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The question is about the proof that $$\displaystyle \mathbb{R} ^{ \omega }$$ is not metrizable in the box Topology. The test is using a series method to show this. For the record: Let X be a Topological space. Let $$\displaystyle A \subset X$$. If there is a sequence of points of A converging to x then $$\displaystyle x \in \bar{A}$$. The converse holds if X is metrizable. Click to expand... The proof that $$\displaystyle \mathbb{R} ^{ \omega }$$ is not metrizable in the box Topology is as follows: We shall show that the sequence lemma (above) does not hold for $$\displaystyle \mathbb{R} ^{ \omega }$$. Let A be the subset of $$\displaystyle \mathbb{R} ^{ \omega }$$ consisting of those points all of whose coordinates are positive. $$\displaystyle A = \{ (x_1, ~ x_2, ~ \text{...}) | x_i > 0 ~ \forall ~ i \in Z_+ \}$$ Let $$\displaystyle \textbf{0}$$ be the origin in $$\displaystyle \mathbb{R} ^{ \omega }$$, that is, the point (0, 0, ...) each of whose coordinates is zero. In the box Topology, $$\displaystyle \textbf{0}$$ belongs to $$\displaystyle \bar{A}$$; for if $$\displaystyle B = (a_1, ~ b_1) \times (a_2, ~b_2) \times \text{ ...}$$ is any basis element containing $$\displaystyle \mathbb{0}$$ then B intersects A. For instance, the point $$\displaystyle \left ( \dfrac{1}{2} b_1 , ~ \dfrac{1}{2} b_2, \text{ ...} \right )$$ belongs to $$\displaystyle B \cap A$$. But we assert that there is no sequence of points of A converging to $$\displaystyle \mathbf{0}$$. For let $$\displaystyle ( \textbf{a}_n )$$ be a sequence of points of A, where $$\displaystyle \textbf{a}_n = (x_{1n}, ~ x_{2n}, \text{ ...} )$$. Every coordinate $$\displaystyle x_{in}$$ is positive, so we can construct a basis element B' for the box Topology on $$\displaystyle \mathbb{R} ^{ \omega }$$ by setting $$\displaystyle B' = ( - x_{11}, x_{11} ) \times ( -x_{22}, x_{22} ) \times \text{...}$$ Then B' contains the origin $$\displaystyle \mathbf{0}$$, but it contains no member of the sequence $$\displaystyle ( \textbf{a}_n )$$; the point $$\displaystyle \textbf{a}_n$$ cannot belong to B' because its nth coordinate $$\displaystyle x_{nn}$$ does not belong to the interval $$\displaystyle ( - x_{nn}, x_{nn} )$$. Hence the sequence $$\displaystyle ( \textbf{a}_n )$$ cannot converge to $$\displaystyle \textbf{0}$$ in the box Topology and thus $$\displaystyle \mathbb{R} ^{ \omega }$$ cannot be metrized in the box Topology. Click to expand... I can understand this proof. But, if no member of the sequence $$\displaystyle ( \textbf{a}_n )$$ is in B' then why not just choose another basis element? Let $$\displaystyle x_{in} = 1/n$$ and B'' = (-1, 1) x (-1, 1) x ... I don't see why we can't just choose a basis element that will work. Clearly I'm missing something fundamental. Thanks! -Dan

MacstersUndead said: B'' bounds $$\displaystyle (a_n)$$ and is a neighborhood of 0, but we need to show that "every neighbourhood of 0 contains all but finitely many points of the sequence." (Henno Brandsma from the link above) Click to expand... Ahhhh! "every". That's the word I was missing! Thanks! -Dan
Going to give helping a college try. I didn't take differential geometry. I think using that $$\displaystyle x_n = 1/n$$ converging to 0 is throwing us off, but I'll use it to work with an example. If you let $$\displaystyle x_{in} = 1/n$$ then you have points such as $$\displaystyle a_1 = (x_{11}, x_{21}, ...) = (1, 1, 1, ...)$$. B' = (-1, 1) x (-1/2, 1/2) x ... B'' = (-1, 1) x (-1, 1) x ... From here, $\Bbb R^{\omega}$ in the box topology is not metrizable , my initial reaction that it parallels Cantor's famous Diagonalization Proof was justified. B' is a construction to show that $$\displaystyle (a_n)$$ cannot converge to 0 in the box topology. B'' bounds $$\displaystyle (a_n)$$ and is a neighborhood of 0, but we need to show that "every neighbourhood of 0 contains all but finitely many points of the sequence." (Henno Brandsma from the link above) I think that's the key here. We only need to construct the counterexample B' to show there is at least one neighborhood that contains no point of the sequence $$\displaystyle (a_n)$$