Let $X$ be a subset of $\Bbb{R}$, and let $f:X\to\Bbb{R}$ 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 consisting of elements of $X$, the sequences $(f(x_{n}))_{n=0}^{\infty}$ and $(f(y_{n}))_{n=0}^{\infty}$ are also equivalent.

My solution (Edit)

I am mainly interested in the implication $(b)\Rightarrow(a)$.

Let us suppose that $(b)$ holds and $(a)$ does not hold.

The following statements are equivalent

  • $f:X\to\mathbb{R}$ is uniformly continuous

  • for every $\varepsilon > 0$, there exists a $\delta > 0$ such that for every $x,y\in X$, \begin{align*} |x - y| < \delta \Rightarrow |f(x) - f(y)| \leq \varepsilon \end{align*}

Based on it, we are going to demonstrate the proposed statement by contradiction.

In other words, let us assume there exists a $\varepsilon > 0$ such that for every $\delta > 0$ there are $x,y\in X$ \begin{align*} (|x - y| < \delta)\wedge (|f(x) - f(y)| > \varepsilon) \end{align*}

In particular, for each $\delta = 1/n$, there are $x_{n},y_{n}\in X$ satisfying \begin{align*} |x_{n} - y_{n}| \leq 1/n\quad\wedge\quad|f(x_{n}) - f(y_{n})| > \varepsilon \end{align*}

Taking the limit, we get that $\displaystyle\lim_{n\rightarrow\infty}(x_{n} - y_{n}) = 0$, but $f(x_{n})$ and $f(y_{n})$ are not equivalent, contradicting the given assumption.

Could someone help me to grasp this difference properly?

  • $\begingroup$ The definition of uniform continuity is: for all $\epsilon>0$ there exists $\delta > 0$ such that for all $x,y$ if $|x - y| < \delta$ then $|f(x) - f(y)|<\epsilon$ To negate this you flip the quantifiers and negate the implication inside: there exists $\epsilon >0$ such that for all $\delta > 0$ there exists $x,y$ such that $|x - y| < \delta$ and $|f(x) - f(y)| \geq \epsilon$. I believe this is what you were using. $\endgroup$ – James May 13 at 2:37
  • $\begingroup$ it should be $|f(x) - f(y)| \ge \varepsilon$ instead of $>\varepsilon$ (and so on). What do you mean by "deny the definition of uniform continuity"? You have proved that if $f$ is not uniformly continuous, then there exists a pair of equivalent sequences whose image is not equivalent. $\endgroup$ – user251257 Jun 11 at 18:53
  • $\begingroup$ Thanks for pointing it out. I forgot to edit this part. I have removed it. $\endgroup$ – KnowledgeSeeker Jun 11 at 19:00

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