Is space $C[0,1]$ is Hausdorff?

Question

Is space $C[0,1]$ is Hausdorff?

I just started studying metric spaces. I encountered the Hausdroff space definition. I wanted to construct a space which does not have that property.

I thought of $C[0,1]$ as candidate. Because we can define 2 distant function which agree at single point.

SO It is not possible to haved isjoint open balls.

Is my argument correct?

Any Help will be appreciated

Answer 1

The set $C[0,1]$ by itself is not a topological space. If you are taking about this set with the standard metric $d(f,g)=\sup \{|f(x)-g(x)|:0\leq x \leq 1\}$ then it is Hausdorff because any metric space is Hausdorff. [ If $x \neq y$ and $r=d(x,y)$ then $B(x,\frac r 2)$ and $B(y,\frac r 2)$ are disjoint open sets containing $x$ and $y$ respectively.