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It is very know that $f(x,y)=\frac{xy}{x^{2}+y^{2}}$ to not null points and $f(0)=0$ as function of $\mathbb{R}^{2}$ into $\mathbb{R}$, both with supremum metric, is not continuous at $0$.

Before get familiarized with the function I proved it to be continuous at $0$, but I am not finding the error in my proof, there it goes:

Once the domain and the codomain are both metric space, if $(x_n)_n$ converges to $0$ then $(f(x_n))_n$ converges to $0$ for every sequence $(x_n)_n$ is a equivalent of definition of continuity at that point. So let $(x_n)_n$ be a sequence on $\mathbb{R}^{2}$, we claim that for every open sphere with radius $\mu$, there exists a natural number $N$ such that for every $n>N$ then $|f(x_n,y_n)|<\mu$.

In fact $$|f(x_n,y_n)|=|\frac{x_ny_n}{x_n^{2}+y_n^{2}}|≤|y_n|$$ but $|y_n|<|(x_n,y_n)| <\frac{\mu}{2}$ for $n>N $ because $(x_n)_n$ converges to $0$, so the same $N$ used in convergence of $(x_n)_n$ may be used to the convergence of $(f(x_n))_n$ to $0$. this would show that $f$ is continuous at $0$, what is absurd.

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The problem is this step $$ \left| \frac{x_ny_n}{x_n^2 + y_n^2} \right| \leq \left| y_n \right| $$

This is not true when $x_n$ and $y_n$ are small. For example, if $x_n=y_n=\frac{1}{4}$, we have the left hand side is $\frac{1}{2}$ but right hand side is $\frac{1}{4}$.

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    God you is right, thank you very much!2017-02-17