86.186.142.172 (talk) 16:21, 4 March 2013 (UTC)[reply]
For a building example, see Kowloon_Walled_City, which was essentially one building at the end. It was built without plan, and famous for the complicated routes and "hacked" connections. SemanticMantis (talk) 00:20, 5 March 2013 (UTC)[reply]
Thank you.—Wavelength (talk) 03:34, 6 March 2013 (UTC)[reply]
"Large" microchips such as CPUs are easily more complex than a large city's road network. Roger (talk) 12:29, 8 March 2013 (UTC)[reply]

Limit points in the cartesian plane

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Am I correct in thinking that in the Cartesian plane when any Set A (for example A = { x | x = ( 0 , 1 / n ) where n ∈ ℕ } is assigned a set A' of limit points these points are all those an infinitesimal distance from the points in set A (e.g. (0,0) is a limit point of the example above?--Gilderien Chat|List of good deeds 16:08, 3 March 2013 (UTC)[reply]

I think you get the idea, but perhaps you should rephrase it in terms of open sets. In the Cartesian plane you can replace the open sets with open metric ε-balls in the formulation (which are open sets). So a point p is a limit point of A if every ε-ball, ε > 0, around p contains a point of A other than p. See Limit point. So if a point q is at a positive distance of at least δ from every point of A (except perhaps q itself), no matter how small, it is not a limit point of A because you can chose ε = δ/2 for the ε-ball around q. That open set contains no points in A except possibly q itself. YohanN7 (talk) 18:07, 3 March 2013 (UTC)[reply]
Thank you, that goes a long way to alleviating my confusion. So does that mean the example I gave above has 0 as its only limit point? The definition you gave seems to imply that every point in a continuous set is itself a limit point for example in x^2 + y^2 > k^2 would this imply that every point on the plane except those in or on a circle of radius k around (0,0) is a limit point of the set? --Gilderien Chat|List of good deeds 18:58, 3 March 2013 (UTC)[reply]
Yes, limit points can be in the set too. So the limit points of are all points except those strictly inside the circle of radius k around the origin—the points on the circle are limit points. —Bkell (talk) 19:02, 3 March 2013 (UTC)[reply]
And therefore whether x is a limit point of A is independent of whether x ∈ A? Thank you.--Gilderien Chat|List of good deeds 19:17, 3 March 2013 (UTC)[reply]
I am under impression that the YohanN7's formulation excludes from the A' set separated points of A (although it allows inner or border-but-not-separated points), while Gilderien's formulation allows all of them into A'. Am I right? For example let's take the set A defined as union of a square without one side, a line segment without one of its end plus a single point:
A = [0, 1)×[0, 1] ∪ [2, 3)×{0} ∪ (4, 4)
What is A' then (according to both formulations, if not identical)? --CiaPan (talk) 21:42, 3 March 2013 (UTC)[reply]
From what I understand, YohanN7's formulation would include internal points in the set as well, for example for the set S = { x | 0 < x < 1 } there would exist a set S' = { x | 0 ≤ x ≤ 1 } which would essentially be { 0 , S , 1 }. --Gilderien Chat|List of good deeds 21:50, 3 March 2013 (UTC)[reply]
Okay, but my question relates to separate points. For a finite set A, say containing 3 points, is A' equal A or is A' empty? --CiaPan (talk) 22:05, 3 March 2013 (UTC)[reply]
A - whether q is part of A is independent of whether it is part of A' .--Gilderien Chat|List of good deeds 22:34, 3 March 2013 (UTC)[reply]
The subtle point here in the definition of a "limit point" is whether a neighborhood of the point p must contain points of the set other than p. There is no general agreement on this matter. So there are various other terms that are often used in this setting. For example, here are definitions from Tom Apostol, "Mathematical Analysis":
  • A point p is an adherent point of the set S if, for all ε>0, the ε ball around p contains points of S.
  • A point p is an accumulation point of the set S if, for all ε>0, the ε ball around p contains points of S\p.
By "limit point" some authors mean an "adherent point", and others mean an "accumulation point" (in this sense). Wikipedia (AFAIK) generally means "accumulation point". Sławomir Biały (talk) 23:18, 3 March 2013 (UTC)[reply]
Sometimes it just happens that a relatively easy concept becomes confusing because there are slightly different conventions and different terminology applies, or worse, the same terminology for slightly, but distinctly, different concepts. It's annoying, but unavoidable. Just look before you leap!
My formulation was meant to be a Cartesian plane version of "Let S be a subset of a topological space X. A point x in X is a limit point of S if every neighbourhood of x contains at least one point of S different from x itself.", which is the formulation in the wikipedia article Limit point. YohanN7 (talk) 11:29, 4 March 2013 (UTC)[reply]