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Let $f_n$ and $g_n$ be uniformly convergent sequences of real-valued functions on a metric space E. Show that then $f_n \cdot g_n$ does not uniformly converge to $f\cdot g.$

I can find two examples that proves the statement here. How can I show it by definition?

Suppose that $f_ng_n \to fg$ then for all $\epsilon>0$ I must find an $N$ such that $|(f_ng_n)(x) - (fg)(x)| < \epsilon$ for any $n>N$ which will lead to a contradiction.

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What you are trying to prove is not true. There exist uniformly convergent sequences $f_n$ and $g_n$ such that their product sequence $f_n\cdot g_n$ converges uniformly to $f\cdot g$. Take $f_n$ and $g_n$ to be constant sequences, for example.

What your examples show is that the proposition in the linked question is false. The negation of that proposition is true, but it is different from the proposition you posted.

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  • $\begingroup$ But in my book it says that the product $f_n(x) \dot\ g_n(x)$ does not converge uniformly to $f(x) \dot\ g(x)$. Since I already have an example that proves that, I was then looking for a proof by definition. $\endgroup$
    – user104235
    Nov 16, 2013 at 2:13
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    $\begingroup$ @user104235 The product does not always converge uniformly. It does sometimes. $\endgroup$
    – Potato
    Nov 16, 2013 at 5:10

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