Proof of “If $f$ is continuous on a closed interval $[a,b]$, then f is uniformly continuous on $[a,b]$.”

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I am having trouble understanding the proof of theorem $19.2$ in "Elementary Analysis-The Theory Of Calculus" By Kenneth A. Ross, here it is:

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I don't understand the purpose of replacing $\delta$ with $\frac{1}{n}$, I know that it is mathematically correct since $\frac{1}{n}>0$ when $n>0$ but I can't see how this added any value to the proof.

The second question, does $\lim_{k\Rightarrow \infty} y_{n_k}=x_0$ stem from the fact that $|x_n-y_n|<\delta$ which implies they are convergent to the same limit, hence all their subsequences converge to this same limit as well?

Best Answer

I don't understand the purpose of replacing $\delta$ with $\frac{1}{n}$

This is needed to move from statements about any positive real $\delta$ to get sequences. We take the sentence "For each $\delta > 0$ ..." and plug in $\delta = 1, \frac{1}{2}, \frac{1}{3}$ to show that the sequences $\{x_n\}$ and $\{y_n\}$ with property $|x_n - y_n| < \frac{1}{n}$ exist. Leaving it as $\delta$, we would have no relation to the sequence index, and having differences smaller than any constant $\delta$ doesn't make the difference converge.

Does $\lim_{k \to \infty} y_{n_k} = x_0$ stem from the fact that $|x_n-y_n| < \delta$ which implies they are convergent to the same limit, hence all their subsequences converge to this same limit as well?

Almost. But again, $|x_n - y_n| < \delta$ doesn't describe convergence; we need the difference to get smaller as a function of $n$, which is true in $|x_n-y_n| < \frac{1}{n}$.

To be more complete about it, the convergence of $\{x_{n_k}\}$ means that for any $\epsilon > 0$ we can find $K \in \mathbb{N}$ such that $k>K$ implies $|x_{n_k} - x_0| < \frac{\epsilon}{2}$. So whenever $k$ is large enough that $k > K$ and $n_k > \frac{2}{\epsilon}$ then by the triangle inequality,

$$ |y_{n_k} - x_0| \leq |y_{n_k} - x_{n_k}| + |x_{n_k} - x_0| < \frac{1}{2/\epsilon} + \frac{\epsilon}{2} = \epsilon $$

which shows $\lim_{k \to \infty} y_{n_k} = x_0$.