Why Is Uniform Continuity Proven by Contradiction?

In summary: So we say "let's pick an \epsilon>0, and see what happens when we take n large enough."In summary, the proof for uniform continuity on a closed bounded interval states that if f is not uniformly continuous on I, then there exists an epsilon greater than 0 and two sequences (x_n) and (u_n) in I such that |x_n-u_n|<1/n and |f(x_n)-f(u_n)| is greater than or equal to epsilon for all n. By the Bolzano-Weierstrass Theorem, a subsequence of (x_n) converges to an element z in I, and since I is closed, the corresponding subsequence (u_n_k) also converges
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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:

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?
 
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Let [tex] (x_{n_k}) [/tex] converge to z. Then for each [tex]\epsilon>0[/tex], there exists a K such that [tex] |x_{n_k}-z| < \epsilon/2 [/tex] for all k>K, by definition of convergence. Now, by assumption, we have that [tex] |x_n - u_n| < 1/n [/tex]. Therefore, we may pick M>K such that [tex] |x_{n_k} - u_{n_k}| < \epsilon/2 [/tex] for all k>M. This implies that for each [tex]\epsilon>0[/tex], there exists N (=M) such that [tex]|u_{n_k}-z| \le |u_{n_k} - x_{n_k}| + |x_{n_k}-z| < \epsilon/2 + \epsilon/2 = \epsilon [/tex]. It's rather intuitive, if you think about it hard enough. If two sequences get arbitrarily close to each other as [tex]n\to \infty[/tex], and moreover, one of the sequences converges to an element z, then the other sequence will as well.

The reason that the distance between the two sequences is less than 1/n is because that's how we make them (i.e., it's an assumption). We assume that f is NOT uniformly continuous. Uniform continuity means that for every [tex]\epsilon > 0 [/tex], and every two sequences [tex]x_n, u_n[/tex], we can find an N such that [tex] |x_n - u_n| < 1/n [/tex] implies [tex]|f(x_n)-f(u_n)|< \epsilon[/tex] for all n>N. The converse of this statement (that is, NON-uniform continuity), says that we CAN find an [tex]\epsilon>0[/tex] such that no matter how close the sequences get, the differences of their images under f are still going to be greater than [tex]\epsilon[/tex]. The 1/n measures how close the two sequences get, and we want to see what happens when they get arbitrarily close.
 
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Related to Why Is Uniform Continuity Proven by Contradiction?

1. What is the Uniform Continuity Theorem?

The Uniform Continuity Theorem is a mathematical theorem that states that a function is uniformly continuous if and only if it preserves Cauchy sequences. This means that the function's output values do not vary wildly when the input values are close together.

2. How is the Uniform Continuity Theorem different from the Continuity Theorem?

The Continuity Theorem only requires that a function be continuous at every point in its domain, while the Uniform Continuity Theorem requires that the function be continuous over the entire domain. In other words, the Uniform Continuity Theorem is a stronger condition than the Continuity Theorem.

3. How is the Uniform Continuity Theorem used in real-world applications?

The Uniform Continuity Theorem is used in many real-world applications, such as in physics, engineering, and economics. It is used to prove the existence and uniqueness of solutions to differential equations, and to analyze the convergence of numerical methods for solving equations.

4. Can the Uniform Continuity Theorem be applied to all types of functions?

Yes, the Uniform Continuity Theorem can be applied to all types of functions, as long as they satisfy the conditions of the theorem. This includes polynomial functions, trigonometric functions, and even more complex functions.

5. What are the main implications of the Uniform Continuity Theorem?

The main implication of the Uniform Continuity Theorem is that it allows us to make precise statements about the behavior of a function over its entire domain, based on its behavior at a few specific points. This makes it a powerful tool for studying the properties of functions and solving mathematical problems.

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