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Let $C[a,b]$ denote all continuous functions $:[a,b] \rightarrow \mathbb{R}$, with the metric

$$d(f,g)=\sup |f(x) - g(x)| \text{ for } a\leq x \leq b.$$

Show that $d(f,g)$ is a metric space.

I have started with this: Since continuos function is bounded, so i use the same proof from it: " $B[a,b]$ denote for all bounded functions $:[a,b] \rightarrow \mathbb{R}$, with

$d(f,g)=\sup |f(x) - g(x)|$ for $a\leq x \leq b$ "

to prove those (continuos function is metric space). Is it right? Or there is another way?

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It sounds like you already know that the set of bounded functions, $B[a,b]$, is a metric space under the metric $d$. In that case, you made a very nice observation that since continuous functions are bounded, $C[a,b]$ is contained in $B[a,b]$. And any subset of a metric space is a metric space! Well done.

You can also show it directly -- nonnegativity and symmetry are almost immediate, and then you only have to show triangle inequality.

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  • $\begingroup$ for the direct proof, i think the proof is like bounded function with metric d? $\endgroup$ – Reza Habibi Sep 13 '16 at 9:55
  • $\begingroup$ @RezaHabibi Yes -- it's going to be the exact same proof :) $\endgroup$ – 6005 Sep 13 '16 at 9:56
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if C(X,Y) denote all continuous functions :X→Y, Y complet metric space, X compact

, with the metric

d(f,g)=sup|f(x)−g(x)| for x in X. so C(X,Y) with uniform norm d, complet metric space or Banach space? and if Y only metric space, do you know show c(X,Y) Banach space?

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  • $\begingroup$ Please use MathJax to format. $\endgroup$ – Saad Jan 30 '18 at 11:43

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