Let D be a non empty set and let F be a scalar field. Then the set of all functi

Question

Let D be a non empty set and let F be a scalar field. Then the set of all functions defined on D with values in F is a vector space over F with the addition and scalar multiplication of functions defined point wise. This space is denoted by       

(F^D)   . Prove that       (F^D)    is finite dimensional if and only if D is finite.

This is my original approach.

Assume that       (F^D)     is finite dimensional. Then       (F^D)    must have a basis. Let       (B=f_{1},...,f_{n})    be a basis fof the vector space       (F^D)    which is n dimensional according to its basis. Let       (a_{1},...,a_{n} in F)   be scalers, and let       (gin F^D)    be an arbitrary function in       (F^D)    . Then g can be represented by a linear combination of the basis vectors for       (F^D) . That is       (g=a_{1}f_{1}+a_{2}f_{2}+.....+a_{n}f_{n} )      (din D)    be arbitrary, then       (g(d)=a_{1}f_{1}(d)+a_{2}f_{2}(d)+....+a_{n}f_{n}(d))     This is where I am stuck. I am not sure how I can show that D is finite with this approach. Maybe I should try another approach by using the contrapositive. In this case I would assume that D is infinite and then I would need to show that       (F^D)    is infinite dimensional.  But I am not sure if this is any easier. Any help, comments, or advisewould be greatly appreciated. Thank you.

Explanation / Answer

For the contrapositve approach :

Define f_{x}(y) = 1 if y=x and 0 otherwise then this defines a function from D to F.

f_{a} is not equal to f_{b} for a not equal to b (because they take different values at a)

{f_{x}} where x runs over all of D is infinite collecton of functions from D to F and you can check they are linearly independent.

Hence proved.

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