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Let $X$ be a subset of $\textbf{R}$, and let $f:X\rightarrow\textbf{R}$ be a uniformly continuous function. Let $(x_{n})_{n=0}^{\infty}$ be a Cauchy sequence consisting entirely of elements in $X$. Then $(f(x_{n}))_{n=0}^{\infty}$ is also a Cauchy sequence.

MY ATTEMPT

Since $f$ is uniformly continuous, for every $\varepsilon > 0$ there is a $\delta > 0$ such that \begin{align*} |x - y| \leq \delta \Longrightarrow |f(x) - f(y)| \leq \varepsilon \end{align*}

On the other hand, since $x_{n}$ is Cauchy, for every $\delta > 0$, there is a natural number $N\geq 0$ such that \begin{align*} i,j\geq N \Longrightarrow |x_{i} - x_{j}| \leq \delta \end{align*}

Hence if we substitute $x = x_{i}$ and $y = x_{j}$, it results that for every $\varepsilon > 0$, there is a natural number $N\geq 0$ such that \begin{align*} i,j\geq N \Longrightarrow |x_{i} - x_{j}| \leq \delta \Longrightarrow |f(x_{i}) - f(x_{j})| \leq \varepsilon \end{align*} and we conclude that $(f(x_{n}))_{n=0}^{\infty}$ is also Cauchy, and we are done.

Could someone please verify if my arguments proceed?

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  • $\begingroup$ The proof works $\endgroup$ – DiegoG7 Apr 30 at 17:54

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