Prove the set of continuous real-valued functions on the interval $[0, 1]$ is a subspace of $\mathbb R^{[0, 1]}$. My attempt(I don't know analysis, so I just looked up the properties of continuous functions):
Let $U$ be the set of continuous real valued functions. Let $u, v \in U$. Since the sum of continuous functions is continuous, $u + v$ is continuous. Since any polynomial is continuous, $a(u + v)$ is continuous. So, $a(u + v) \in U$. Since $0(u + v) = 0, 0 \in U$.
I am trying to learn how to do subspace proofs. Seeing a correct proof at least once would be very helpful. Thanks.
 A: In order to verify that $U$ is a subspace of a vector space $V$, you need to check:
(1) if $u,u'\in U$ and $\alpha,\beta$ are scalars, then $\alpha u + \beta u'\in U$, and
(2) $0\in U$ (or, equivalently, that $U$ is nonempty, because together with (1), $0\in U$ is equivalent to $U\neq\emptyset$ (why?))
Thus, in this case, your proof is essentially correct - a linear combination of continuous real-valued functions is a continuous real-valued functions, and the zero function is a continuous real valued function. Note that there is nothing special about the domain $[0,1]$ here (except that it is a topological space, but don't worry about this if you don't know what that is). However, it is important that the codomain $\mathbb{R}$ is a field so it makes sense to add and scale continuous functions into $\mathbb{R}$, and thus have a vector space structure. For example, with the roles of $[0,1]$ and $\mathbb{R}$ switched (i.e., functions from $\mathbb{R}$ into $[0,1]$), the problem wouldn't make sense because there would be no vector space structure.
Hope that helps!
