Proving Function Space as vector subspace I was doing some vector space questions and stumbled upon a question with a solution provided.
Solution
I did not understand the need to substitute the functions of $f(3)$ with $f(1)$ as they don't serve any purpose in proving the vector subspace.
 A: To show that $W$ is a subspace of $V$, we have to show that it is closed under addition of elements (we have to show we cant add elements of a subspace to get out of our subspace; our subspace is closed.
First, it shows that it is indeed a subspace by showing that it contains the zero vector as $0(x)=0$ so $0(3)=0(1)$. 
Now we have to show that it is closed under addition and scalar multiplication. So given two functions in $W$(call them $A$ and $B$, we have to show that $A+B$ is in $W$ and multiplying them by scalars also results in a function in $W$. So your source goes from $3$ to $1$ and shows they map to the same element so that it lies in $W$. To do so, it uses the fact that your individual functions have that property. 
So:
$f+g$ is in $W$ if $(f+g)(3)=(f+g)(1)$. to show this, your source breaks it down into individual functions. So $(f+g)(3)=f(3)+g(3)=f(1)+g(3)=f(1)+g(1)=(f+g)(1)$
for scalar multiplication (any scalar $c$):
$cf(3)=cf(1)$ 
your link clutters everything up by showing everything at once. Remember that our vectors are actully functions not the real numbers we plug into them
