# Prove that $f:X\rightarrow\Bbb{R}$ is uniformly continuous iff it maps equivalent sequences onto equivalent sequences

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?

• 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. – James May 13 at 2:37
• 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. – user251257 Jun 11 at 18:53
• Thanks for pointing it out. I forgot to edit this part. I have removed it. – KnowledgeSeeker Jun 11 at 19:00