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Prove that $d$ is a metric on the set $X$.

$d_u(f,g) = \sup\{|f(x) - g(x)|: x \in I \}, X = C(I)$ the set of all continuous functions from the closed bounded interval $I = [a,b]$ to $\mathbb{R}$

I don't understand what difference using the supremum of $|f(x) - g(x)|$ makes. $|f(x) - g(x)|$ will always be a real number so its seems taking the suprmemum of it is pointless. Am I missing something here?

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up vote 2 down vote accepted

The problem is that $|f(x) - g(x)|$ is a function of $x$; for each $x$, it returns the difference between $f$ and $g$ at that point. You want it to be a single number for all $x \in I$, that's why you take the supremum. You define the distance between two functions to be the farthest they are ever separated.

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Cheers. Is it the curly brackets $sup\{...\}$ that indicate that I am taking the supremum of a set of values? –  dukenukem Sep 23 '12 at 15:10
    
@dukenukem: Pretty much. Sometimes the supremum is written a little differently, like this: $\displaystyle \sup_{x\in I}|f(x)-g(x)|$. But it's the same thing. –  Javier Badia Sep 23 '12 at 15:21
    
Would taking the max be the same as taking the supremum in this case> –  dukenukem Sep 24 '12 at 6:45
    
@dukenukem Yes, since you are considering continuous functions on a closed and bounded interval, taking the max is the same as taking the supremum (this is the extreme value theorem). –  lazyhaze Sep 24 '12 at 10:23
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