Metrization of R^(omega)

Question

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

Answer 1

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

Answer 2

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 \(\displaystyle (a_n)\)