1. Let $X, Y $ be metric spaces. $(f_n), (g_n): X \rightarrow \Bbb{R} $ are sequences of functions such that $(f_n) $ converges uniformly to a function $f$, and $(g_n)$ converges uniformly to a function $g$, prove that $(f_n)(g_n) $ doesn't converge uniformly to $fg$.

A simple counterexample, and as it is established in another post, Does $\{f_ng_n\}\to fg$ uniformly?, if $f_n(x) = g_n(x) = x + \frac{1}{n}$, then $f_n(x) g_n(x)$ does not converge uniformly to $x^2$, but it converges pointwise.

As I understand, if $fg$ is bounded, then $f_n(x) g_n(x)$ does converge uniformly to $fg$, and I have proven that.

Suppose that I would like to prove the exercise $1$, NOT by counterexample. I don't understand why it is sufficient to prove that $f_n(x) g_n(x)$ converges pointwise to $fg$. Answering my own question, and I have seen another post, Negation of uniform convergence , that pointwise convergence is the negation of uniform convergence, but

Couldn't there exist a product $f_n(x) g_n(x)$ that converges to another function different than $fg$? Probably not, but It's not that clear why not. Maybe it is related to the fact that as $f_n, g_n$ are uniformly convergent to $f,g$ respectively, then we know that $f_n, g_n, f, g$ are continuous, so the products $f_n(x) g_n(x)$, $fg$ would have to be continuous?

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    $\begingroup$ If $\{a_{n}\}, \{b_{n}\}$ are given sequence of real numbers s.t. $a_{n}\to a$ and $b_{n}\to b$, then $a_{n}b_{n}\to ab$. So $f_{n}g_{n}$ should converge to $fg$ point wisely if we consider $a_{n}=f_{n}(x), b_{n}=g_{n}(x)$ for fixed $x$. $\endgroup$ – Seewoo Lee Nov 10 '16 at 7:13
  • $\begingroup$ The second post you link to doesn't say what you think. Pointwise convergence is not the negative of uniform convergence—indeed it is implied by uniform convergence. That answer says that a statement in the OP is the negation of pointwise convergence, but that statement is not about uniform convergence. $\endgroup$ – Greg Martin Nov 10 '16 at 7:58

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