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 Theorem: A function that is uniformly continous on a bounded subset of the line,E, has a continous extension on the closure of E. Proof: We begin by showing that for each $$y\in E' , \lim_{\substack{x\rightarrow y}} f(x)$$ exists. Consider $$x_n \rightarrow y$$ with $$x_n \in E$$ for all n. We show $$f(x_n)$$ is Cauchy and therefore convergent b/c the line is complete. Fix $$\epsilon > 0$$ because f is uniformly continous there is a $$\delta > 0$$ so that when $$d(x_n,x_m) < \delta$$ we also have $$d(f(x_n),f(x_m)) < \epsilon$$. Because $$(x_n)$$ is convergent it is cauchy so we see that $$f(x_n)$$ is cauchy and therefore convergent. Taking any other $$y_n \rightarrow x$$ then by the triangle inequality: $$d(y_n,x_m) \leq d(y_n,x) + d(x_n,x)$$. So the LHS can be made small for all but finitely many terms. Put $$\lim_{\substack{n\rightarrow\infty}} f(x_n) = A$$. Using the triangle inequality again $$d(f(y_n),A) \leq d(f(y_n),f(x_n)) + d(f(x_n),A)$$. Fix $$\epsilon > 0$$ then there is a $$\delta > 0$$ so that when $$d(x_n,y_m)<\delta$$ we also have $$d(f(x_n),f(y_n))<\frac{\epsilon}{2}$$. It is clear that $$d(f(y_n),A) < \epsilon$$ for all but finitely many n. We have shown that $$\lim_{\substack{x\rightarrow y}} f(x)$$ exists for all $$y \in E'$$. So define our continous extension as $$g(x)=f(x)$$ for $$x \in E$$ and $$g(y)=\lim_{\substack{x\rightarrow y}} f(x)$$ for $$y \in E'$$. We now prove g is continous on E. Take a sequence $$x_n \rightarrow x$$ and WLOG $$x_n \in E'$$ for all n. For each of these $$x_n$$ there is a $$y_k \rightarrow x_n$$ with $$y_k \in E$$. Define $$(s_p)$$ as in the lemma above. Then $$s_p \rightarrow x$$, so $$g(s_p) \rightarrow g(x)$$ because $$s_p \in E$$. By the triangle inequality $$d(g(x_m),g(x_n)) \leq d(g(x_m),g(s_m)) + d(g(s_m),g(s_n)) + d(g(s_n),g(x_n))$$ Using the triangle inequality on the first term on the RHS we obtain $$d(g(x_m),g(s_m)) \leq d(g(y_i^m),g(s_m)) + d(g(y_i^m),g(x_m))$$ and since $$y_i^m \rightarrow x_m, y_i^m \in E$$ we also have that $$g(y_i^m) \rightarrow g(x_m)$$. Since the LHS is independent of i we see that $$d(g(x_m),g(s_m)) \leq d(g(y_i^m),g(s_m))$$. Now because g is uniformly continuous on E and $$y_i^m,s_m \in E$$ we can make the RHS and thus the LHS small if m is large enough. This is because the sequences $$(y_k^n)$$ were created as above in the lemma, that is $$d(y_k^n,y_j^n) \leq \frac{1}{n}$$ for all k,j. Similarly we see that $$d(g(s_n),g(x_n))$$ goes to 0. Since $$g(s_n)$$ converges it is cauchy and the middle term goes to zero as well. Since these three values on the RHS goto zero as $$m,n \rightarrow \infty$$ so does the LHS and we have that $$g(x_n)$$ is cauchy. Referring to the lemma above we see that $$g(x_n) \rightarrow \lim_{\substack{p\rightarrow \infty}} g(s_p)=g(x)$$ Hence g is continuous on the closure of E. QED