Continuous Extension Theorem in real analysis How to prove:
if $f: (a,b) \rightarrow R$ is uniformly continuous, then $f$ can be extended to a continuous function $F: [a,b] \rightarrow R$.
It's suffice to show that $f$ can  be extended to a continuous function $G: [a,b) \rightarrow R$.
I saw similar theorem (Tietze extension theorem) in topology books, but I'm not very familiar with lots of topology concepts. Also I've read one proof for this, it uses Cauchy sequence and Cauchy Completeness Theorem, which I'm also not that comfortable with. Can you think of any other proof for this using methods from uniform continuity, limits, etc. in real analysis? Thanks! :)
 A: There are two parts to this statement. The first half does not require completeness. 
Claim If $f:(a,b)\to\mathbb{R}$ is uniformly continuous, $\limsup_{x\to a^+} f \leq \liminf_{x\to a^+} f < \infty$. 
Sketch of proof First we show that $\limsup f$ must be bounded. Suppose it were not. Let $\delta$ be some small number such that $\delta < b-a$. Consider $x_0 = a + \delta$, and $y_0 = f(x_0)$. Construct a sequence $(x_i,y_i)$ in the following manner. Pick $x_i < a + \frac12 (x_{i-1} - a)$ such that $y_i = f(x_i) > 2 y_{i-1}$. (Notice that if this sequence terminates after finitely many terms and cannot be continued, you have will have that $f|_{(a,\frac12(x_N + a))}$ is bounded.) However, we have that $|x_{i+1} - x_{i}| < \frac{\delta}{2^i}$ while $|y_{i+1} - y_{i}| > 2^i y_0$. So it is easy to see that by taking $i\nearrow \infty$ you get a contradiction to uniform continuity. 
Similarly $f$ must be bounded below. Now suppose the $\limsup f > \liminf f$ are not equal. Then by a similar argument, you can construct an alternating sequence $x_1, x'_1, x_2, x'_2$ such that $2 |x'_i - a| < |x_i - a| < \frac12 |x'_{i-1} - a|$ and $f(x_i) \to \limsup f$ while $f(x'_i) \to\liminf f$. Then you have that for $0 < \epsilon < \limsup f -\liminf f$, for any $\delta$, there exists some $x_i$ such with a point $x'_i$ within $\delta$ of it taking value more than $\epsilon$ away under $f$, contradicting uniform continuity. Q.E.D. 
Like Nate said in his comments, now you are required to use the completeness of $\mathbb{R}$. By using the completeness you can say that since $\limsup f \leq \liminf f$, the limit $\lim_{x\to a^+}f$ exists, so setting $f(a) = \lim_{x\to a^+} f$ you get a continuous function. 
For illustration, imagine instead of you have the function $f:(0,1)\to \mathbb{R}^2\setminus \{ 0\}$ given by 
$$ f(s) = (s \cos s, s\sin s) $$
This function is uniformly continuous on $(0,1)$, but does not have continuous extension to $[0,1)$ with codomain $\mathbb{R}^2\setminus \{ 0\}$. 
A: If you have Cauchy's criterion for functions at your disposal there is a simpler proof: Given an $\epsilon>0$, by uniform continuity there is a $\delta>0$ with $|f(x)-f(x')|<\epsilon$ for all $x, x'\in\ ]a,a+\delta[\>$. Since $\epsilon$ was arbitrary it follows by Cauchy's criterion that $\lim_{x\to a+} f(x)=:f(a)$ exists, and similarly at $b$.
A: $\Rightarrow$ Suppose that $f$ is uniformly continuous on $I=(a,b)$. Let $(x_n)$ be a sequence in $(a,b)$ that converges to $a$. Then since $(x_n)$ is a convergent sequence, it must also be a Cauchy sequence. which implies that $(f(x_n))$ is also a Cauchy sequence, and so $(f(x_n))$ must converge in $\mathbb{R}$, that is $\lim _{n\to \infty} f(x_n)=L$ for some $L\in \mathbb{R}$.
Let $y(n)$ be any other sequence converging to $a$. then $\lim _ {n\to \infty} (x(n)-y(n))=0$
 So by using uniform continuity of $f$ we have
  $\lim _{n\to \infty} f(y_n)=\lim [f(y_n)-f(x_n)+f(x_n)]=\lim [f(y_n)-f(x_n)]+\lim f(x_n)=0+L=L$.
So for every sequence $(y_n)$ in $(a,b)$ that converges to $a$, we have that $(f(y_n))$ converges to $L$. Therefore by the Sequential Criterion for Limits, we have that $f$ has the limit L at the point $a$. Therefore, define $f(a)=L$ and so $f$ is continuous at $a$. Using same argument for $b$ we can prove the claim of theorem.
