prove that $f_n$ conerges uniformly in $X$"

Please, help me to find a solution to this question:

"If a sequence of continuous functions $f_n:X \rightarrow \mathbb R$ converges uniformly in a dense set $D \subset X$, prove that $f_n$ conerges uniformly in $X$".

Obs:

Once $f_n$ converges uniformly in $D$, then $f_n$ is a Cauchy sequence, i.e., given $\varepsilon >0$, there exists $n_0 \in \mathbb N$, such that $m,n>n_0 \Rightarrow |f_m(d)-f_n(d)|<\varepsilon$, for every $d \in D$ Besides that, $f_n$ is continuous and every point of $X$ is the limit of a sequence of points pertaining to $D$. Then, $d_n \rightarrow x \Rightarrow |f(d_n) \rightarrow f(x)|$. How to conclude that there exists $n_0 \in \mathbb N$, such that $m,n>n_0 \Rightarrow |f_m(x)-f_n(x)|<\varepsilon$?

Hint: For $x\in X$, pick $d\in D$, and pick $n,m > N_{\epsilon/3}$, so large that $$|f_m(x) - f_m(d)| < \frac{\epsilon}{3}$$ $$|f_n(x) - f_n(d)| < \frac{\epsilon}{3}$$ And also so that $$|f_m(d) - f_n(d)| < \frac{\epsilon}{3}$$ (Why can we choose $d\in D$ so that this happens?)
Now, with these choices of $n$, $m$, and $x$, can you conclude that $|f_n(x) - f_m(x)| < \epsilon$?
• Hi, Nicholas. $|f_m(d)-f_n(d)|<\frac{\varepsilon}{3}$ is true because $f_n$ converges uniformly on $D$. But why is true that exists $n_0 \in \mathbb N$ such that $m>n_0 \Rightarrow |f_m(x)-f_m(d)|<\frac{\varepsilon}{3}$? – Walter r Apr 24 '14 at 1:20
• Actually, we don't even need $m>n_0$ for this to be true. What do you know about $f_m$? – Nicholas Stull Apr 24 '14 at 1:48
• In other words, the functions in the sequence $\{f_n\}$ have a certain property that tells us that, given $x\in X = \overline{D}$, there exists $d\in D$ so that $|f_m(x) - f_m(d)|< \frac{\epsilon}{3}$. We will need to use this. – Nicholas Stull Apr 24 '14 at 1:53
Fix $m$ and $n$, both greater than $n_0$, and let $x \in X$. You can pick $d \in D$ such that $|f_n(d) - f_n(x)| < \varepsilon$ and $|f_m(d) - f_m(x)| < \varepsilon$. Then you get $|f_m(x) - f_n(x)| \leq |f_m(x) - f_m(d)| + |f_m(d) - f_n(d)| + |f_n(d) - f_n(x)| < 3\varepsilon$.