Let $X$ be a subset of $\textbf{R}$, and let $f:X\rightarrow\textbf{R}$ be a function. Then the following two statements are logically equivalent:
(a) $f$ is uniformly continuous on $X$
(b) Whenever $(x_{n})_{n=0}^{\infty}$ and $(y_{n})_{n=0}^{\infty}$ are two equivalent sequences of elements of $X$, the sequences $(f(x_{n}))_{n=0}^{\infty}$ and $(f(y_{n}))_{n=0}^{\infty}$ are also equivalent.
MY ATTEMPT
I am mainly interested in proving the implication $(\Rightarrow)$.
Since $f$ is uniformly continuous on $X$, for every $\varepsilon > 0$, there exists a $\delta > 0$ s.t. that for every $x,y\in X$
\begin{align*}
|x – y| \leq \delta \Longrightarrow |f(x) – f(y)| \leq \varepsilon
\end{align*}
Moreover, since $x_{n}$ and $y_{n}$ are equivalent, for every $\delta > 0$, there exists a natural $N\geq 0$ such that
\begin{align*}
n\geq N \Longrightarrow |x_{n} – y_{n}| \leq \delta
\end{align*}
Consequently, if we substitute $x = x_{n}$ and $y = y_{n}$, it results that
\begin{align*}
n\geq N \Longrightarrow |x_{n} – y_{n}| \leq \delta \Longrightarrow |f(x_{n}) – f(y_{n})| \leq \varepsilon
\end{align*}
And the proposed resul holds.
Best Answer
Let's say $f$ "has property U" if whenever $x_n$ and $y_n$ are equivalent sequences in the domain of $f$, $f(x_n)$ and $f(y_n)$ are also equivalent. (This is not standard terminology, I just made it up on the spot for ease of writing.)
Here are some strategies to prove the desired implication:
In my experience many people write proofs by contradiction for implications that are really just slightly obfuscated proofs by contraposition, because they show $p \wedge \neg q \Rightarrow \neg p$: in other words the thing they contradict is the actual assumption of the implication, rather than an actual unrelated fact, which means they could've just done a proof by contraposition in the first place.
Meanwhile, in this particular problem, trying for a direct proof leaves you with an awkward assumption to use; you are left to to somehow use sequences to "probe" the modulus of continuity of $f$, and it's not particularly obvious how to do that.
So I will explain how to use proof by contraposition here.
Assume that $f$ is not uniformly continuous, then there exists $\varepsilon > 0$ such that for all $\delta>0$ there exist $x,y$ with $|x-y|<\delta$ and $|f(x)-f(y)|>\varepsilon$. Instantiate such a $\varepsilon$, then use the statement with $\delta=1/n$ to furnish some particular $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|>\varepsilon$. Then $x_n$ and $y_n$ are equivalent sequences whereas $f(x_n)$ and $f(y_n)$ are not equivalent sequences.