How to show that this subset is/is not a subspace I have this practice problem which is

Consider the vector-space $U=F(-\infty,\infty)$,which is the vector-space of all functions from $-\infty$ to $\infty$, and the subset $V$ defined as:
$V:= \{\textbf{f} \in V \mid \textbf{f}(1)=0\}$
  which is the set of functions which assigns the value of $0$ to $1$. Prove wether or not this subset is a subspace of $U$.

Thats part a) part be is the same but where $\textbf{f}(0)=1$. I am only asking for guidance for part a) which is what I quoted above... 
My initial thought was to check the subspace criterion which is to check that it was closed under addition, and then check for scalar multiplication. I also had the thought of checking to make sure there is a zero vector. 
But my problem that I am having is that I am trying to figure out how to check these conditions for a space for which is defined by functions instead of the typical vectors, but also how to account for the fact that $\textbf{f}(0)=1$ for all functions $\textbf{f} \in V$ also I should note that there is no specification on $\textbf{f}$ having to be continuous either.
If anyone could offer some guidance that would be greatly appreciated. 
 A: As you said one has to check the axioms, namely:


*

*Closure under addition

*Closure under scalar multiplication

*Neutral element


I will show the first as the rest is the same:
let $f,g$ in $V$. Then $f(1)=0=g(1)$.
Thus $$(f+g)(1)=f(1)+g(1)=0+0=0$$ and so $f+g$ satisfies the requirements to be in $V$, i.e. is in $V$. Consequently $V$ is closed under addition.
A: Part a) $V=\{f \in U \mid f(1)=0\}$


*

*Closed under addition: let $f,g \in V$, i.e. $f(1)=g(1)=0$. Then, $(f+g)(1) = f(1)+g(1) = 0+0 = 0$. Thus, $f+g \in V$.

*Closed under scalar multiplication: let $f \in V$ and $k \in F$ where $F$ is the field of scalars, i.e. $f(1)=0$. Then, $(kf)(1) = k(f(1)) = k(0) = 0$. Thus, $kf \in V$.

*Contains $\mathbf{0}$: $\mathbf{0}(1) = 0$, so $\mathbf{0} \in V$.


Thus $V$ is a subspace of $U$.
Part b) $V=\{f \in U \mid f(0)=1\}$


*

*Closed under addition: let $f,g \in V$, i.e. $f(0)=g(0)=1$. Then, $(f+g)(0) = f(0)+g(0) = 1+1 = 2$. Thus, $f+g \notin V$.


Thus $V$ is not a subspace of $U$.
