Proving that $B:=\{f(x)\in C[a,b]:f(a)=0\}$ is close set

Question

Let $B$ be a group of all the continuous functions in the interval $[a,b]$ such that $f(a)=0$. Prove that $A$ is close group in the metric space $C[a,b]$

My attempt:

Metric space $C[a,b]$ defined by $d(x(t),y(t))=\max\limits_{a\leqslant t\leqslant b} \mid x(t)-y(t)\mid$

I think I need to prove that $B^c$ is open group, I am not really know what to do here.

Answer 1

The fact that $B$ is a group shouldn't be difficult: if $f,g \in B$, then $f(a) + g(a) = 0 + 0 = 0$ and $f+g$ is obviously continuous. Next, if $f \in B$ then $-f \in B$ because it is continuous and $(-f) (a) = - (f(a)) = -0 = 0$, and $0$ is obviously the neutral element. That addition is associative follows from the fact that the addition of functions is associative in general.

To see that $B$ is continuous, let $(f_n) \subset B$ be a sequence of functions such that $f_n \to f$ in the topology of the distance $d$, with $f \in C[a,b]$. You want to show that $f \in B$, but that is obvious: $f(a) = \lim \limits _n f_n(a) = \lim \limits _n 0 = 0$, so $f$ satisfies the properties that define $B$, so $f \in B$. Therefore, $B$ contains all its limit points, therefore it is closed in $C[a,b]$.

Answer 2

If $f\notin B$ then all $g$ with $d(f,g)<|f(0)|$ are also $\notin B$.

Answer 3

Let $E_a:C([a,b])\to\mathbb{R}$ defined as $E_a(f)=f(a).$ You can see that $E_a$ is linear : $$E_a(f+g)=(f+g)(a)=f(a)+g(a)=E_a(f)+E_a(g),E_a(\lambda f)=(\lambda f)(a)=\lambda f(a)=\lambda E_a(f)$$ and that $E_a$ is continuous : $$|E(f)|=|f(a)|\leq ||f||_{\infty},$$ where $||f||_\infty=\max\limits_{x\in[a,b]}|f(x)|$ defines the same topology as your distance on $C([a,b]).$

Thus, as $\{f\in C([a,b]\,;\,f(a)=0)\}=E_a^{-1}(\{0\})$ then you get your result.