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As stated in the title, prove that there exists $g\in C[0,1]$-continuous function on $[0,1]$- such that $g(x_1)\ldots g(x_n) \neq g(y_1)\ldots g(y_n)$ for all $x_1 \neq y_1$, $\ldots$, $x_n \neq y_n$.

I think this is true, but actually I don't know how to construct such a function. Does anybody have idea?

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    What do you mean by $g(x_{1})...g(x_{n})$? – 2017-02-27
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    This is the product of real numbers $g(x_1), g(x_2),\ldots, g(x_n)$. – 2017-02-27
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    You should use $\cdots$ to denote products, just as you used $\ldots$ in this comment – 2017-02-27
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    As written, there is no such function. You can take $x_{1}, x_{2}$ arbitrarily but distinct and set $y_{1} = x_{2}$ and $y_{2} = x_{1}$. – 2017-02-27
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    Thank you so much, I'm sorry for a silly question – 2017-02-27

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