Why are function spaces generally infinite dimensional The other day, I was trying to explain some concepts in Fourier analysis and wavelets to my girlfriend (an electrical engineering student) and obviously, the concept of Lebesgue integration came up in the form of the function space $L^{2}(\mathbb{R})$ and I tried to explain that they are infinite dimensional spaces of functions that satisfy
$$
\int^{\infty}_{-\infty} \vert f(t) \vert^{2}\, dt < \infty
$$
with a norm given by
$$
\lvert\lvert f(t)\rvert\rvert=\left( \int^{\infty}_{-\infty} \vert f(t) \vert^{2}\, dt \right)^{\frac{1}{2}}$$
and an inner product $ \langle \cdot, \cdot \rangle: L^{2} \to \mathbb{R}$  which allows us the notion of orthogonality such that given $f,g \in L^{2}(\mathbb{R})$
$$
\langle f,g \rangle=\int^{\infty}_{-\infty} f(t)g(t) \, dt =0 \implies f \perp g
$$
I think she understood all of that as well as anyone first exposed to the concepts would but she seemed to have a hard time understanding the "infinite dimensional" part. I guess it is something that always seemed intuitive to me but my explanation was that if you imagine a monotone sequence of functions $f_{n}(t)$ which are pointwise convergent to some function $f(t)$, then we have that 
$$
\lim_{n \to \infty}\int^{\infty}_{-\infty} \lvert f_{n}(t)-f(t) \rvert \, dt =0
$$
and that in the general case, one could imagine that each term in the sequence represents a scaled element of a basis (though in general, they are not orthogonal) in a Banach space. I also explained that if you imagine a Hilbert space (take $L^{2}$ for the sake of simplicity) then every function can be decomposed into a sum of scaled orthogonal basis functions such that given $f(t) \in L^{2}(\mathbb{R})$ and an orthogonal basis $e_{i}$
$$
f(t)=\sum_{n=0}^{\infty} \langle f,e_{i} \rangle e_{i}
$$
such that 
$$
 \left( \int^{\infty}_{-\infty}\left\vert f(t)-\sum_{i=0}^{\infty} \langle f,e_{i} \rangle e_{i} \right \vert^{2} \, dt \right)^{\frac{1}{2}}=0$$
and I tried to explain that if we let $i$ approach some large but finite number that we would get closer to the function we desired to approximate but it would still be a different function. I also tried to give her the standard example of the space $L^{2}([a,b])$ being the completion of $C([a,b])$ with respect to the $L^{2}$ norm via a sigmoid function converging to the Heaviside function as an example but she still seems to be struggling with the notion and I really want her to get it. 
Basically, I'm asking whether someone can offer me a more intuitive explanation or an alternative perspective or any other commentary on why function spaces are infinite dimensional.
 A: $\newcommand{\Reals}{\mathbf{R}}$Here's an indirect approach (not unrelated to littleO's comment): Let $X = \{1, 2, 3, \dots, n\}$ be a finite set of $n$ elements. The set of all real-valued functions on $X$ is $\Reals^{n}$ by viewing an ordered $n$-tuple $(x_{1}, x_{2}, x_{3}, \dots, x_{n})$ as the function $f:X \to \Reals$ defined by $f(i) = x_{i}$ for $i = 1, \dots, n$. (The space of all mappings $X \to Y$ is sometimes denoted $Y^{X}$; this is consistent with using $\Reals^{n}$ to denote the set of all functions from an $n$-element set $X$ to $Y = \Reals$.)
Given that the set of all functions on a finite set can be identified with a finite-dimensional Cartesian space, it's hardly surprising the spaces of (smooth/continuous/measurable) functions on an interval are infinite-dimensional.
This analogy runs deeper in ways useful to (e.g.) electrical engineers: If $K:[a, b] \times [a, b] \to \Reals$ is a (continuous, say) "kernel", there is a linear operator $I_{K}$ on $C([a, b])$ defined by
$$
(I_{K} f)(x) = \int_{a}^{b} K(x, y) f(y)\, dy,
$$
which should remind one of matrix multiplication
$$
Ax = \sum_{j=1}^{n} A_{ij} x_{j}.
$$
In this setting, the Dirac $\delta$-function induces the kernel $K(x, y) = \delta(x - y)$, which formally defines the identity operator:
$$
(I_{K} f)(x) = \int_{a}^{b} \delta(x - y) f(y)\, dy = f(x).
$$
A: So the whole space of functions from $\mathbb{R}$ to $\mathbb{R}$ really does contain "uncountable vectors". But we usually deal with much smaller subspaces of that (in the sense of both dimension and cardinality). These are still infinite dimensional, though. To see it, notice that if $f_i$ are nonzero functions and have disjoint support then $\sum_{i=1}^n c_i f_i$ is zero if and only if all $c_i$ are zero. Yet we can make infinitely many such functions. For example, $f_n(x)=1_{(\sum_{k=1}^{n-1} 1/k^2,\sum_{k=1}^n 1/k^2)}(x)$ is a sequence of linearly independent functions in $L^p([0,2])$ for any $p$ you like.
A: Functions on a finite set would be characterized by a finite set of parameters.
The idea of characterizing polynomials of all orders on a non-trivial interval of $\mathbb{R}$ using a finite set of parameters is immediately objectionable. You just know that cannot be right, which rules out the idea for more general functions that include such polynomials on a finite interval.
To make this precise, start by looking at a finite interval $[-R,R]$ where $R > 0$. The functions $\{1,x,x^2,x^3,\cdots\}$ are linearly independent on any interval $[-R,R]$ because any finite linear combination of such powers that is identically $0$ is a polynomial on $[-R,R]$ that has infinitely many zeros in $[-R,R]$ and, hence, must have all $0$ coefficients. So the set of polynomials on $[-R,R]$ is an infinite-dimensional space of functions. If you extend these functions to $\mathbb{R}$ by settings them to $0$ outside of $[-R,R]$, they remain independent on $\mathbb{R}$. The set of integrable functions is even larger.
