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This question already has an answer here:

Suppose $(X_{i}, \Vert \cdot \Vert_i)_{i\in I}$ are all normed spaces over the same field $\Phi= \mathbb{R}, \mathbb{C}$ and suppose $X= \prod_{i \in I} X_i$ is the product space. I want to show that the product topology on $X$ is normable iff $I$ is finite. My work in the '$\Leftarrow$'-direction, is to show that the bases for the two topologies are equal and so therefore generate the same topology.

In the other direction however, I seem to get stuck, since I think the right way to go about it is to suppose that $I$ is infinite, but that the product topology on $X$ is normable and try to reach a contradiction, but I can't figure out where the contradiction is supposed to come from?

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marked as duplicate by levap, Community Apr 24 '16 at 11:20

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Suppose for example $\displaystyle \mathbb R ^\infty = \prod_{i=1}^\infty \mathbb R$ is normable. Then there is an open neighborhood (the unit ball) $B$ of $\overline 0$ such that $\displaystyle \bigcap_{n=1}^\infty\frac{1}{n} B = \{\overline 0\}$. You should prove this if it is not obvious to you. But every neighborhood of the origin in $\mathbb R ^\infty$ contains some set $U = \displaystyle (-\varepsilon,\varepsilon)^N \times\prod_{i={N+1}}^\infty \mathbb R$ and we have $\displaystyle \bigcap_{n=1}^\infty\frac{1}{n} U = \{0\}^N \times\prod_{i={N+1}}^\infty \mathbb R \ne \{ \overline 0 \}$.

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I think you can prove it by showing if the index is infinite, then any non empty open set is not bounded (for definition of "bounded" in this case see here

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