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Please tell me where my reasoning goes wrong:

Using multivariable calculus and other methods, one can easily show that the surface area of a sphere is equal to $4\pi r^2$ and I will consider this a fact. Now let's imagine a sphere with radius $R$. Let's center it at the origin of the standard 3-dimensional space so that its top point is located at $(0,0,r)$ and its bottom point is centered at $(0,0,-r)$. Now we will remove the bottom hemisphere and let the top hemisphere stay. The surface area of the top part is equal to half the whole sphere so it is equal to $2\pi r^2$. Now consider the point $(0,0,-r)$ and call it $P$. It is clear that every point on the top hemisphere can be connected using straight lines to our point. All points of the top hemisphere pass through the equator of our sphere which is a circle centered at the origin defined by $x^2+y^2=r^2$ which has area $\pi r^2$. All points on the hemisphere correspond bijectively to points on the circle. This implies that there are equal points on the sphere and our circle which implies that the hemisphere has surface area $\pi r^2$. But we know that the surface area of the hemisphere is $2\pi r^2$. This is only true if and only if $2=1$.

This is clearly wrong, where have I gone wrong? I think it has to do something with an infinite set being equal to its subset or something. I think an area is nothing more than the sum of infinitely many one-dimensional points and if two figures have equal points, their area should be equal.

A picture to help you picture what I mean: enter image description here

David C. Ullrich
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Sina Babaei Zadeh
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    you could find a bijective correspondence between points in a circle of radius $1$ and points in a circle of radius $2$, but that does not mean they have the same area – J. W. Tanner Jan 03 '20 at 02:00
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    The conclusion "which implies that the hemisphere has surface area $\pi r^2$" is wrong because the assertion "an area is nothing more than the sum of infinitely many one-dimensional points" is not correct. – Pedro Jan 03 '20 at 02:10
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    An even more primitive example of bijections not preserving measure is the interval $(0,1)$ to $(0,2)$ by the function $f(x)=2x$. – zhuli Jan 03 '20 at 03:20

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A bijection between sets does not typically preserve area, length, volume or any other measure. To see this take a small ball, say the size of an apple, and put it inside a much larger ball, say the size of the Earth so that they share the same center. Now clearly any line through the center passes though both balls at exactly two points but it would be absurd to try and cover the entire planet with the skin of an apple. So even though the points on the balls have the same cardinality as the other they do not have the same surface area.

CyclotomicField
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