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How to prove this fact:

If $X \subset \mathbb{R}$ is finite, then the space of continuous functions $C(X)$ is finite dimensional (isomorphic to $\mathbb{C}^n$)?

Thanks for help!

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Related:math.stackexchange.com/questions/25983/… –  Jonas Meyer May 12 '12 at 3:05
Thanks for your link! –  Spring Xiao May 12 '12 at 4:36
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2 Answers

up vote 4 down vote accepted

Let $X=\{a_1,\cdots,a_n\}$. Define $f_1,\cdots,f_n$ such that $f_i(a_j)=\delta_{ij}$. Then $\{f_1,\cdots,f_n\}$ forms a basis of $C(X)$. Indeed, if $f\in C(X)$, then $f=\sum\limits_{i=1}^n f(a_i)f_i$ and if $g=x_1f_1+\cdots+x_nf_n=0$, for $x_1,\cdots,x_n\in\mathbb{R}$, then $0=g(a_i)=x_1f_1(a_i)+\cdots+x_nf_n(a_i)=x_i$, for $i=1,\cdots,n$, so $x_1=\cdots=x_n=0$.

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Thank you very much! –  Spring Xiao May 12 '12 at 2:34
@SpringXiao don't forget to check with the "green mark" the accepted answer, to mantain the policy of ratings working =) –  matgaio May 12 '12 at 2:42
Ok! But why Can't display Latex properly? –  Spring Xiao May 12 '12 at 2:55
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Since $X=\{x_1,\ldots,x_n\}$ is finite, each of its points is open in the relative topology. This implies that every function $X\to\mathbb{C}$ is continuous. So $$C(X)=\mathbb{C}^X=\{f:X\to\mathbb{C}\}\simeq\{f:\{1,\ldots,n\}\to\mathbb{C}\}=\mathbb{C}^n. $$

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