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February 7

Moore plane and Local compactness

Hi. I've been thinking about this for a few days now, and I can't figure out what makes the Moore plane not be locally compact. It seems to me that every point has a compact neighborhood, and the space is T2, so I don't see what the problem is. Can someone help me see what I'm missing? It must have something to do with neighborhoods of points on the x-axis, but I don't see that closures of basic neighborhoods there are anything but compact.

Thanks in advance for any helpful hints. -GTBacchus^(talk) 00:11, 7 February 2009 (UTC)[reply]

The closures of basic neighbourhoods are not compact. One way of seeing this is that if they were, their boundaries would be compact. But the boundary of such a neighbourhood is just a circle with one point removed in Euclidean space. Algebraist 02:45, 7 February 2009 (UTC)[reply]

The basic open neighborhoods of points on the x-axis do not have compact closure. This is easier to see if you transform the plane by the complex map z → –1/z. This map restricts to a self-homeomorphism of the open half-plane y > 0, and maps the point (0,0) to a new "point at infinity". The images of the open discs tangent to (0,0) are regions of the form y > h in the upper half-plane, so these are the basic open neighborhoods of the "point at infinity". It is easy to see that such a neighborhood does not have compact closure. For example, the closed set {y ≥ 1} ∪ {∞} can be covered by the open set {y > 2} ∪ {∞} together with an infinite number of discs covering the strip 1 ≤ y < 2, and this cover has no finite subcover.

If you don't like the transformation, just picture an open cover of a closed disc tangent to (0,0) consisting of a smaller open disc tangent to (0,0), together with an infinite number of open discs not tangent to the x-axis covering the moon-shaped complementary region. Jim (talk) 02:44, 7 February 2009 (UTC)[reply]

I thank you both. That's very helpful. I think what I was missing is that the point on the x-axis is actually an interior point of a tangent circle, and not part of the boundary after all.

Cheers! -GTBacchus^(talk) 04:41, 7 February 2009 (UTC)[reply]

I need the next in this sequence

In the sequence, ones, tens, hundreds, thousands, millions,(?), what are the next set of numbers refered to? Please do not tell me billions (it's only a multiple of millions) because that is not what I'm looking for. I need to know what what comes after *illions. Thank you.68.167.68.167 (talk) 02:07, 7 February 2009 (UTC)[reply]

I'm not entirely sure I understand your question, but I suggest looking at the names of large numbers article. It has information on all of the large numbers that have standard english names. Jim (talk) 02:17, 7 February 2009 (UTC)[reply]

Application of F-statistic and Mallows Cp

Hi, I'm looking for some referable references that indicate the followings: 1) F-statistic in comparing two models should be employed only where models are nested. 2) Also, Mallows Cp is applicable when the models are linear. Any suggestion is highly appreciated. Re444 (talk) 10:55, 7 February 2009 (UTC)[reply]

Graphical Transformations

I know that to sketch $y^{2}=f(x)$ , you first sketch $y=f(x)$ then delete the part below the x-axis and then reflect the remaining graph in the x-axis. My first question is, why does this approach not work for $f(x)=x$ ? My second, how do you sketch $y^{3}=f(x)$ ? My third, will the approach used for $y^{2}=f(x)$ and $y^{3}=f(x)$ work for $y^{2n}=f(x)$ and $y^{2n+1}=f(x)$ ? Thanks 92.4.224.144 (talk) 21:43, 7 February 2009 (UTC)[reply]

That doesn't work for any graph (except f(x)=1 or 0, I guess). It will give you a rough approximation, but not an accurate graph. That method actually gives you |y|=f(x). To get y²=f(x) you would need to scale the y-axis correctly (points move towards 1 (or -1 for negative points), points further away from 1 move more, proportionally, than those nearer. Points on the x-axis stay where they are). --Tango (talk) 21:58, 7 February 2009 (UTC)[reply]

To draw

y^{2}=f(x)

, I would first draw

y={\sqrt {f(x)}}

and then

y=-{\sqrt {f(x)}}

. —Bromskloss (talk) 22:27, 7 February 2009 (UTC)[reply]

I know that it won't give an accurate graph but I'm really only looking for a quick and simple method that produces an approximate sketch, with just the main shape and features of the graph; for my purposes it doesn't need to be to scale. For the few graphs I have tried sketching using this method and then checked using Grapher, it meets my criteria. I was just wondering how it could be adapted for higher powers of y. 92.4.224.144 (talk) 23:52, 7 February 2009 (UTC)[reply]

Well, if you only want something that approximate then for odd powers leave the graph alone and for even powers do what you did for squares. It's really inaccurate, though. For example, for f(x)=x, your method ends up with two straight lines, the correct graph has two curved lines. I consider whether something is straight or not to be a pretty key feature... --Tango (talk) 00:16, 8 February 2009 (UTC)[reply]

Graphs can be useful for developing an intuition for how a function behaves, but I agree with Tango that your approach is too inaccurate to be useful (although I suppose that's up to you to decide depending on what you are using the graphs for). However, it may be worthwhile to try the re-scaling that Tango suggests in his first post. I'd suggest taking 4 or 5 functions

f(x)

and transforming their graphs in the way you described, and also creating accurate representations of the graphs (using a graphing calculator, say, if you're having difficulty), and compare how the two (the inaccurate and the accurate) look. Also try this for

y^{2}

vs.

y^{4}

and y vs.

y^{3}

vs.

y^{5}

. If you don't already have an intuition for the re-scaling operation that Tango described, this should help make it more clear. Hopefully afterwards you will be better able to roughly sketch those graphs yourself without detailed computations or calculators. Eric. 131.215.158.184 (talk) 09:50, 8 February 2009 (UTC)[reply]