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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.

Kenneth.K
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    Why do you believe that continuity extends to end points? The function $f(x)=\frac1x$ is continuous on the open interval $(0,1)$. Regardless of however one chooses to define it at $x=0$, the function is discontinuous there. – Mark Viola Dec 10 '15 at 04:55
  • Are we possibly assuming that $f$ is continuous on the closed interval, and then asking whether uniform continuity on the open interval implies uniform continuity on the closed interval? – Brian Tung Dec 10 '15 at 04:58
  • Well, my professor said that "uniform continuous functions extend to accumulation points of domain" and I couldn't get the point. I am trying to understand the notion of "extend". – Kenneth.K Dec 10 '15 at 04:59
  • @Kenneth.K In a second I will update my answer with this. By the way this is an example of the so called "XY problem".http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem –  Dec 10 '15 at 05:08

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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 $\hat{f}\restriction\operatorname{dom}(f)=f$.