Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

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 $x_1=\cdots=x_n=0$.

share|improve this answer
    
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
add comment

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 $$C(X)=\mathbb{C}^X=\{f:X\to\mathbb{C}\}\simeq\{f:\{1,\ldots,n\}\to\mathbb{C}\}=\mathbb{C}^n. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.