A question on set C[0,1] Let $A=\{f \in C[0,1]:f(x)\neq 0 \text{ for all } x\in [0,1]\}$ where $C[0,1]$ is the set of continuous functions $f:[0,1]→\mathbb R$ with sup norm. Is set $A$ open? Is $A$ closed?


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*My work: As zero function doesn't belong to $A$, it can not be closed. To check that whether it is open, I consider a continuous function $\|\cdot\|_\infty$ which maps $A$ onto $(0,a]$ where $a$ is finite real number suggesting that $A$ is not open too. Am I correct? 

 A: Proof Outline:
($A$ is open) Choose any function $f$ in $A.$ Consider the graph of the function $f.$ This graph doesn't cross the $x$-axis (why?) and this graph doesn't get arbitrarily close to the $x$-axis (why?), so you can put an $\epsilon$-tube about the graph that doesn't contain any points of the $x$-axis. Since this $\epsilon$-tube contains all functions in $A$ whose sup-distance from $f$ is less than $\epsilon$ (why?), it follows that there exists an $\epsilon$-ball $B(f,\epsilon)$ about $f$ such that $B(f,\epsilon) \subseteq A,$ which tells us . . .
($A$ is not closed) If $A$ were closed, and assuming the previous result that $A$ is open, then both $A$ and $C[0,1]-A$ would be nonempty open subsets of $C[0,1]$ whose union is $C[0,1],$ which contradicts the fact that $C[0,1]$ is connected (indeed, $C[0,1]$ is even convex).
A: Here are hints for each part of your question:
Is $A$ open? 
For any $f$ in $A$, we know $F$ is continuous on the compact set $[0,1]$, so $f$ attains a minimum, i.e. there is an $x$ in $[0,1]$ such that $f(x)\leq f(c)$ for all $c\in[0,1]$. Further, we have $f(x)>0$. What can you say about the ball of radius $\frac{f(x)}{2}$ centered at $f$?
Is $A$ closed?
What can you say about the sequence of constant functions in $A$, $\{f_n\}$, with 
$$
f_n(x)=\frac{1}{n}
$$
for all $x$ in $[0,1]$? Does this sequence converge? If so, to what? 
A: Let $f$ be some member of this set of functions. A continuous function on a compact domain attains its maximum, so there is some point at which $1/|f(x)|$ is as big as it gets.  Let $M$ be the maximum value of $1/|f(x)|$.  Then $|f(x)|\ge 1/M$ for every $x\in[0,1]$.  So look at the set of all functions at distance less than $1/(2M)$ from $f$ in the sup norm.  The triangle inequality shows that all of those differ everywhere from $0$ by more than $1/(2M)$, so they are nowhere $0$.  Thus some open ball about $f$ is entirely within this set of functions.  Hence it is open.
