It began with my trying to prove that a uniformly continous function on a bounded subset of the line is bounded. I took the hard route cause I couldn't figure out how to do this directly. I prove that if a real function is uniformly continous on a bounded set E then there exists a continous extension on the closure of E. My first post will be this lemma.(adsbygoogle = window.adsbygoogle || []).push({});

Lemma: Let [tex] E \subset X [/tex] If [tex] x_n \Rightarrow x [/tex] and for each n there is a sequence [tex] y_k \Rightarrow x_n [/tex] with [tex] (y_k) \in E[/tex] then [tex] s_p=y_p^p [/tex] (the latter being the pth term of the pth sequence) is a sequence contained within E that converges to x.

Proof: Consider subsequences of each [tex] y_k^n [/tex] to obtain a new collection of [tex] y_m^n [/tex] with the following property:

[tex] d(y_m^n,y_j^n) \leq \frac{1}{n} [/tex] for all m,j

Put [tex] s_p = y_p^p [/tex] so that [tex] s_p \in E [/tex] and let [tex] \epsilon > 0 [/tex]. Then by the triangle inequality

[tex] d(s_p,x) \leq d(s_p,y_m^p) + d(y_m^p, x_n) + d(x_n,x) < \frac{1}{p} + d(y_m^p,x_n) + d(x_n,x)[/tex]

Because the LHS is independent of n, m and [tex] x_n \rightarrow x [/tex]

taking limits as [tex] n,m \rightarrow \infty [/tex] we obtain

[tex] d(s_p,x) \leq \frac{1}{p} + d(x_p,x_n) [/tex]. Taking n and p large enough we easily see that [tex] s_p \rightarrow x [/tex]

PS: this is my first go at Latex. I will be slow to post the rest! Please give me feedback on both readability and the content.

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# Characterization of Uniform Continuity on the line

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