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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.

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    $\begingroup$ Choose $\{0\}$. Or choose a vector $x\neq 0$ and consider its span. $\endgroup$ – Friedrich Philipp Mar 17 '16 at 0:34
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    $\begingroup$ Did you think about taking multiples of a vector? Or taking the space generated by two vectors? Or three? $\endgroup$ – user296602 Mar 17 '16 at 0:34
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    $\begingroup$ 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? $\endgroup$ – Friedrich Philipp Mar 17 '16 at 0:36
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    $\begingroup$ For a simple real example, take the polynomials of degree $\leq n$ in the vector space of all polynomials. $\endgroup$ – Thomas Andrews Mar 17 '16 at 0:39
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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.

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    $\begingroup$ And, of course, there exists $n$ independent elements for any positive integer $n$ precisely because the vector space is infinite-dimensional. $\endgroup$ – Thomas Andrews Mar 17 '16 at 0:41

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