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Box topology does not preserve first countable

  1. Jul 8, 2013 #1
    So, in the topology text I'm reading is mentioned that if each ##X_n## is first countable, then ##\prod_{n\in \mathbb{N}} X_n## is first countable as well under the product topology. And then it says that this does not need to be true for the box topology. But there is no justification at all. Does anybody know a good counterexample?
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  3. Jul 8, 2013 #2
    Let [itex]X_n=\mathbb{R}[/itex] and [itex]X= \prod_{n \in \mathbb{N}} X_n = \mathbb{R}^{\mathbb{N}}[/itex]. Suppose [itex]U_{n \in \mathbb{N}}[/itex] is a countable neighborhood basis for the point [itex]\vec{0} = (0,0, \ldots ) \in X[/itex].

    See if you can think of an open neighborhood of [itex]\vec{0}[/itex] that does not contain any one of the [itex]U_n[/itex]'s. That would contradict the assumption that the [itex]U_n[/itex]'s form a countable neighborhood basis of [itex]\vec{0}[/itex] in the first place and would show that [itex]X[/itex] is not first countable.
  4. Jul 8, 2013 #3


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    As an added hint, recall Cantor's diagonal argument.
  5. Jul 9, 2013 #4
    Thanks a lot guys!

    Here's what I came up with. We can of course assume that ##U_n = \prod_{k\in \mathbb{N}} (a_k^n,b_k^n)##, and that all intervals ##(a_k^n,b_k^n)## are bounded. Now we consider the neighborhood ##\prod_{k\in \mathbb{N}} (a_k^k/2, b_k^k/2)##. This is not included in any ##U_n##. That seems to do it.

    So it's not first countable. But maybe it is sequential? Is there something known about that?
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