How to show two sequences of function uniformly converge to the same limit? I am quite confused with the following question. I am a new to mathematics so I am not too sure. 
Suppose that $(f_n)_{n=1}^\infty$ and $(g_n)_{n=1}^\infty$ are two sequences of functions that converge uniformly to some limits. 
If the supnorm $\|f_n -g_m\|=\sup_y |f_n(y)-g_m(y)|\to 0$ as $n$, $m$ tend to $\infty$, is it true to say that $(f_n)_{n=1}^\infty$ and $(g_n)_{n=1}^\infty$ converge uniformly to the same limit?
I think it is, since I can construct any new sequence from all $(f_n)_{n=1}^\infty$ and $(g_n)_{n=1}^\infty$. The fact that $\sup_y |f_n(y)-g_m(y)|\to 0$ ensure the new sequence to be Cauchy convergent. Hence the new sequence converge to one limit. Is this true?
Please help. Thanks in advance.
 A: First of all welcome you in the world of Mathematics Stack Exchange.
Let domain of all function used here is $D \subset \mathbb{R}$.
$\{f_n(y)\}$ converges uniformly to the function $f(y)$ i.e. for any positive real number $\epsilon$ $\exists$ a natural number $k_1$ s.t. $\forall$ $y \in D$ $| f_n(y) - f(y)| < \frac{\epsilon}{3}$ when $n > k_1$. Thus $\sup_{y}|f_n(y) -f(y)| < \frac{\epsilon}{3}$ when $n > k_1$.
Similarly for the sequence of function $g_n(y)$ you shall get $\sup_{y}|g_n(y) - g(y)| < \frac{\epsilon}{3}$ when $n > k_2 \in \mathbb{N}$
Apply triangular inequality and get $|f(y) - g(y)| < |f(y) - f_n(y) + f_n(y) - g_m(y) + g_m(y) - g(y)| < |f(y) - f_n(y) | + | f_n(y) - g_m(y)| + |g_m(y) - g(y)|$.
Thus $\sup_{y}|f(y) - g(y)| < \sup_{y}|f(y) - f_n(y) | + \sup_{y}| f_n(y) - g_m(y)| + \sup_{y}|g_m(y) - g(y)|$ $\forall$ $y \in D$.
You have the inequality $\|f_n - g_n\| = \sup_{y} |f_n(y) - g_n(y)| < \epsilon$ when $n , m \rightarrow \infty$. Now replace $\epsilon$ by $\frac{\epsilon}{3}$ get $\sup_{y} |f_n(y) - g_n(y)| < \frac{\epsilon}{3}$ when $n , m > k_3 \in \mathbb{N}$ 
Take $k = \max\{k_1, k_2, k_3\}$ and apply all the inequalities togther. So you shall get
$$\sup_{y}|f(y) - g(y)| < \sup_{y}|f(y) - f_n(y) | + \sup_{y}| f_n(y) - g_m(y)| + \sup_{y}|g_m(y) - g(y)| < \epsilon$$
when $n , m >k$
$\epsilon$ is any positive number. So you may consider $f(y) = g(y)$.
