How do I know if $A_1=\{f\in C[0,1]: \inf_{0\leq t \leq 1}f(t)>0\}$ set is open or not? Is the set  $$A_1=\{f\in C[0,1]: \inf_{0\leq t \leq 1}f(t)>0\} $$ open or closed in the space$(C[0,1],\|\cdot\|_\infty)$?
I tried to take $(f_n)_{n\in \mathbb{N}}$ series such that $\forall n\in \mathbb{N}: f_n \in A_1$ and $\inf_{0\leq t \leq 1}f_n(t)\longrightarrow 0$. So I can show (I guess) that it isn't closed, but is it open? 
 A: The set is open.
Hint:
1) Define $\delta:=\inf_{0\leq t\leq 1} f(t)$ and show that $\{g\in C[0,1]~:~\|f-g\|_\infty<\delta\}\subset A$.
2) For each $f\in C[0,1]$ there exists $t_{min}\in[0,1]$ such that $f(t_{min})=\inf_{0\leq t\leq 1} f(t)$, since $f$ is contiuous and $[0,1]$ is compact.
A: 
$A_1$ is not  closed
  Take $f_n \equiv \frac{1}{n}$ see that, $f_n\in A_1$ and $f_n\to0$   then $A_1$ is not closed 
$A_1$ is opened Let $f\in A_1$ that then take $\delta =\min f>0.$ Then  $$B(f,\delta/2) \subset A_1$$

Indeed assume,  $$\delta/2> \|f-g\|_\infty \ge |f(x)-g(x)|~~~~\forall~~x\in [0,1].$$ then for all $x\in [0,1]$ we have
$$\delta/2>  |f(x)-g(x)|\implies g(x)\ge -\delta/2+f(x) \ge-\delta/2+\delta= \delta/2>0 $$
That thus $$\min g >\delta/2>0$$
That is $g\in A_1$ hence this shows that $$B(f,\delta/2) \subset A_1$$ that is $A_1$ is open.
A: Notice that the function $F : C[0,1] \to \mathbb{R}$ defined as $F(f) = \displaystyle\inf_{t\in[0,1]} f(t)$ for $f \in C[0,1]$ is continuous.
Indeed, let $\varepsilon > 0$ and set $\|f - g\|_\infty < \frac{\varepsilon}2$. We have $-\frac{\varepsilon}2 < f(x) - g(x) < \frac{\varepsilon}2$ for all $x \in [0,1]$.
So $g(x) - \frac{\varepsilon}2 < f(x)$ for all $x \in [0,1]$. We conclude $$\inf_{t \in [0,1]} g(t) -\frac{\varepsilon}2 < f(x)$$ for all $x \in [0,1]$. Taking the infimum again, we get $$\inf_{t \in [0,1]} g(t) -\frac{\varepsilon}2 \le \inf_{t\in[0,1]}f(t)$$ so:
$$\inf_{t \in [0,1]} g(t) - \inf_{t\in[0,1]}f(t) \le \frac{\varepsilon}2$$
Similarly we get $$-\frac{\varepsilon}2 \le \inf_{t \in [0,1]} g(t) - \inf_{t\in[0,1]}f(t)$$
so $$\left|\inf_{t \in [0,1]} g(t) - \inf_{t\in[0,1]}f(t)\right| \le \frac{\varepsilon}2 < \varepsilon$$
Now, your set $A_1$ is simply the preimage of an open set by a continuous function $F$:
$$A_1 = F^{-1}\big(\langle 0, +\infty\rangle\big)$$
Hence $A_1$ is open.
We can automatically conclude that $A_1$ is not closed, since the only sets which are both open and closed are $\emptyset$ and $C[0,1]$. This follows because every normed space is connected.
