Show $\prod_{N} \mathbb{R}$ with the box topology is not metrizable.

The Box Topology on $\prod_{j \in J} X_j$ ($X_j$ topological spaces) is generated by the basis $\left\{\prod_{j \in J} U_j \; \Big| \; U_j \text { is open in } X_j \right\}$.

I made an attempt, and found the following:

Consider the contrary. Suppose that there is a countable basis at $0$ say $ (U_i ) $ where $U_i =V_i^1 \times V_i^2 \times \cdots \times V_i^j \times \cdots$. The set $W =\left(\frac{1}{2} V_1^1 \right)\times \left(\frac{1}{2} V_2^2 \right)\times \cdot \times \left(\frac{1}{2} V_i^i\right) \times \cdots$ is open but there exists no $n$ such $U_n\subset W$, so we are done. $ \blacksquare $

StackExchange is a great place for comments on this solution and alternate solutions!


Be careful: Simply putting a factor $\frac12$ before the neighborhood does not necessarily make it strictly smaller. If $V_i^i=X_i$, then $\frac12V_i^i=X_i=V_i^i$. But you can assume that the first one of the local base elements, $U_1$, is bounded, because there is a bounded neighborhood of $0$ ("bounded" in the sense that each projection is bounded). Furthermore, we can assume that $U_1\supset U_2\supset...$ and that all of them are open and, as you correctly noted, are of the box-form.
Now it remains to show that indeed $V_i^i\not\subseteq\frac12 V_i^i$. But since $V=V_i^i$ is bounded, the connected component $C$ of $V$ containing the $0$ is just an interval $(a,b)$ with $-\infty<a<0<b<∞$. Then $\frac12b\in V\setminus\frac12V$
This way, $W=\frac12V^1_1×\frac12V^2_2×...$ is the neighborhood not containing any $U_i$.

  • $\begingroup$ No, please don't delete your comment! It was very helpful. I will upvote it when my vote lock gets removed. Actually, Lord_Farin edited my question, but the original question asked for both "alternate proofs" and "comments on my proof". $\endgroup$ – Ahaan S. Rungta Nov 1 '13 at 23:00
  • 1
    $\begingroup$ Will reward this the bounty in 23 hours. $\endgroup$ – Ahaan S. Rungta Nov 11 '13 at 17:49
  • $\begingroup$ @AhaanRungta: Thanks a lot! $\endgroup$ – Stefan Hamcke Nov 11 '13 at 19:55
  • $\begingroup$ No problem; thank you! $\endgroup$ – Ahaan S. Rungta Nov 11 '13 at 20:45
  • $\begingroup$ Done. Thank you again! $\endgroup$ – Ahaan S. Rungta Nov 12 '13 at 17:47

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