# Open Close Sets On Metric Space Help

I am having a tough time, trying to understand $C([0,1])$ the sup metric space. I can even prove that it is complete. However, I encountered the following question and I really really have no idea what this is asking for. OR put it in another way, I had a hard to visualize it. Anyway, I 'd appreciate if you can give me a little little hint. Or anything that I can start with.

To determine whether the following set is open or closed in the space $C([0,1])$

a. $\{f\in C([0,1])\;;\;f(1/2)<6\}$;

b.$\{f\in C([0,1])\;;\;f(1/2)\leq 6\}$;

c. $B(0,1)$.

I don't know what does $f(1/2)<6$ looks like here, are they points? Or functions? I am so confused...Oh and I know a set is open if it doesn't contain any boundary points. And it's closed if it contains all of its boundary points. But, where is the boundary in the example above??

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When there is a strict inequality (resp. large), it is natural to try to prove that the set is open/not closed (resp. closed/not open). Of course, this is not a theorem. But in most standard well behaved spaces, it is a good indication. – 1015 Mar 22 '13 at 3:46

Here is another way: Use continuity with the appropriate function.

Let $\phi: C[0,1] \to \mathbb{R}$ be $\phi(f) = f(\frac{1}{2})$. It is straightforward to show that $\phi$ is continuous (in fact it is Lipschitz continuous with rank one).

Then the sets are a) $\phi^{-1} (\infty,6)$, and b) $\phi^{-1} (\infty,6]$. Since $(\infty,6)$ is open and $(\infty,6]$ is closed, this answers a), b).

For c), let $\eta(f) = \|f\|$. It should be clear that $\eta$ is continuous (in fact, as Julien pointed out, it is Lipschitz continuous with rank one), and $B(0,1) = \eta^{-1} (-\infty, 1)$. Since $(-\infty, 1)$ is open, this answers c).

The following rough illustration may help understanding the sets in question (I am taking the range of the functions to be $\mathbb{R}$):

The blue lines represent continuous functions on $[0,1]$. In a) & b) a function cannot cross the vertical ray at $\frac{1}{2}$, in b) the function can pass through the open circle. In c) a function must not enter the hatched area. In all cases, the norm of the function is the maximum distance from the horizontal zero line.

In all cases, the boundary of the sets in questions are collections of functions.

For a) & b), the boundary is the collection of functions $f$ that satisfy $f(\frac{1}{2}) = 6$.

For c), the boundary is the collection of functions $f$ that satisfy $\max_{t \in [0,1]} |f(t)| = 1$, ie. $f(t) \in [-1,1]$ for all $t$ and $f(t_0) \in \{-1,+1\}$ for at least one $t_0$.

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Nicely done, +1. Maybe instead of "it should be clear", you could say (in the same spirit as in your argument for a. and b.) that $\eta$ is $1$-Lipschitz by the triangular inequality of a norm. – 1015 Mar 22 '13 at 4:14
@julien: Thanks! I included your suggestion. – copper.hat Mar 22 '13 at 4:17

I think that by $\{ f(1/2)<6\}$ what it actually means is the set $$\{f\in C[0,1] \text{ s.t. } f(1/2)<6 \}.$$ You can also prove that a set is closed $A$ if you take a Cauchy sequence $(a_n)\subset A$ and prove that its limit is also in A. A set, $A$, is open if for every element $x \in A$ there is a neighborhood, $x \in U$, such that $U \subset A$.

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