connectedness and disconnectednes of special functions space Let $t_0 \in [0,1]$ fixed point. show that:


*

*$ \{ f | f(t_0) \neq 0 \} \subset C([0,1],\mathbb{R})$ is disconnected

*$ \{ f | f(t_0) \neq 0 \} \subset C([0,1],\mathbb{C})$ is path-connected


where C(X,Y) stands for all continuous functions from X to Y 
Thanks
 A: For the first, take $A_1=\{f \mid f(t_0) >0\}$ and $A_2=\{g \mid g(t_0) <0\}$ and prove this is a separation.
For the second, I give you a hint: you can "go around" $0$ :).
A: Define $F_x:(C[0,1],\mathbb{R}) \to \mathbb{R}, F_x(f) := f(x). $
$F_x$ are bounded linear functionals (we are using the sup metric, so $|F_x(f)| \leq \sup|f| = |f|$), so in particular it is continuous.
So, the sets $A := \{f : f(x) < 0 \}  B:= \{f : f(x) > 0 \}$ are disjoint.
And further $F_x^{-1}(-\infty,0) = A, F_x^{-1}(0,\infty) = B $ are open, so we are done with the first part.
Second part. 
Let $f,g \in A := \{l \in (C[0,1],\mathbb{C}) : l(z) \neq 0 \}$.
Let $K = \{q \in (C[0,1],\mathbb{C}) : \forall z \in [0,1] q(z) = k, k \neq 0  \}$. 
We know that $K \cong \mathbb{C} \backslash\ \{ 0\}$. So they are both path connected.
Let $a := f(z), b := g(z)$.
Let $h \in A $
Define $H_l : [0,1] \to (C[0,1],\mathbb{C})$ where $ H_l(t)(\omega) := at + (1-t)l(\omega) $
As $h \in A:$
$H_h(t)(z) = at + (1-t)h(z) = at + (1-t)a = a \neq 0.$
So $f,g$ are path connected to $K$ in $A$, and $K$ is path connected, so $A$ is path connected QED
