How Can a Set Fail the Scalar Identity Axiom in Vector Spaces?

[elements]

So I understand how to prove most of the axioms of a vector space except for axiom 10, I just do not understand how any set could fail the Scalar Identity axiom; Could anybody clarify how exactly a set could fail this as from what I know that anything times one results in itself

1u = u
1(x,y,z)=(x,y,z)
1(1,2,3) = (1,2,3)
1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1V ≠ V

 

[Mark44]

So I understand how to prove most of the axioms of a vector space except for axiom 10, I just do not understand how any set could fail the Scalar Identity axiom; Could anybody clarify how exactly a set could fail this as from what I know that anything times one results in itself

1u = u
1(x,y,z)=(x,y,z)
1(1,2,3) = (1,2,3)
1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1V ≠ V

It can happen if you have a vector space with an unusual kind of scalar multiplication. Keep in mind that a vector space consists of a set of vectors over some field (often, the field of real numbers $\mathbb{R}$ or the field of complex numbers $\mathbb{C}$), together with operations for vector addition and for multiplication by a scalar.

If scalar multiplication is defined like this
$k \cdot <x, y> = <kx, 0>$
then there is no scalar k for which $k \cdot <x, y> = <x, y>$.