I hope I'm not the only one who recognizes that you are more interested in this conversation than me. :grin: — Leontiskos
A circle is a plane figure bounded by one curved line, and such that all straight lines drawn from a certain point within it to the bounding line, are equal. The bounding line is called its circumference and the point, its centre. — Circle | Wikipedia
Euclid says: not a circle. The great circle is not a plane figure. — fdrake
Why do you think this? — Leontiskos
A circle is a plane figure bounded by one curved line, and such that all straight lines drawn from a certain point within it to the bounding line, are equal. The bounding line is called its circumference and the point, its centre. — Circle | Wikipedia
And what is "the great circle"? — Leontiskos
The great circle is the circle I've highlighted on the surface of the sphere. Since the circle is confined to the surface of the sphere, it is not a plane figure. — fdrake
But why accept the counterexample? We proved our conjecture— now it is a theorem. I admit that it clashes with this so-called ‘counterexample’. One of them has to give way. But why should the theorem give way, when it has been proved? It is the ‘criticism’ that should retreat. It is fake criticism. This pair of nested cubes is not a polyhedron at all. It is a monster, a pathological case, not a counterexample. — Lakatos, as quoted in Russell
The cross-section of a sphere is a circle. — Leontiskos
Well who said anything about cross sections? — fdrake
You'll now need to tell me in what circumstances can you take a cross section of a volume and have it work to produce a circle. Let's assume that you can take any volume and any cross section and that will produce a circle... — fdrake
The cross-section of a sphere is a circle. — Leontiskos
Why are you doing this sort of thing? — Leontiskos
but it does not satisfy Euclid's definition of one verbatim — fdrake
You could also have ardently insisted that indeed, the great circle was not a great circle because it was not a plane figure. — fdrake
I think it does. You've only asserted otherwise, you haven't shown it. — Leontiskos
but it does not satisfy Euclid's definition of one verbatim — fdrake
I think it does. You've only asserted otherwise, you haven't shown it. — Leontiskos
As for me, I mean a set of points equidistant from a point. — fdrake
What do you think a plane figure is? — fdrake
that circle instead would be a closed curve inside a subset of R3 — fdrake
He doesn't need to. The sphere is a 2-manifold, and his great circle is a set of points on that manifold. There are no planes here, nothing else, only the points on the surface of the ball.
You are imagining the sphere embedded in the usual 3d Euclidean space. Now, imagine it isn't. There is no point the points on this great circle are equidistant from. — Srap Tasmaner
Planes and points cannot be stipulated to exist or not exist. — Leontiskos
Your word "imagine" is on point given my earlier claim — Leontiskos
But don't you need to specify coplanar? If we're in 3d space, you've defined a sphere, in 4th I guess some sort of hypersphere, I don't know, blah blah blah. — Srap Tasmaner
We took our definition from Euclid, and the term there means a figure that lies entirely on a flat plane. — Leontiskos
Do you think the "great circle" (which you have yet to define) lies in three dimensional space rather than two dimensional space? That ambiguity is why I asked you to be more clear about what you were depicting in the first place. — Leontiskos
If we define "distance" in the commonly accepted way, then there aren't. Are we disagreeing on something more profound than that? — Leontiskos
I'm enjoying this discussion. — Banno
And you are ignoring the fact that I used it twice. — Srap Tasmaner
I had thought the example, Euler’s formula, a bit obtuse. But perhaps Lakatos chose it so as to minimise the number of auxiliary hypotheses that his students could produce.It is a lot like something from Proofs and Refutations. — fdrake
Either you consider the sphere as embedded in 3-space, and the cross section plane isn't "flat" in some sense - it's at an incline. Or you consider the surface as a 2-dimensional object, in which case there's not even a plane to think about. Pick your poison. The latter is the original counter example and is much stronger, the former is easier to remedy. — fdrake
But clearly that's not true, as the definition you provided doesn't match something you clearly recognised as a circle! — fdrake
How will you parry my counterexample? — fdrake
If it makes you uncomfortable, stop tromping on my bridge. — Banno
But it does match it, as I've already noted. Your mere assertions are getting old. — Leontiskos
A set of coplanar points equidistant from a point in the plane of coplanarity. — fdrake
The cross-section of a hollow sphere will be a circle regardless of whether I imagine a point at the center or not. — Leontiskos
I don't know why I'm participating in this. — Srap Tasmaner
―― I don't know why I'm participating in this. — Srap Tasmaner
Does that point need also to be coplanar? Is there a counterexample I'm missing? — Srap Tasmaner
The great circle is the circle I've highlighted on the surface of the sphere. Since the circle is confined to the surface of the sphere, and the surface of the sphere is not a plane, it is not a plane figure. — fdrake
We could also define a circle as the cross-section of a sphere, but I was only saying that every (planar) cross-section of a sphere will in fact fulfill the definition I already set out. — Leontiskos
No.You artificially inserted an extraneous conversation into your own thread and then invited me here, remember? — Leontiskos
I'm glad you dropped in, at Leon's invitation, I think?―― I don't know why I'm participating in this. — Srap Tasmaner
Cheers. I'm glad someone looked at the Russell article.I'm gonna bugger off now too. — fdrake
It would if you give yourself the liberty of hammering the cross section down onto a flat plane. — fdrake
I'm glad someone looked at the Russell article. — Banno
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