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Let $X$ be a non empty set and let $C[a,b]$ denote the set of all real or complex valued continuous functions on $X$ with a metric induced by the supremum norm.

How to find open and closed balls in $C[a,b]$? Can we see them geometrically? For example what is an open ball $B(x_0;1)$ i.e. ball centered at $x_0$ with radius $1$ in $C[a,b]$. I can visualize them in $\mathbb R^n$ but when it comes to functional spaces I have no clue how to identify them?

Thanks for helping me.

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I assume $X$ is the interval $[a,b]$? Given a continuous function $f:[a,b]\to\Bbb R$ I would visualize the open ball of radius one around $f$ to be the set of all functions whose graphs exist in the strip between the graphs of $y=f(x)+1$ and $y=f(x)-1$. –  anon May 22 '12 at 19:11
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2 Answers

up vote 10 down vote accepted

Yes, you should think of it just like you think of any other metric space. Every norm $\|\cdot\|$ induces a metric $d(x,y) := \|x-y\|$.

In your example, $$ B(x_0, 1) = \{ f: X \to \mathbb R \Big \vert \|f - x_0\|_\infty < 1 \}$$

In the $\sup$-norm, these $f$ are all functions that are never further away from $x_0$ at any given $x$ in $\mathbb R$. This is what it looks like: enter image description here

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Thanks for answering but still i am not able to see it geometrically. –  srijan May 22 '12 at 19:17
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@srijan Does the picture help? –  Matt N. May 22 '12 at 19:27
    
N I am very much thankful to you. Now things are transparent. –  srijan May 22 '12 at 19:30
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Note also: the graphs of the functions inside the open/closed ball can intersect the black curve, and the boundary of the closed ball (the sphere) are those functions inside whose graphs touch (at least once) one of the two red curves that bound the strip. –  anon May 22 '12 at 19:30
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@srijan Glad I could help : ) –  Matt N. May 22 '12 at 19:31
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So I'm new here and sorry if I have this wrong, but apparently I'm not meant to respond to other answers; yet I also need 50 reputation to add a comment, so I can't do that? Anyway, I believe the above answers are incorrect. Just take the center of an open ball to be the zero function (radius 1) and $f$ to be $2/\pi \arctan(x)$. Then $f(x)$ is always within a distance of 1 from the center (strictly) in R, but $f$ is a distance of 1 from the zero function (exactually 1, so not in the open ball) in the function space, because the supremum of $\{f(x)-0 | x \in R\}$ is indeed 1. In fact f is on the boundry in the function space despite its values never touching 1 or -1 in R.

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OP's question was specifically about $C[a,b]$ not $C(\mathbb R)$ which indeed behaves a bit differently. –  kahen Sep 6 '13 at 19:37
    
$\mathbb{R}$ behaves differently because it isn't bounded; hence it isn't compact. Since $[a,b]$ is compact, any function attains its supremum value somewhere on the interval, thus we won't run into this problem. –  Goos Sep 6 '13 at 19:48
    
I agree with the comments give above. –  srijan Sep 10 '13 at 10:23
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