uniform continuity problem with open and closed set let say we have a $f: (a,b) \to R   $  is uniformly continuous on $(a,b)$
lets show the  f is continuous on [a,b]
I'm not sure how to approach this question 
since worst case for [a,b] is that a and b is an isolated point 
but even though a,b is the isolated point its continuous 
so  isolated(a) + (a,b) + isolated(b) = continuous + uniform continuous + continuous 
= continuous , I do not think its right... how to approach this?
 A: The actual theorem is that if $f$ is uniformly continuous on $(a,b)$ then it can be extended to a continuous function on $[a,b]$. Indeed Cauchy sequences seem to be the easiest approach, as Deven Ware suggests.
A: Let $D$ be dense in a metric space $(X,d)$.
Let $(Y,d')$ be a complete metric space.
Let $f\colon (D,d)\to(Y,d')$ be uniformly continuous.
Let $x\in X$. Then there is a sequence $(x_n)$ in $D$ with limit $x$, which is therefore a Cauchy sequence by the triangle inequality.
For any $\epsilon>0$, there is a $\delta>0$ such that whenever $d(p,q)<\delta$, $d'(f(p),f(q))<\epsilon$. Since $(x_n)$ is a Cauchy sequence, there is an $N\in\Bbb N$ such that when $p,q\ge N$, $d(x_p,x_q)<\delta$ and thus $d'(f(p),f(q))<\epsilon$. Therefore $(f(x_n))$ is also a Cauchy sequence. Since $(Y,d')$ is complete, this sequence converges. By interspersing sequences, it's easy to see that all sequences converging to $x$ give the same limit. Let $g$ be the function derived from $f$ in this fashion. We must still show that $g$ is continuous. I believe a somewhat trickier interspersing argument will get the job done, but it't time for me to go home now.
