Given: $f: X \longrightarrow Y$ is uniformly continuous on $X$, $(x_n)_n \in X $ is a Cauchy sequence.

Question: What can you say about the sequence ${f(x_n)}$ ?

My attempt:

Since $f$ is uniformly continuous on $X$, then $f$ is uniformly continuous at each $x \in X$, thus for an $\varepsilon > 0$, there's a $\delta > 0$ such that $|f(x_n) - f(t)| < \epsilon$ for |$x_n - t| < \delta$, for all $x_n,t \in X$. Take $\epsilon = 1$, so $|f(x_n)| < 1+ |f(t)|$. Hence $f(x_n)$ is bounded.

And since $x_n$ is Cauchy then it is convergent, so suppose that $x_n \to t$ as $n \to \infty$, and so $f(x_n) \to f(t)$ as $n \to \infty$.

Therefore, the sequence $f(x_n)$ is Cauchy.

Is my proof correct? How can it be improved?

  • 1
    $\begingroup$ The fact that $x_n$ is Cauchy doesn't have to mean that $x_n$ is covergent, unless $X$ is a complete metric space space. Moreover, for a general metric space $Y$, norm $|y|$ is not defined, what is defined is only the distance between points $d(y_1,y_2)$. Similarily, for a general $Y$ you cannot make substraction $f(x_n)-f(x)$, that requires at least affine structure. $\endgroup$ – Adam Latosiński Apr 6 at 9:35

There are some problems with your proof. From the expressions you write it seems that you assume that $X$ and $Y$ are normed vector spaces (you can substract the elements, you can take their norm). Moreover you assume that $X$ is a complete space (only then Cauchy property implies convergence).

In general case, we only need $X$ and $Y$ to be metric spaces, to make sense of Cauchy property and uniform continuity. So let $d_X(x,x')$ be a metric on space $X$, and $d_Y(y,y')$ be a metric on space $Y$.

Uniform continuity of $f$ means that $$ \forall \epsilon>0 \;\exists \delta>0 \;\forall x,x'\in X : (d_X(x,x')<\delta)\Rightarrow(d_Y(f(x),f(x'))<\epsilon) $$ If $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence, then $$ \forall \delta>0 \;\exists N \;\forall n,m>N : d_X(x_n,x_m) < \delta$$ So if we choose $x=x_n$, $x'=x_m$ with $n,m>N$ where $N$ is given by the second condition, from the first condition it follows that $$ \forall \epsilon>0 \;\exists N \;\forall n,m>N : d_Y(f(x_n),f(x_m))<\epsilon $$ which means that $(f(x_n))_{n\in\mathbb{N}}$ is a Cauchy sequence.


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