Consider a sequence of continuous functions $f_n : (0,1) \rightarrow \mathbb R $ such that $f_n$ converges uniformly to $f$ on $[a,b]$ for any $0\le a \le b \le 1$. Prove that f is continuous on (0,1)
Proof: I know that if $f_n : [0,1] \rightarrow \mathbb R$ and $f: [0,1] \rightarrow \mathbb R$, then $f$ is continuous on the interval [0,1]. However, in this problem, I have to prove that $f$ is continuous on on the (0,1) and $f_n$ is defined on the open interval. So how do i approach this problem?
I can show that $f$ is continuous on $[a,b]$, and however, i got stuck how to prove that on open interval. Or can i just show that $f$ is continuous on $[a,b]$, then since $[a,b] \in (0,1)$, then $f$ is continuous on $(0,1)$.
Hints and helps are greatly appreciated!