If $f$ is uniformly continuous on $(a,b)$, is $f$ uniformly continuous on $[a.b]$?
Since $f$ is uniformly continuous on the open interval and $a$, $b$ are accumulation points of the open interval, $f$ is continuous on $a$ and $b$. However, I wonder if $f$ is uniformly continuous on $[a,b]$? My professor said that we can use Tietze theorem to extend the uniformly continuous from open interval to closed interval. I can only prove continuous on the closed interval.
This is not true. Consider:
$$f(x)=\begin{cases} 1 \text{ for } x\in (0,1]\\ 0 \text{ for } x=0\end{cases}$$
However in your comment you ask what does "uniform continuous functions extend to accumulation points of domain" mean. This just means that if $x$ is an accumulation point of the domain of $f$ then there is a unique uniformly continuous function $\hat{f}$ on $\operatorname{dom}(f) \cup \{x\}$ such that $\hat{f}\restriction\operatorname{dom}(f)=f$.