Uniform continuous function proof Suppose that $f$ is defined and continuous on $[a,b]$. Prove that $f$ is uniformly continuous on $(a,b)$ if and only if it is uniformly continuous on $[a,b]$
I know this is true because the open interval is the` sub interval of the closed one, but I don't know how to prove it formally.
 A: When a function $f:\ [a,b]\to{\mathbb R}$ is continuous then it is automatically uniformly continuous on $[a,b]$, as well as on any subinterval thereof.
Now comes the interesting part of the intended exercice: Assume that the function $f: \ ]a,b[\ \to{\mathbb R}$ is uniformly continuous on the open interval $\ ]a,b[\ $. Prove that it can be extended to a continuous function $\tilde f:\ [a,b]\to{\mathbb R}$.
It suffices to prove that  $\tilde f(a):=\lim_{x\to a+} f(x)$ exists. 
(i) The fast track argument: Given an $\epsilon>0$ there is a $\delta>0$ with $|f(x)-f(y)|<\epsilon$ for all  $x, y\in \ ]a,b[\ $ with $|x-y|<\delta$. In particular we have
$$|f(x)-f(y)|<\epsilon\quad \forall x,\ y\in\ ]a,a+\delta[\ .$$
This is the "Cauchy condition for functions" and already implies the existence of $\tilde f(a)$. 
(ii) In terms of sequences we  argue as follows: Given an $\epsilon>0$ there is a $\delta>0$ such that $$x, y\in \ ]a,b[\quad \wedge \quad |x-y|<\delta\quad\Longrightarrow\quad |f(x)-f(y)|<\epsilon\ .\tag{1}$$ Consider the sequence $x_n:=a+{b-a\over 2n}$. There is an $n_0$ with $x_n<b$ and $|x_m-x_n|<\delta$ for all $m$, $n>n_0$. It follows that $$|f(x_m)-f(x_n|<\epsilon\qquad\forall m,\ n>n_0\ ,$$
which proves that $\bigl(f(x_n)\bigr)_{n\geq1}$ is a Cauchy sequence. Therefore the $\lim_{n\to\infty} f(x_n)=:\eta$ exists. Now $(1)$ implies
$$|f(x)-f(y)|<\epsilon\quad \forall x,\ y\in\ ]a,a+\delta[$$
and therefore
$$|f(x)-f(x_n)|<\epsilon\qquad \forall x\in \ ]a,a+\delta[\ ,\quad \forall n>n_0\ .\tag{2}$$
Keeping $x$ fixed and letting $n\to\infty$ in $(2)$ we obtain
$$|f(x)-\eta|\leq\epsilon\qquad\forall x\in \ ]a,a+\delta[\ .$$
This proves the claim with $\tilde f(a)=\eta$.
