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I wanted to prove that the box topology on $\Bbb R^\omega$ is not Frechet-Urysohn. I know that is not first countable, but it is clearly not enough and I cannot find a subspace that is not sequential..

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up vote 1 down vote accepted

Let $A=\{x\in\square\Bbb R^\omega:\forall n\in\omega(x_n\ne 0)\}$, and let $z$ be such that $z_n=0$ for each $n\in\omega$. Clearly $z\in\operatorname{cl}A$, but there is no sequence in $A$ converging to $z$; the argument for this is the same as the argument showing that $\square\Bbb R^\omega$ is not first countable.

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