Proof of two lemma for Heine–Cantor theorem, with some difference in definition.

Question

Here is a definition comes from my calculus book:

For convenience, let's agree to say that the interval $[a,b]$ has the property $P_{\epsilon}$ if there exist sequences $x_{1}, x_{2}, x_{3}, ...$ and $y_{1}, y_{2}, y_{3}, ...$ with $$x_{n},y_{n} \in [a,b],\quad|x_{n} - y_{n}| \lt \frac{1}{n},\quad |f(x_{n}) - f(y_{n})| \gt \epsilon$$ for all indices n.

If I change the last inequality to $$|f(x_{n}) - f(y_{n})| \ge \epsilon,$$

and call it $\tilde{P_{\epsilon}}$, then it's more convenient to prove both lemma. The definition of f is uniformly continuous on $[a,b]$ is equivalent to [a,b] has $\lnot \tilde{P_{\epsilon}}$, both are the same: $$\forall \epsilon \gt 0, \exists \delta \gt 0, \forall x,y \in [a,b] : |x - y| \lt \delta \Rightarrow |f(x) - f(y)| \lt \epsilon.$$

I prove the following lemma using this new definition $\tilde{P_{\epsilon}}$.

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To prove $lemma_{1}$:

if $f$ is not uniformly continuous on [a,b], then [a,b] has the property $\tilde{P_{\epsilon}}$ for some $\epsilon \gt 0$.

Assume $f$ is not uniformly continuous on [a,b], which implies $\lnot(\lnot \tilde{P_{\epsilon}}) = \tilde{P_{\epsilon}}$.

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To prove $lemma_{2}$, which is designed for proving the Heine–Cantor theorem:

Let $f$ be continuous on $[a,b]$. If $[a,b]$ has the property $\tilde{P_{\epsilon}}$, then at least one of the subintervals $[a,c], [c,b]$, which $c = \frac{a + b}{2}$, has the property $\tilde{P_{\epsilon}}$.

Suppose the $lemma_{2}$ is false. which means:

$[a,b]$ has $\tilde{P_{\epsilon}}$, and both $[a,c], [c,b]$ don't have $\tilde{P_{\epsilon}}$, which means $f$ is uniformly continuous on $[a,c]$ and $[c,b]$.

Now, given an arbitrary $\epsilon \gt 0$, compute $\frac{\epsilon}{2}$, and since $f$ is continuous at c, there exists three integers $p, q, r \gt 0$:

$$\forall x,y \in [a,c]: |x-y| \lt \frac{1}{p} \Rightarrow |f(x)-f(y)| \lt \epsilon. \\ \forall x,y \in [c,b]: |x-y| \lt \frac{1}{q} \Rightarrow |f(x)-f(y)| \lt \epsilon. \\ |x - c| \lt \frac{1}{r} \Rightarrow |f(x) - f(c)| \lt \frac{\epsilon}{2}.$$

Assume $x,y \in [a,b]$, $|x - y| \lt \frac{1}{s}$, $s = \max\{p,q,r\}$. Since the last possible case: $$x \in [a,c], y \in [c,b], |x - c| \lt \frac{1}{r}, |y - c| \lt \frac{1}{r},$$

which implies: $$|f(x) - f(c)| \lt \frac{\epsilon}{2}, \quad |f(x) - f(c)| \lt \frac{\epsilon}{2}$$

For any $\epsilon \gt 0$, there exists $\delta = \frac{1}{s}$, $s = \max\{p,q,r\}$, for any $x,y \in [a,b]$, $$|x - y| \lt \delta \Rightarrow |f(x) - f(y)| \lt \epsilon,$$ which contradicts the assumption: $lemma_{2}$ is false.

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Please help me to check whether there is an error in the proof.

By the way, I use $\tilde P_{\epsilon}$ is because the negation of the $P_{\epsilon}$ will end up with: $$|f(x_{n}) - f(y_{n})| \le \epsilon.$$

But I'm not sure: $$|f(x_{n}) - f(y_{n})| \le \epsilon \Rightarrow |f(x_{n}) - f(y_{n})| \lt \epsilon.$$

Answer 1

I think lemma 2 incorrect. If $f$ is continuous on the bounded, closed interval $[a, b]$, then it is uniformly continuous on $[a, b]$. So there cannot be an $\epsilon > 0$ such that $P_{\epsilon}$ is true on $[a, b]$.

Should the interval be open instead?

EDIT: Discussion with the OP shows these lemmas are used in a proof of the Heine-Cantor theorem, which is not yet available at that point in the book he is studying.

Regarding your question about the use of strict inequalities: for inequalities of the form $\forall \epsilon > 0, \exists \delta > 0, Q < \epsilon$, where $Q$ does not contain $\epsilon$ nor $\delta$, you can proceed in 2 steps:

1. Choose a $0 < \epsilon_1 < \epsilon$ and prove the existence of $\delta$ such that the final inequality is of the form $Q \leq \epsilon_1$.
2. Since $\epsilon_1 < \epsilon$, the same value of $\delta$ can be used to deduce that $Q < \epsilon$.

Because "not $P_{\epsilon}$" is of the form mentioned above, replacing the $| f (x_n) - f(y_n) | < \epsilon$ with $| f (x_n) - f (y_n) | \leq \epsilon$ will not invalidate the result (but you are right about which is the logical negation of $P_\epsilon$).

Looking at the proofs of the lemmas: they are correct. You might want to go into greater detail in the proof of lemma 1. For example:

Since $f$ is not uniformly continuous on $[a, b]$, there exists $\epsilon > 0$ such that $\forall \delta > 0, \exists x, y \in [a, b], | x - y | < \delta$ and $| f (x) - f (y) | \geq \epsilon$. Take such an $\epsilon_0 > 0$ and define $x_n$ and $y_n$ as follows:

for each $n \geq 1$, let $\delta_n = 1 / n$, and choose $x_n$ and $y_n$ in $[a, b]$ such that $| x_n - y_n | < \delta_n$ and $| f (x_n) - f (y_n) | \geq \epsilon_0$. Such $x_n$ and $y_n$ exist because $f$ is not uniformly continuous on $[a, b]$.

The series $x_n$ and $y_n$ built above verify $P_{\epsilon_0}$, which is the result.