Proving a sequence is Cauchy from uniform continuity

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?

• 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. – 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.