Equivalent forms of uniform continuity for a real function

Question

Suppose $D\subseteq \mathbb{R}$ and $F: D \rightarrow \mathbb{R}$

$(1) F$ uniformly continuous on $D$

$(2)$ Given $\epsilon >0$, there exists $\delta >0$ such that if $A\subseteq D$ and $\operatorname{diam}(A)< \delta$, then $\operatorname{diam}(F(A))<\epsilon$

$(3)$ if $\{x_n\}$ and $\{y_n\}$ are sequences in $D$ and $||x_n-y_n|| \rightarrow 0$, then $||F(x_n)-F(y_n)||\implies 0$

I Need to Show: $(1)\implies (2) \implies (3) \implies (1)$

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My thoughts

[$(1) \implies (2)$]

Let $\epsilon >0$, $f$ uniformly continuous so $\exists \delta >0$ such that $|f(x)-f(y)|<\epsilon/2$ for $x,y \in D$ and $|x-y|< \delta$.

For $u,v \in f(D)$, $\exists x,y \in D$ such that $u = f(x)$ and $v=f(y)$

$|x-y| \le \operatorname{diam}(D)< \delta$ ; $|f(x) -f(y)|< \epsilon/2$ ; $|u-v|< \epsilon /2$

$\operatorname{diam}(f(D)) = \sup\{|u-v|: u,v \in f(D)\}\le \epsilon/2$

$\operatorname{diam}(f(D)) < \epsilon$ $_\square$

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$[(2) \implies (3)]$

Not sure how to go about this step...

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[$(3) \implies (1)$]

I know that if $D \subseteq \mathbb{R}$, so a function $F:D \rightarrow \mathbb{R}$ is uniformly continuous iff for every $\{x_n\},\{y_n\}$ such that

$\lim_{n \rightarrow \infty} ||x_n - y_n|| = 0$, then

$\lim_{n \rightarrow \infty} ||F(x_n) - F(y_n)|| = 0$,

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Any help is greatly appreciated!

Answer 1

Your proof of $(1) \to (2)$ is fine, it's a consequence of the definition of the sup

norm. Hint for $(2) \to (3)$: Consider what it means for $||x_n- x|| \rightarrow 0$. It means

that for every natural number N, there exists an $n\geq N$ such that $||x_n-x|| < \epsilon $.

Assume now that $x_n$,$y_n$ are sequences of points in D. Then they are functions from the natural numbers into D whether they converge to points in D or not.

Consider by (2), diam(A) < $\delta $ for some $\epsilon >0$. This means where d is the metric in D and any natural number $n > N$, $d(x_n,y_n)$= $||x_n- y_n||\rightarrow 0$. So since diam (F(A) < $\epsilon$, what does this say about the sup norm induced metric in F(A) for $||F(x_n)-F(y_n)||$?

$(3)\to (1)$ looks fine,but I'm not 100 percent on that.

Hope that helps!

Answer 2

How does this look?

(2) $\Rightarrow$ (3)

Let $\epsilon >0$ be given, and consider the $\delta >0$ furnished by (2). Since $|x_{n}-y_{n}| \rightarrow 0$:

There exists $N \in \mathbb{N}$ such that $n≥N$ implies $|x_{n}-y_{n}|<\delta$. Hence if we let $$A_{n}=\left \{ x_{n},y_{n} \right \}$$

It follows that diam$A_{n}$=sup$\left \{ |x_{n} -x_{n}|,|x_{n}-y_{n}|,|y_{n}-y_{n}| \right \}<\delta$ for $n≥N$. Therefore, by (2):

$$\text{diam} F(A_{n})=\sup\left \{ |F(x_{n}) -F(x_{n})|,|F(x_{n})-F(y_{n})|,|F(y_{n})-F(y_{n})| \right \}<\epsilon$$ for all $n≤N$. In particular $|F(x_{n})-F(y_{n})|<\epsilon$ for all $n≤N$. This means $|F(x_{n})-F(y_{n})| \rightarrow 0$.