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I know that the space of continuous bounded function is complete in the sup norm. However, I am confused that if $X\subset R$ is not compact, then is the space of continuous function on $X$, denoting by $C(X)$ still complete?

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    With respect to what norm? – 2017-02-11
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    If $X$ is not compact, the "supremum norm" is not defined on _all_ of $\mathscr{C}_{\Bbb R}(X).$ You need to restrict to $\mathscr{C}_{\Bbb R}^b(X),$ continuous and bounded. – 2017-02-11
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    If $X$ is not bounded, $t \mapsto t$ is an unbounded continuous function; if $X$ is not closed then for a cluster value $a$ of $X$ not belonging to $X,$ $t \mapsto \dfrac{1}{t - a}$ is continuous and unbounded. So really you need to restrict the space. – 2017-02-11

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The issue is not whether the norm is complete but whether it is defined at all. If $f$ is a continuous function on $X$ that is not bounded, then the sup norm of $f$ is infinite (by definition of $f$ not being bounded), so the sup norm does not even define a norm on all of $C(X)$.

However, this isn't an insurmountable problem for some purposes. You can still talk about convergence of sequences, or Cauchy sequences, or the topology induced by a "norm", even if that "norm" is sometimes infinite. And in fact the sup "norm" on $C(X)$ is complete in the sense that every Cauchy sequence converges, even if you allow unbounded functions: the proof is exactly the same as it is when you restrict to bounded functions.

(Note, however, that the topology induced by a possibly infinite "norm" can be quite pathological compared to what you are used to in genuine Banach spaces. For instance, scalar multiplication is not continuous on $C(X)$: if $f$ is an unbounded continuous function on $X$, then $cf$ is also unbounded for any nonzero scalar $c$, and so has infinite norm. This means that $cf$ does not converge to $0\cdot f=0$ as $c$ approaches $0$!)