The Klein bottle and projective plane | Algebraic Topology | NJ Wildberger 

Hi Loic, Yes that is right I believe: the projective plane can be viewed as a Mobius band with a disk glued to the boundary. And any closed disk can be topologically collapsed to a single point.

@@njwildberger Many thanks for answering me! I though a bit more about it, and I think now that the distinction has to be made about removing a closed or an open disk from the projective plane. Removing an open disk yields the Möbius band (which has a boundary), whereas removing a closed disk (or a point, which is also a closed set) yields a boundary-less Möbius band, or, equivalently, a punctured projective plane.

S^1 with antipodal points identified is still S^1; not true in one higher dimension.

On all things numbers. O,9 0×9=0 0÷9=9 0,9 1,9 1+9=10 1×9=9 10+9=19 1+9=10 1+0=1. 2,9 2+9=11 2×9=18 18+11=29 2+9=11 1+1=2 11+2=13 3,9 3×9=27 3+9=12 12+27=39 r13 4,9 4×9=36 4+9=13 13+36=49 4+9=13 13+4=17=8+9. 5,9 5×9=45 5+9=14 45+14=59. 6,9 6×9=54 6+9=15 54+15=69 7,9 7×9=63 9+9=16 63+16=79 8,9 8×9=72 8+9=17 72+17=89 8+9=17 1+7=8. 17+8=25,18+7. Derive sequence in n+1,9=a. 09 19 29 39 49 59 69 79 89=a. Derives in addition =b=x,y. 9 10 11 12 13 14 15 16 17=b=x+y 29 40 41 52 43 73 84 96=a+b (84 96 118)(347 ave=5) 84+12 96+22. 89+17/106 Vertical addition 18 Addition again 9 1 2 3 4 5 6 7 8 0+9=9 4+5=9. 9+10=19 1+9=10. 18+11=29 2+9=11. 27+12=39 3+9=12. 36+13=49 4+9=13. 45+14=59 5+9=14. 54+15=69 6+9=15. 63+16=79 7+9=16. 72+17=89 8+9=17. 81+18=99 9+9=18. Conclusion, Good place to start cycling from for p. P in inverses and conjugates and out right p. Looking like a spiral cone. Maybe find a connection with geometry that is highly intuitive. Like a triangle. Vertical vertical points. 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9. Horizontal points . 1,0, 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0. Diagonally. 0,9 1,8 2,7 3,6 4,5 5,4 6,3 7,2 8,1.9,0 maybe that works? Triple checked seems to be a provable and proper triangle obeying the ro9. dy/dx=1=9/9. What is the hypotenuse? From 0,9 to 9,0 90-9=81=9? x²+y²=h² x²+2xy+y²=h². Magnitude of x,y=9,9. So 81+2×81+81=2×162= 324. 324=2²×3⁴=4×81, sqrt= 2×9=18. Mistake somewhere? 9,9,18. This going to be fun. We are going to strike an arc outside the hypotenuse from zero,zero. We want this because a square inscribed by a circle has the same arc but just ×4 to make a circle. Note 1×2×3=6 1+2+3=6. 6=6 in multiplication and addition. 1+2=3 3+3=6 6+6=12 12+3=15. 1+5=6 15+6=21 2+1=3 2×9+3=21 2+1=3. Pretty good case for a core to the number system and fundamental model of mathematics. On all things numbers. O,9 0×9=0 0÷9=9 0,9 1,9 1+9=10 1×9=9 10+9=19 1+9=10 1+0=1. 2,9 2+9=11 2×9=18 18+11=29 2+9=11 1+1=2 11+2=13 3,9 3×9=27 3+9=12 12+27=39 r13 4,9 4×9=36 4+9=13 13+36=49 4+9=13 13+4=17=8+9. 5,9 5×9=45 5+9=14 45+14=59. 6,9 6×9=54 6+9=15 54+15=69 7,9 7×9=63 9+9=16 63+16=79 8,9 8×9=72 8+9=17 72+17=89 8+9=17 1+7=8. 17+8=25,18+7. Derive sequence in n+1,9=a. 09 19 29 39 49 59 69 79 89=a. Derives in addition =b=x,y. 9 10 11 12 13 14 15 16 17=b=x+y 29 40 41 52 43 73 84 96=a+b (84 96 118)(347 ave=5) 84+12 96+22. 89+17/106 Vertical addition 18 Addition again 9 1 2 3 4 5 6 7 8 0+9=9 4+5=9. 9+10=19 1+9=10. 18+11=29 2+9=11. 27+12=39 3+9=12. 36+13=49 4+9=13. 45+14=59 5+9=14. 54+15=69 6+9=15. 63+16=79 7+9=16. 72+17=89 8+9=17. 81+18=99 9+9=18. Conclusion, Good place to start cycling from for p. P in inverses and conjugates and out right p. Looking like a spiral cone. Maybe find a connection with geometry that is highly intuitive. Like a triangle. Vertical vertical points. 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9. Horizontal points . 1,0, 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0. Diagonally. 0,9 1,8 2,7 3,6 4,5 5,4 6,3 7,2 8,1.9,0 maybe that works? Triple checked seems to be a provable and proper triangle obeying the ro9. dy/dx=1=9/9. What is the hypotenuse? From 0,9 to 9,0 90-9=81=9? x²+y²=h² x²+2xy+y²=h². Magnitude of x,y=9,9. So 81+2×81+81=2×162= 324. 324=2²×3⁴=4×81, sqrt= 2×9=18. Mistake somewhere? 9,9,18. Equilateral triangle masquerading as a right triangle? This going to be fun. We are going to strike an arc outside the hypotenuse from zero,zero. We want this because a square inscribed by a circle has the same arc but just ×4 to make a circle. Note 1×2×3=6 1+2+3=6. 6=6 in multiplication and addition. 1+2=3 3+3=6 6+6=12 12+3=15. 1+5=6 15+6=21 2+1=3 2×9+3=21 2+1=3. Pretty good case for a core to the number system and fundamental model of mathematics.

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