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I'm wondering if there is an error in this question and maybe the solution.

  1. Let X be a set and F a field. Prove that M_fin(X, F) is a subspace Of M(X, F).

Note:M_fin(X,F) The space of all functions from the set X to the field F, which have finite support This is what my professor claimed.

Let V=M(X,F),W=M_fin(X,F) and let u,v in W were u and v are vectors. then u=f,v=g where f:X->F and g:X->F there are sets A,B subset X such that if x in X/A,f(x)=0 and if x in X/B g(x)=0. Claim (f+g)(x)=0 if x in X/(A union B) x/(A union B) = f+g(x)=0

My question is wouldn't this fail if g is the additive inverse? then we would get that (f+g)(x)=0 then this would be a contradiction. Most importantly the second property of c being an arbitrary element, then cu must be in W but if c=0?

Thank you so much for reading my question.

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    The empty set is a finite set. So if $f(x)=0$ for all $x$, then $f\in M_{\text{fin}}(X,F)$. No contradiction with the additive inverse, either. – 2017-01-31
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    Thank you Dr.Shilfrin, It is a great honor to have you respond to my post. – 2017-01-31

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