Suppose $f:(a,b) \rightarrow \mathbb{R}$ is uniformly continuous. Show $f$ can be extended uniquely to a continuous function on $[a,b]$. The proof I have sketched below seems a bit messy - is it correct and is there a simpler way to prove this? Thank you - self studying so not always clear if proofs are right.
Suppose $x_n \rightarrow b$ and consider the sequence $f(x_n)$. Because $f$ is uniformly continuous on $(a,b)$ we have for every $\epsilon>0$ $\exists \delta >0$ such that $|x-y|<\delta \rightarrow |f(x)-f(y)| < \epsilon$. Since $x_n$ converges to $b$ it is cauchy so that for every $\delta >0$ there exists $N>0$ so that when $n, m > N$, $|x_n - x_m|<\delta \rightarrow |f(x_n)-f(x_m)| < \epsilon$ so $f(x_n)$ is Cauchy and converges to a limit which we can set as $L=f(b)$.
Suppose there is another sequence $y_m \rightarrow b$ with limit of $f(y_m)=L'\neq L$. Then $\exists \epsilon > 0$ such that $|L-L'|>\epsilon$. Since $f$ is uniformly continuous we can find a $\delta'$ such that $|x_n-y_m| < \delta' \rightarrow |f(x_n)-f(y_m)| < \frac{\epsilon}{3}$. We can guarentee the existence of such a $\delta'$, for the convergence of $x_n, y_m \rightarrow \exists N_x, N_y > 0$ such that $|x_n - y_m| \leq |x_n-b| + |y_m-b| < \frac{\delta'}{2} + \frac{\delta'}{2} = \delta'$ whenever $n>N_x, m>N_y$. We can find a $\delta_x$ and corresponding $N_x'$ such that $n>N_x' \rightarrow |x_n - b| < \delta_x \rightarrow |f(x_n) - L|< \epsilon/3$ and a similar $\delta_y, N_y'$ for $y_m$. Setting $N=max(N_x, N_x', N_y, N_y')$ and $\delta = min(\delta',\delta_x, \delta_y)$. Then $n,m>N \rightarrow \epsilon < |L-L'| \leq |f(x_n) -L| + |f(x_n)-f(y_m)|+ |f(y_m) - L'| < \epsilon$ which is a contradiction and establishes uniqueness of the extension to $(a,b]$. Similarly we can extend $f$ to $[a,b]$.