This is what I have so far:


We must show that $f$ is uniformly continuous, hence satisfy that

$\forall \epsilon > 0, \exists \delta > 0, \forall x,y \in X$ with $d(x,y) < \delta \Rightarrow \rho(f(x),f(y)) < \epsilon$.

Let $\epsilon > 0$ and $x,y \in X$. Since each $f_n$ is uniformly continuous we have that

$\exists \delta_1$ such that $\rho(f_n(x),f_n(y)) < \frac{\epsilon}{3}$, whenever $d(x,y) < \delta_1$.

We also have that $f_n \to f$ uniformly, then

$\exists N_1 \in \mathbb{N}, \forall x \in X$, such that $\rho(f_n(x),f(x)) < \frac{\epsilon}{3}$ whenever $n\geq N_1$.

$\exists N_2 \in \mathbb{N}, \forall y \in X$, such that $\rho(f_n(y),f(y)) < \frac{\epsilon}{3}$ whenever $n\geq N_2$.

By the triangle inequality

$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f(y))$

Applying again the triangle inequality yields

$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f_n(y)) + \rho(f_n(y),f(y))$


$\rho(f(x),f(y)) \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$.

I know there is something weird (incomplete) since I didn't use the $\delta_1$ and also I didn't specify how to choose the $\delta$. My problem is how to relate the $\delta$ with the $\delta_1, N_1$ and $N_2$.

Any help would be appreciated. Thanks.


1 Answer 1


Your proof is mostly correct, but you don't need both $N_1$ and $N_2$. Since the convergence is uniform, one $N$ works for all $x\in X$. And when you finally get to work and start writing $d(f(x),f(y))\leq\cdots$, you don't specify what $n$ is (does any $n\in\mathbb N$ work?). Also you originally let $x,y\in X$ be arbitrary, but then specify that $d(x,y)<\delta_1$, which isn't good practice when writing up a formal proof. Here's how you could clean things up a bit:

Let $\varepsilon>0$ be given. Since $f_n\to f$ uniformly, there is some $N\in\mathbb N$ such that $d(f_n(x),f(x))<\varepsilon/3$ for any $n\geq N$ and $x\in X$. Fix $n\geq N$. Since $f_n$ is uniformly continuous, there is some $\delta>0$ such that for any $x,y\in X$ with $d(x,y)<\delta$ we have $d(f_n(x),f_n(y))<\varepsilon/3$. Now for any $x,y\in X$ with $d(x,y)<\delta$, we have \begin{align} d(f(x),f(y))&\leq d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(y),f(y))\\ &<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon \end{align} and the result follows.

  • $\begingroup$ Thank you very much. You make it very clear for me. $\endgroup$ Jun 14, 2017 at 19:27
  • $\begingroup$ This problem has two parts, this one was the first.Can you verify the second part of this proof here: math.stackexchange.com/questions/2325178/… $\endgroup$ Jun 16, 2017 at 15:41
  • $\begingroup$ @RichardClare I'll take a look at it. $\endgroup$
    – Aweygan
    Jun 16, 2017 at 16:27

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