# C($\mathbb{R}$) is complete under the discrete metric.

Examine if the space $$X=C(\mathbb{R})$$ endowed with the discrete metric $$d_0$$ is complete or not.

We know that $$(X,d_0)$$ is complete if every Cauchy sequence is finally constant. We want to construct a function $$f(x_n),Cauchy$$ such that $$f(x_n)\rightarrow f(x_0)$$ and also $$f$$ is continuous.

Is this gonna work?

If you endow any set with that discrete metric, it becomes a complete metric space, because any Cauchy sequence is constant then, if $$n$$ is large enough.
• I don't know if i used the right word because i tranlated the problem from Greek. But it basically asks you to prove that C($\mathbb{R}$),$d_0$) is complete or not. – argiriskar Oct 11 '18 at 11:36
• This post answers your question. In order for a metric space to be complete, every Cauchy sequence must converges (with respect to the metric defined on the space). Every Cauchy sequence in your space must be constant for $n$ large enough, so it converges. – nicomezi Oct 11 '18 at 11:42