# Infinite-dimensional Vector space that has a finite-dimensional subspace

If I have an Infinite dimensional vector space how do I find a finite dimensional subspace?

I know plenty infinite dimensional vector spaces for instance All the continuous functions from ℝ to itself. It makes sense that there exists finite-dimensional subspaces but I just don't know how to show they exists.

• Choose $\{0\}$. Or choose a vector $x\neq 0$ and consider its span. – Friedrich Philipp Mar 17 '16 at 0:34
• Did you think about taking multiples of a vector? Or taking the space generated by two vectors? Or three? – user296602 Mar 17 '16 at 0:34
• I think it is really bad behaviour to downvote such questions. The OP does not seem to understand, so he/she asks. What's the point? – Friedrich Philipp Mar 17 '16 at 0:36
• For a simple real example, take the polynomials of degree $\leq n$ in the vector space of all polynomials. – Thomas Andrews Mar 17 '16 at 0:39

Consider for instance the space $V=C(\mathbb{R},\mathbb{R})$ (the space of continuous functions on $\mathbb{R}$ to itself).
One way to get a finite-dimensional subspace $S$ is to pick $n$ functions $f_1,f_2,\ldots,f_n\in V$, and then to consider their span: $$S = \text{span}(f_1,f_2,\ldots,f_n) = \{c_1f_1+\cdots+c_nf_n\ |\ c_1,c_2,\ldots,c_n\in\mathbb{R}\}.$$ If the chosen functions are linearly independent, this will give you a finite subspace of dimension $n$. For instance, $\text{span}(x^2,\cos(x))$ is a $2$-dimensional subspace.
• And, of course, there exists $n$ independent elements for any positive integer $n$ precisely because the vector space is infinite-dimensional. – Thomas Andrews Mar 17 '16 at 0:41