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 complexnumber May16-10 07:04 AM

Determine if this subset is compact

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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 May16-10 10:53 AM

Re: Determine if this subset is compact

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 May17-10 06:12 AM

Re: Determine if this subset is compact

Quote:
 Quote by Landau (Post 2719565) 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?

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