Are the following sets closed in $C[0,1]$? I am trying to figure out if the following two sets are closed in $C[0,1]$ with $||f||_{\infty}$ and $||f||_1$.

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*$A = \{f \in C[0,1]: f(\frac{1}{2}) = 0\}$

*$B = \{f \in C[0,1]: f(x) \geq 0, \forall x \in [0,1] \}$
Here $C[0,1]$ denotes the set of all real-valued continuous functions over the interval $[0,1]$. For $A$, I do think it is closed with respect to $||f||_{\infty}$. The reason is that since $C[0,1]$ is Banach with the infinity norm, then any subset of it is closed if and only if it is a complete metric space, with the metric induced the norm of $C[0,1]$. However, I am not too sure if $A$ is closed with respect to $||f||_1$. I want to say it is not closed but I do not know how to form my argument. I cannot use the contrapositive of the statement I mentioned above since $C[0,1]$ with $||f||_1$ is not Banach. I think $B$ is closed with respect to the infinity norm, but not the 1-norm.
Any help would be appreciated!
Thanks guys.
 A: $A$ is closed in $\| \cdot\|_\infty$  (since uniform convergence implies pointwise convergence) but it is not closed in $\| \cdot\|_1$: Consider functions $f_n$ that satisfy
$f_n(0)=f_n(1/2-1/n)=1$ and $f_n(1/2)=0$ and $f_n(1/2+1/n)=f_n(1)=1$ and  interpolate linearly between the specified points. These $f_n$ converge in  $\| \cdot\|_1$ to the constant function $1$.
$B$ is closed in $\| \cdot\|_\infty$ and in  $\| \cdot\|_1$. It suffices to prove that $B^c$ is open in $\| \cdot\|_1$.
Suppose $f \notin B$. Then for some $\epsilon>0$ and $x_0 \in [0,1]$, we have $f(x_0)<-2\epsilon$, so by continuity, there exists $\delta>0$ such that $f(x)<-\epsilon$ whenever $|x-x_0|<\delta$.
This implies that the $\epsilon \delta$ ball centered at $f$ in the norm $\| \cdot\|_1$ is disjoint from $B$.
A: A is closed in $∥\cdot∥_{\infty}$ because of the uniform convergence theorem.
If you know about bump functions and how you can get smooth functions very close to the indicator functions, I can layout a way to find non-continuous examples in the closure for the $||\cdot||_{1}$.
Let I(A) be an indicator function for a set A.
You can consider the sequence of functions $g_{n} = I([0,\frac{1}{n}])$ each function $g_{n}$ is in the closure of both of the sets under $||\cdot||_{1}$ because of bump functions. So the discontinuous function $g(x)=0$ for $x\neq 0$ and $g(0)=1$ is in the closure of both the sets.
If you don't know about bump functions then we can make this construction $h_{n}(x) = 1$ for $x\in [0,\frac{1}{n}]$. $h_{n}(x) = 1 - n(x - \frac{1}{n})$ for $x\in [\frac{1}{n},\frac{2}{n}]$. $h_{n}(x) = 0$ for $x\in [\frac{2}{n}, 1]$. And consider the sequence for  $n>3$.
