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 Aug 8 comment Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$ @EricWofsey: OK, I agree with you if "infinite subsequence" means "image of a map $\omega \to \omega_1$. But I interpreted it as "an infinite interval of $\omega_1$", where Brent's argument doesn't work. So: you're definitely right, the OP should clarify what he means :-) Aug 8 comment Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$ You're right, I misread Brent's answer. Mea culpa. Aug 8 comment Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$ @EricWofsey: I think you misinterpreted the last paragraph. He's saying that if $C$ contains only 0s on an infinite interval $I \subset \alpha$, then, for every sequence $\sigma$ in $\{0, 1\}^I$, you get a "parallel" subset $C_\sigma$ where the sequences now show $\sigma$ in restriction to $I$. Now, all these subspaces $(C_\sigma)$ should intuitively have the same measure, and as there are many of them, $0$ is the only possibility... Aug 8 comment Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$ There is a well-defined probability measure on $\{0,1\}^{\omega_1}$, thanks to Kolmogorov's extension theorem. I don't think that "having an infinite subsequence of 1" will define a measurable subset, though (if only because $\omega_1$ has uncountably many such subsequences). It might be possible to circumvent this difficulty (showing that a bigger set is measurable and has measure 0, for instance), but I somehow doubt it. Aug 8 comment Probability of an infinite subsequence in a randomly generated sequence of order type $\omega_1$ Note that the OP is talking about coin tosses indexed by a countable ordinal (which might very well be $\neq \omega$). In this situation, having an infinite sequence of heads isn't the same thing as having eventually only heads. For instance, on the ordinal $\omega \cdot 2 = \omega + \omega$, having heads for all $\alpha < \omega$ and then anything you want gives you an infinite subsequence of heads... Jul 15 comment An equivalent definition of the rotation number of a circle homeomorphism I took the liberty of rewriting the question (which I like a lot). I hope you don't mind. Jul 15 revised An equivalent definition of the rotation number of a circle homeomorphism rewording of the question Jul 15 answered Smooth isometric embeddings of Riemannian manifolds Jul 9 answered Curvature flow for convex planes curves Jul 9 revised Curvature flow for convex planes curves deleted 80 characters in body Jul 9 comment Understanding definition of Riemann Integral E may come before I in your name, but it's the other way around in Bernhard's :-) Jul 9 comment Curvature flow for convex planes curves Je ne vois pas comment un phénomène ou un exemple réel pourrait expliquer un résultat mathématique. Cherches-tu une application de cette théorie ou s'agit-il d'autre chose ? [Je veux bien traduire la question en anglais, mais il faudrait déjà que je sois sûr de bien la comprendre...] Jun 25 comment Find whats's wrong with this proof about the reflexive vector space @BrianM.Scott: I guess some people use this word to talk about the purely algebraic fact that the canonical morphism $X \to X^{**}$ is an isomorphism if $X$ is a finite-dimensional vector space (Bourbaki does in the module setting, for an example), but I agree the vocabulary comes from the topological, infinite-dimensional case. Jun 25 comment Algorithm to find non-zero matrix $N$ such that $N \times M = 0$ Note that if $M$ is in echelon form, the question is easy. So Gauss's algorithm (en.wikipedia.org/wiki/Gaussian_elimination) will allow you to construct such an algorithm. Jun 25 comment Find whats's wrong with this proof about the reflexive vector space @BrianM.Scott: your first comment seems off-topic to me. The original question deals with "naked" vector spaces, without topology. Jun 25 comment Find whats's wrong with this proof about the reflexive vector space Of course, since $X^*$ is in a sense always strictly larger than $X$ when $\dim X$ is infinite, there's no hope for preserving a meaningful notion of reflexivity (at least purely algebraically, with topological vector space, the notion is highly interesting). Jun 25 comment Find whats's wrong with this proof about the reflexive vector space If $X$ is the set of real sequences with only finitely many terms, it is quite easy to identify $X^*$ to the set of all real sequences, which is strictly larger (and it's a general property of infinite-dimensional spaces, cf. for instance math.stackexchange.com/questions/1297845/…). Jun 19 awarded Nice Answer Jun 16 answered Does there exist a double cover with trivial deck transformation group? Jun 10 comment Coordinate systems on manifolds @Will: you can always put some Coordinates (they're useful for computations), but you cannot ask them to preserve the metric (=to be isometries). If you have a chart which is an isometry, then your manifold is locally Euclidean, which is a very exceptional case. In Riemannian Geometry, it is sometimes important to use charts with extra properties (so that the problem you're studying has a nice expression), but these "extra properties" are always weaker than being an isometry because the manifold is not flat.