I'm self studying real analysis from Wade's "An Introduction to Real Analysis" and I've come across a proof that I don't understand. I was hoping that some might be able to walk me through it. The theorem is as follows

Theorem. Suppose that $I$ is a closed, bounded interval. If $f:\rightarrow\mathbb{R}$ is continuous on $I$, then $f$ is uniformly continuous on $I$.

Proof. Suppose to the contrary that $f$ is continuous but not uniformly continuous on $I$. Then there is an $\varepsilon_0>0$ and points $x_n, y_n \in I$ such that $|x_n-y_n|<\frac{1}{n}$ and $$|f(x_n)-f(y_n)|\geq \varepsilon_0\;\;\;\;n\in\mathbb{N}$$

By the Bolzano-Weierstrass Theorem and the Comparison Theorem the sequence $\{x_n\}$ has a subsequence, say $x_{n_k}$, which converges as $k\rightarrow\infty$, to some $x \in I$. Similarly the sequence $\{y_{n_k}\}_{k\in\mathbb{N}}$ has a convergent subsequence say $y_{n_{k_j}}$, which converges as $j\rightarrow \infty$, to some $y \in I$. Since $x_{n_{k_j}} \rightarrow x$ as $j\rightarrow \infty$ and $f$ is continuous it follows from above that $|f(x)-f(y)|\geq \varepsilon_0$; that is $f(x)\neq f(y)$. But $|x_n-y_n|<\frac{1}{n}$ for all $n \in \mathbb{N}$ so the Squeeze Theorem implies $x=y$. Therefore, $f(x)=f(y)$, a contradiction.

Why in this proof do we need to take a sub-subsequence, why wont subsequences suffice? I have seen slightly different proofs of this theorem which use only subsequences and the triangle inequality. If someone could help me I would be most grateful.

  • 3
    $\begingroup$ That's because the inequality is true for terms of the sequences with the same index. In other words, you cannot say anything like $$|f(x_{n_k} ) - f(y_{n_j})| \ge \varepsilon_0$$ $\endgroup$ – Crostul Jul 18 '16 at 16:09
  • $\begingroup$ Thanks for your reply. I'm still slightly unsure of why this proof needs sub-subsequences however. Many thanks $\endgroup$ – mark Jul 18 '16 at 18:26

Under the assumption that $f$ is not uniformly continuous on $I$, we have $\epsilon_0$ and sequences $\{x_n\}$ and $\{y_n\}$ such that $|x_n-y_n|<\frac{1}{n}$ for all $n \in \mathbb{n}$, but $|f(x_n)-f(y_n)| \geq \epsilon_0$.

Note that, by framing the inequality in this way, we require $x_n$ and $y_n$ to have the same index; so, in particular, we cannot say $|f(x_n)-f(y_m)| \geq \epsilon_0$ unless $n=m$. Therefore, we cannot simply take any convergent subsequence of each, say $\{x_{n_k}\}$ and $\{y_{m_k}\}$, since then we're not able to guarantee that $n_k=m_k$ for each $k \in \mathbb{N}$. However, once we choose $\{x_{n_k}\}$, we can look at the sequence $\{y_{n_k}\}$. This may or may not converge. If it converges, we are good. If not, we can use Bolzano-Weierstrass once more to furnish a convergent subsequence of it (i.e. a sub-subsequence of $\{y_n\}$), $\{y_{n_{k_j}}\}$.

Since subsequences of convergent sequences are also convergent, we now know that both $\{x_{n_{k_j}}\}$ and $\{y_{n_{k_j}}\}$ converge. Since they have the same index, we can use the inequalities above and proceed with the proof.

Feel free to ask for more clarification if anything is unclear.

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  • $\begingroup$ Ah I see. That makes sense now. Thank you. $\endgroup$ – mark Jul 18 '16 at 18:49
  • $\begingroup$ Well written (+1) $\endgroup$ – Mark Viola Apr 30 '17 at 16:18

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