Determine if this subset is compact

[complexnumber]

Homework Statement

Let $$(X,d) = (C[0,1], d_\infty)$$, $$S_1$$ is the set of constant
functions in $$B(0,1)$$, and $$S_2 = \{ f \in C[0,1] | \norm{f}_\infty = 1\}$$.

Are $$S_1$$ and $$S_2$$ compact?

Homework Equations

The Attempt at a Solution

I am trying to use the Arzela - Ascoli theorem. For $$S_1$$, the set of functions with value in the ball (assuming that's what the question meant) $$B(0,1)$$ are bounded. They are also equicontinuous at all $$x \in [0,1]$$. How do I show if the subset is closed or not?

For $$S_2$$, how does the norm $$||f||_\infty = 1$$ determine if the set is closed, bounded and equicontinuous? What is the norm $$||f||_\infty = 1$$ defined as?

 

[Landau]

Well, first you have to understand the notation and definitions. $d_\infty$ is just the metric induced by the supremum norm:

$$\|f\|_\infty:=\sup_{x\in[0,1]} |f(x)|$$

 

[complexnumber]

Well, first you have to understand the notation and definitions. $d_\infty$ is just the metric induced by the supremum norm:

$$\|f\|_\infty:=\sup_{x\in[0,1]} |f(x)|$$

$$S_1$$ is not closed because the function $$f = 0$$ is a limit point
outside $$S_1$$. Therefore $$S_1$$ is not compact.


For $$S_2$$, the metric space $$d_\infty(f,g) := \norm{f - g}_\infty$$ means that it is bounded, however it does not make $$S_2$$ equicontinuous. Is the subset closed?