The leap to infinite dimensions Extending this question, page 447 of Gilbert Strang's Algebra book says

What does it mean for a vector to have infinitely many components?  There are two different answers, both good:


1)  The vector becomes $v = (v_1, v_2, v_3 ... )$
2)  The vector becomes a function $f(x)$.  It could be $\sin(x)$.

I don't quite see in what sense the function is "infinite dimensional". Is it because a function is continuous, and so represents infinitely many points? The best way I can explain it is:


*

*1D space has 1 DOF, so each "vector" takes you on "one trip"

*2D space has 2 DOF, so by following each component in a 2D (x,y) vector you end up going on "two trips"

*...

*$\infty$D space has $\infty$ DOF, so each component in an $\infty$D vector takes you on "$\infty$ trips"


How does it ever end then?  3d space has 3 components to travel (x,y,z) to reach a destination point.  If we have infinite components to travel on, how do we ever reach a destination point? We should be resolving components against infinite axes and so never reach a final destination point.

One thing that might help is thinking about the vector spaces you already know as function spaces instead.  Consider $\mathbb{R}^n$.  Let $T_{n}=\{1,2,\cdots,n\}$ be a set of size $n$.  Then $$\mathbb{R}^{n}\cong\left\{ f:T_{n}\rightarrow\mathbb{R}\right\} $$ where the set on the right hand side is the space of all real valued functions on $T_n$.  It has a vector space structure since we can multiply by scalars and add functions.  The functions $f_i$ which satisfy $f_i(j)=\delta_{ij}$ will form a basis.
So a finite dimensional vector space is just the space of all functions on a finite set.  When we look at the space of functions on an infinite set, we get an infinite dimensional vector space.
