I would like to show that $C[0,1] \setminus X = \{f(x) \in C[0,1] : f(x) < x\}$, where our metric is $d_{\infty}(f,g) = \text{sup}\{|f(x)-g(x)| : x \in [0,1]\}$, is open which would imply $X$ is closed but am I unsure as to how to go about showing this.
Any tips would be much appreciated.
The set $Y=\{f\in C[0,1]:\textrm{ }f(x)<x\textrm{ } \forall x\in [0,1]\}$ it's open because, take $f$ on this set, then $f(x)<x$ $\forall x\in [0,1]$ take $\epsilon=\min_{[0,1]}x-f(x)$, then suppose that $d_{\infty}(f,g)<\frac{\epsilon}{2}$ then $$ |f(x)-g(x)|<\frac{\epsilon}{2}, \forall x\in [0,1] $$
therefore $g(x)-x=g(x)-f(x)+f(x)-x<\frac{\epsilon}{2}-\epsilon<0$ then $g(x)<x$ $\forall g\in B(f,\frac{\epsilon}{2})$, thus $B(f,\frac{\epsilon}{2})\subset Y$
Take some $f\in C[0,1]\setminus X$. Then $f(x)<x$ on $[0,1]$. Since the identity function $i:x\mapsto x$ is continuous we can consider $\epsilon=d_\infty(f,i)$. Now consider $B(f,\frac{\epsilon}{2})=\{g\in C[0,1]:\sup\{|f(x)-g(x)|:x\in[0,1]\}<\frac{\epsilon}{2}\}$ and show that every $g$ in $B(f,\frac{\epsilon}{2})$ is also in $C[0,1]\setminus X$.
