In Boas' Mathematical Methods an example says that the set of functions of the form $f(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3$ with $f(1)=1$ is not a vector space because if we add two of the polynomials then the value of the sum at x=1 is 2, which is not an element of the set. I feel like I'm really close to understanding this but that it's not quite clicking yet.
Is it because $f(1)+g(1)=2$ $\notin$ $\{1, x, x^2, x^3\}$?
Could we just set $a_{0}=2$?
On
Not exactly.
Let $V$ be the space you defined in your post. For example, if we define $f(x) = 1$ and $g(x) = x$, then both $f$ and $g$ are elements of the space $V$. Let's check that $g$ is really an element of your space $V$: First $$ g(x) = x = 0 + 1 \cdot x + 0 \cdot x^2 + 0 \cdot x^3 \; , $$ so it has the right form, and $$ g(1) = 1 $$ clearly holds too.
Now, a vector space has the property, that if two elements $f$ and $g$ are elements of the vector space, then $f+g$ must also be an element of the vector space.
But this fails to be true here, because $$ (f + g)(1) = f(1) + g(1) = 1 + 1 = 2 \neq 1 \; , $$ so $f + g \notin V$, and we conclude that $V$ is not a vector space.
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You want every element of your vector space satisfies $f(1)=1$ Now if $f$ and $g$ are members of your vector space you like to have $(f+g)$ also to be a member.
You check $(f+g)(1)$ and you see it is not $1$, so $(f+g)$ does not satisfy the condition of being a member.
That is your space is not a vector space because it is not closed under addition.
It is because $f(1)+g(1)$ is $2$... but not for the reason you think. The problem is that when you combine two 'vectors', then the thing you get must also be a 'vector'. In this case, 'vector' means: 3rd degree polynomial that equals $1$ when evaluated at $1$. But if you add two such polynomials, their sum won't equal $1$ when evaluated at $1$, so the sum is not a 'vector'.