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Show a set is a subspace of vector space, should I prove X(0)=Y(0)?

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No. You should shows that if you add two functions from $U$ that the result is also in $U$, and that if you take a function $f$ in $U$ and create the new function $x \mapsto \lambda f(x)$, then the new function is also in $U$. I have no idea what you mean by $X(0)= Y(0)$. – copper.hat Sep 22 '13 at 20:51
"$X(0)=Y(0)$" makes no sense here. $X$ are $Y$ are sets, not functions. – Stefan Smith Sep 22 '13 at 22:48

You must prove:

$$\begin{align*}(1)&\;\;\;0\in U\;,\;\;\text{and here "zero" is the neutral element in}\;\;\mathcal F(X,\Bbb F)\;,\;\text{i.e. the zero function}\\ (2)&\;\;\;f,g\in U\;,\;\;a,b\in\Bbb F\implies af+bg\in U\;,\;\text{meaning}:\;\;af(u)+bg(u)=0\;\;\;\forall\;u\in U\end{align*}$$

Try to show the above two points and you're done.

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To show that an object is a subspace, you first must determine if it is a subset (this is usually easy). Then you must show that it is a vector space in its own right. You can do this from the axioms, but if you have already discussed the useful theorem that [[a subset $S$ is a subspace if $ax+by\in S$ whenever $x,y\in S$ and $a,b\in \mathbb F$]] then you should use that.

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