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For any two topological spaces $X$ and $Y$, consider $X \times Y$. Is it always true that open sets in $X \times Y$ are of the forms $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$?

I think is no. Consider $\mathbb{R}^2$. Note that open ball is an open set in $\mathbb{R}^2$ but it cannot be obtained from the product of two open intervals.

Idonknow
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    You are right. This is good example. – user52045 Nov 18 '13 at 10:43
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    Your example works, but it still requires to show that an open ball $B(x,\epsilon)$ is indeed open in the product topology on $\Bbb R×\Bbb R$, not just in the metric topology on $\Bbb R^2$. If you know that it is, then you can use it. – Stefan Hamcke Nov 18 '13 at 15:03
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    @Stefan H How we show that open ball is open in the product topology $\mathbb{R} \times \mathbb{R}$? By expressing the open ball as union of basis elements? – Idonknow Nov 18 '13 at 17:02
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    For $(x_1,x_2)\in\Bbb R^2$ and $ε>0$ the box $\left(x-\fracε2,x+\fracε2\right)×\left(x_2-\fracε2,x_2+\fracε2\right)$ contains $(x_1,x_2)$ and is a subset of $B_ε(x_1,x_2)$. Therefore the product topology is finer than the metric topology, hence an open ball is an open set in the product $\Bbb R×\Bbb R$. – Stefan Hamcke Nov 18 '13 at 20:05

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Nice done. The forms $U\times V$ is a base for the topology. That means that every open set in $X\times Y$ is a union of elements of the form $U\times V$.

Haha
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