I don't have time now to provide complete answers of the type you want, but I can give you a little insight.

> Matrices, especially matrix multiplication.

Consider a transformation of space, specifically a shear, or a rotation, or an expansion (or contraction). All of these leave the origin unmoved, and a line will always become a line. They are what we call linear transformations.

So given a point (x,y,z) in 3D space, the new values of x, y, and z are obtained by equations like a * x + b * y + c * z. If you then chase through how to combine two of these transformations, one after the other, the definitions of matrix multiplication drop out.

This, by the way, also explains why you can't always divide by a matrix. Sometimes the transformation collapses the space into a lower dimensional space, and that can't be undone.

> Dot and cross products.

For unit vectors u and v, the dot product tells you how much of u points in the direction of v. Then we want it to be true that doubling the vector length doubles the size of the product. Then the result you ask about - the magnitude of dot product in 3D is a.b.cos(theta) - automatically follows.

With the cross product, that can be defined as the area of a parallelogram that has the vectors as the side. Again, thinking about that interpretation shows how we get the answer you ask about.

> How is a set and "belongs-to" operator defined?

Set theory is often regarded as axiomatic, and so you can go back to the axiomatic definitions. We have a universe of discourse that has a bunch of atomic things. A "set" is then a collection of things, and then each set becomes a new thing in our universe. Thus sets can contain other sets.

So a set is defined by what things are in it, so the "belongs-to" concept is fundamental to how the set in question is specified.

> Why is 0.9999... considered to be "equal" to 1.

People trip up on this in part because they think somehow that something is moving, that 0.999... "approaches" something. But no, when we write "0.999..." we have a representation of a point on the real line. Deciding what it is that you are defining requires that you be clear and detailed about what you think 0.999... means.

So what does it mean? We drop back to saying that it's the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... so we need to ask what real number can possibly be that limit? Let's call the limit L, and ask what it might be.

Well, clearly:

L > 0.9
L > 0.99
L > 0.999
L > 0.9999

... and so on.

Now, for every number X that's less than 1, there'll be something in that sequence that will be bigger than X. That means that the limit can't be less than 1. But equally, none of those numbers is bigger than 1, so the limit can't be bigger than 1.

In fact, the smallest upper bound of { 0.9, 0.99, 0.999, 0.9999, ... } is 1.

And then, why should it not be 1? We have different representations of points elsewhere. 2/3 is also 4/6 and 6/9, and sqrt(8) is the same as 2 * sqrt(2). These are simply different ways of specifying the same place on the real line. Likewise, writing 0.9999... is just a different representation of the point more often written as 1.

You said:

> I understand them to be equal "under the limit",
> but not without.

You tell me what you mean by 0,9999... and then I'll answer your question in more detail.

> This seems to be in my way of understanding Cantor's infinities.

Well, there are two different types of infinities based on set theory, but you're probably asking about cardinal infinities, the infinity of counting stuff - "How many things are in this set."

So here, I have a sack with a collection of cubes, and they're numbered from 1 onwards. Call that collection A.

Now let's take a cube and divide one face into two sections. Flip a coin, and write the result in one of the sections. in the other section, divide into two, flip a coin, write the result in one section, and then lather, rinse, repeat. We can fit infinitely many coin flips onto the cube in this way. In sack B put a cube with every possible result of this process.

I claim that the cubes from sack A and sack B cannot be paired off with nothing left over. I won't prove that here, I'm now out of time.

So, ask questions.