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October 6

Question on topological manifolds

Ok I'm a novice at this so please keep your answers simple. Let R be a square with unit length. Now remove three of its sides, and here's the important part: keep the two end points of the 4th non removed side. Is it a topological manifold with boundary? I've been thinking about what happens to the two end points of the non removed side, and can't seem to stretch any of their neighborhoods into an open ball of the half plane. Can anyone help? Breath of the Dying (talk) 09:40, 6 October 2009 (UTC) Note that in the mathematical literature (especially in the topological literature), a "square" usually refers to the boundary of a "square-shaped plate". A "circle" is defined similarly, and a disk is a "circle together with its interior". --PS T 11:52, 7 October 2009 (UTC)[reply]

If I understand your question correctly, then yes, it is a topological manifold with boundary. Note that a topological manifold (in the simplest sense), is simply a space (well-behaved geometric figures, for instance), each point of which possesses a neighbourhood homeomorphic to an open ball in some Euclidean space. In one dimension, an open "ball" is simply a line segment without endpoints. Thus, in your question, without the end points, your space is a pure topological manifold. In one dimension, however, a boundary can also be defined, and in this case, the boundary consists of the two endpoints of the line segment. Hence, it is a topological manifold with boundary. Note that the real line can be viewed a subset of the plane (and it is a manifold) but does that mean it can be stretched onto an open ball in the plane? The answer is no because it is a one-dimensional manifold. Similarly, the same logic applies to a line segment - just because you cannot stretch any neighbourhood onto some "open ball in the half plane" does not mean it is not a manifold; it just means that it is not a two-dimensional manifold. --PS T 11:03, 6 October 2009 (UTC)[reply]

This was how I also interpreted the question originally, but then I thought Breath might be looking at the square as a 2-manifold with boundary and then only removing parts of the boundary while retaining the interior in its entirety, as otherwise it would seem like a very contrived way of simply asking something about a line segment. —JAO • T • C 11:24, 6 October 2009 (UTC)[reply]

Yes, the OP seems pretty clearly to be describing the open square (0,1)x(0,1) with the closed side [0,1]*{0} added on. That is not a manifold with boundary, due to bad behaviour at (0,0) and (1,0). Algebraist 12:27, 6 October 2009 (UTC)[reply]

(ec) No, it isn't a manifold with boundary. The 4th side and its endpoints are kept. (If they weren't kept it would be a MWB). There aren't any local homeomorphisms at those corners to a halfspace. As our Topological_manifold#Manifolds_with_boundary article says, "If M is a manifold with boundary of dimension n, then Int M is a manifold (without boundary) of dimension n and ∂M is a manifold (without boundary) of dimension n − 1." The boundary is a line segment, which is a manifold with boundary, not a manifold without boundary.John Z (talk) 12:43, 6 October 2009 (UTC)[reply]

I agree it's not a manifold with boundary in the subspace topology of R² (which I'm sure is what the OP meant). But I think I can imagine a topology (if you allow a noncountable basis) that makes it a 1-manifold with boundary. Staecker (talk) 14:00, 6 October 2009 (UTC)[reply]

It's a set with cardinality continuum, so if we're choosing a topology on it at will, we can make it into any manifold we please. Algebraist 14:16, 6 October 2009 (UTC)[reply]

Good point. Staecker (talk) 14:44, 6 October 2009 (UTC)[reply]

Geometry

This was a question which was presented to me in an event I attended, which I couldn't solve.. How do you arrange 7 points in a plane such that when any three of them are chosen, atleast two of them are 1cm apart. I'm interested in finding out the solution to this, assuming it exists.... So can someone help me.... Rkr1991 ^{(Wanna chat?)} 17:58, 6 October 2009 (UTC)[reply]

Maybe this way:

You choose plane:

1. any point.
2+3. two points that their distance from one is $\scriptstyle {\sqrt {3}}$ and from each other 1 cm.
4+5. two points with 1 cm distance from points 1 and 2.
6+7. two points that their distance from 1 and 3 is 1 cm.

--Gilisa (talk) 18:45, 6 October 2009 (UTC)[reply]

I just googled "seven points in the plane". The first hit was this one ~~ Dr Dec (Talk) ~~ 19:27, 6 October 2009 (UTC)[reply]

Googling does not count!--Gilisa (talk) 19:33, 6 October 2009 (UTC)[reply]

:o) You've got me, I'm sorry! ~~ Dr Dec (Talk) ~~ 19:37, 6 October 2009 (UTC)[reply]

Strictly speaking, this is a reference desk, so we are actually supposed to find references for people rather than answering questions from personal knowledge. Therefore, Declan wins! --Tango (talk) 22:51, 6 October 2009 (UTC)[reply]

I', forced to agree for technical reasons!--Gilisa (talk) 12:43, 7 October 2009 (UTC)[reply]

You are forgiven this time! Now I better explain my logic:

We choose three points without two at distance of 1 cm.
Had we choose 1, we can't choose none of 4, 5, 6 or 7 (but we then choose 2 and 3 which are 1 cm from each other).
Have we didn't choose 1, we must now choose one of these groups of three {2,4,5} {3, 6, 7} to take from two points and the distance between them is 1 cm.
--Gilisa (talk) 20:02, 6 October 2009 (UTC)[reply]

See Hadwiger–Nelson problem for an interesting application of this configuraton. Dmcq (talk) 22:39, 6 October 2009 (UTC)[reply]

Since when do centimeters figure into plane geometry? →Baseball Bugs ^{What's up, Doc?} carrots 23:49, 6 October 2009 (UTC)[reply]

I came across this thread earlier, and thought it looked an interesting problem. I've created a diagram of the solution (right), based on the link proveided by Dr Dec. Is the problem sufficiently notable to create an article about? — Tivedshambo (t/c) 11:11, 7 October 2009 (UTC)[reply]

Specifying a particular measurement is an irrelevant distraction. It could just as easily be a mile - or a light year. Plane geometry doesn't talk about specific measurements, only about proportional measurements. →Baseball Bugs ^{What's up, Doc?} carrots 11:50, 7 October 2009 (UTC)[reply]

It's called the Moser spindle and directs to the article I mentioned above. I'm very surprised though that there seems to be no article on matchstick geometry, I'd have though it quite notable. Dmcq (talk) 12:02, 7 October 2009 (UTC)[reply]

I pressed ctrl- a few times and the distances became quite close to 1cm and the text was still quite readable. I don't think one could do that with miles! Dmcq (talk) 12:10, 7 October 2009 (UTC)[reply]

Plane geometry doesn't use specific measurements. The dots could represent galaxies. →Baseball Bugs ^{What's up, Doc?} carrots 12:15, 7 October 2009 (UTC)[reply]

Yes, you've made that point (3 times now). We know it doesn't matter what the unit is, but does it actually make any difference whether the OP says "1cm" or "1 arbitrary unit"? --Tango (talk) 15:44, 7 October 2009 (UTC)[reply]

That's right, Tango. And thanks a lot, guys. And a very heartfelt apology to Baseballbugs for not stating the question properly. Cheers. Rkr1991 ^{(Wanna chat?)} 18:05, 7 October 2009 (UTC)[reply]

Finally, a plane and simple response. :) →Baseball Bugs ^{What's up, Doc?} carrots 23:39, 7 October 2009 (UTC)[reply]

Sorry for the egging too, the devil got into me or whatever the atheist equivalent of that is. Dmcq (talk) 09:40, 8 October 2009 (UTC)[reply]

The atheist equivalent is: "I made a mistake. It was my fault. Sorry." --Tango (talk) 21:35, 8 October 2009 (UTC)[reply]