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I'm having some trouble understanding the proof for uniform continuity. I'm using the book Introduction to Real Analysis by Bartle and Sherbert 3rd Edition, page 138, if anyone has access to it. The Theorem states:
I understand the proof up to the part where it says it is clear that [tex](u_n_k)[/tex] also converges to z. I don't know why showing this inequality shows that the subsequence converges to z. Also why is the difference between the two sequences less then 1/n?
Let I be a closed bounded interval and let f:I->R be continuous on I. Then f is uniformly continuous on I.
Proof(By Contradiction):
If f is not uniformly continuous on I then, by the preceding result(Nonuniform Continuity Criteria), there exists [tex]\epsilon_{0}>0[/tex] and two sequences [tex](x_n)[/tex] and [tex](u_n)[/tex] in I such that [tex]|x_n-u_n|<1/n[/tex] and [tex]|f(x_n)-f(u_n)|\geq\epsilon_{0}[/tex] for all [tex]n\in\mathbb{N}[/tex]. Since I is bounded, the sequence [tex](x_n)[/tex] is bounded; by the Bolzano-Weierstrass Theorem there is a subsequence [tex](x_n_k)[/tex] of [tex](x_n)[/tex] that converges to an element z. Since I is closed, the limit z belongs to I. It is clear that the corresponding subsequence [tex](u_n_k)[/tex] also converges to z, since
[tex]|u_n_k - z|\leq|u_n_k - x_n_k| + |x_n_k - z|[/tex].Now if f is continuous at the point z, then both of the sequences [tex](f(x_n_k))[/tex] and [tex](f(u_n_k))[/tex] must converge to [tex[f(z)[/tex]. But this is not possible since
[tex]|f(x_n)-f(u_n)|\geq\epsilon_0[/tex]
I understand the proof up to the part where it says it is clear that [tex](u_n_k)[/tex] also converges to z. I don't know why showing this inequality shows that the subsequence converges to z. Also why is the difference between the two sequences less then 1/n?