# How to show that $(C^0((a,b)), d_\infty)$ is not a metric space

Let $d_\infty:C^0([a,b]) \times C^0([a,b]) \to [0,\infty)$ be defined as

$$d_\infty(f,g)=\sup\limits_{x \in [a,b]} \left\{ |f(x) - g(x)| \right\}$$

I have already shown that $(C^0([a,b]), d_\infty)$ is a metric space.

How can I show that $(C^0((a,b)), d_\infty)$ is not a metric space? I'm confused as I did not use the compactness of the intervall when proving that $(C^0([a,b]), d_\infty)$ is a metric space.

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Check that $d_\infty(f,g)$ is actually defined for all $f,g\in C((a,b))$. –  daw Apr 9 at 8:11

Hint: If $\displaystyle f : x \mapsto \frac{1}{x-a}$, what is $d_{\infty}(f,0)$?
I think it would be $d_\infty(f,0)=\sup\limits_{x \in [a,b]} \left\{ |\frac{1}{x-a} - 0| \right\}$, but how does this help? –  sj134 Apr 9 at 9:04
It is rather $d_{\infty}(f,0)= \sup\limits_{x \in (a,b)} \left| \frac{1}{x-a}-0 \right|$. Can you evaluate this number? –  Seirios Apr 9 at 9:26
I cannot and I do not know what is the benefit of knowing the number unless the number would be outside of $[0,\infty)$ where it would not be defined... –  sj134 Apr 9 at 9:45
You have to slightly modify the definition of $d_{\infty}$, taking the sup over $(a,b)$ instead of $[a,b]$, otherwise $d_{\infty}$ would not be defined on $C(a,b)$; it is more or less implicit. Then, by definition, a distance has finite values, that is its image is included into $[0,+ \infty)$. Here, $d_{\infty}(f,0)=+ \infty$. –  Seirios Apr 9 at 11:12
So from the definition it would have to be $[0 \le f(x) < \infty)$ but as $\infty \notin [0, \infty)$ it violates the definition of the modulus (and metric axioms)? How is the value/sup of $\frac1{x-a} = \infty$? –  sj134 Apr 9 at 13:20