Need to confirm: Sup Metric $C[0,1]$, question about boundary

Question

For the sup metric, $C[0,1]$. Let $S \subset C[0,1]$ be given by:

$$S=\left\{f:[0,1]\to \mathbb{R} \ : \ 0 \leq f\left(\frac{1}{2}\right)<1\right\}$$

The question is simple: is this set open or closed?

I've had a hard time to visualize a set of functions instead of numbers. Anyway, I think the set is open since it's boundary point or condition is $f(1/2)=1$, which is clearly not in the set $S$. Am I correct? I feel like it could be a tricky question...

I would appreciate if you can confirm my idea. Or correct me if I get it wrong!

Answer 1

It is would be open if both inequalites were strict and close if both were nonstrict, since the evaluation maps $E_x: \mathcal{C}[0,1]\rightarrow\mathbb{R}$ defined by $f\mapsto f(x)$ are continuous.

As shown the set is neither. The constant functon $x\mapsto 1$ is on the boundary of this set and does not lie in it. On the other hand, the constant function $x\mapsto 0$ has no open neighborhood about it.

I erred originally in thinking the inequality was strict on both sides. So, therefore, I complete my answer since I cannot delete it.

Answer 2

It is not a closed set.Hints consider sequence of function $f_n $ in $ C [0, 1] $ defined by $ f_n (x)=1-\frac {1}{n} $for all $x$ belongs to$[0,1]$ Then think its convergence.