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I'm trying to show that $\mathbb {R}^{\omega}$ in the box topology is not first countable. But I cannot come up with a contradiction by assuming that for each $x \in \mathbb {R}^{\omega}$, there is a countable basis. My intuition is that I need to use the fact that a countable product of countable sets is uncountable, so I need to show that for any local basis, I can come up with a basis consisting of the product of countable open intervals in each coordinate. Can anyone help me through?

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    $\begingroup$ Diagonal argument. $\endgroup$ – aes Dec 4 '14 at 5:02
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Use a diagonal argument. Given a putative countable basis at x (wlog x = all zeros), construct a basic open set not containing any of the countable ones by making sure it's smaller than the $i^{\mathrm{th}}$ open set in the $i^{\mathrm{th}}$ coordinate.

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If it is 1st countable then for any subset A with a limit point x then there should exist a sequence $x_n$ converging to x. Now take $A=\{(y_n)_n: y_n>0 \forall n\}$. 0 is a limit point of A but there is no sequence in A converging to 0. If possible let $a_n$ is any such sequence where $a_n=(a_{n1},a_{n2},\ldots)$ then the open set $(-a_{11},a_{11})\times (-a_{22}, a_{22})\times\cdots$ contains no points in the sequence.

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