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 Jan 6 answered $V = W\oplus W^\perp$ Jan 6 comment Integral domain whose irreducible elements are not prime Ah, a domain without non-trivial prime ideals must contain no non-trivial ideals at all (since every ideal is contained in a maximal ideal which is prime) and thus be a field. So we're indeed looking for rings having only non-principal prime ideals. Jan 6 comment Integral domain whose irreducible elements are not prime It's true that an integral domain with no prime ideals is a field (actually, I don't even have proof of this much weaker statement but it seems plausible). But can't it happen that a ring has only non-principal prime ideals therefore having no primes? Jan 6 comment Integral domain whose irreducible elements are not prime @Gaffney: can you give a proof? Jan 6 comment $H^1$ of a constant sheaf I'll just add that the nerve of any covering is indeed a simplex for irreducible spaces since intersection of non-empty opens is again non-empty (a fact that is neatly hidden in your nice answer). Jan 5 comment Periodic solution of differential equation y′=f(y) @Julien: yes, by $S^1$ I mean circle, either by itself or embedded in ${\bf R}^2$ as a unit circle or as the image of $y$, which I denote by $y(S^1)$. The degree of a map is basically how many times it wraps around the circle. E.g. the complex maps $z^n$ restricted to the unit circle have degree $n$, identity has degree one and so do all the embeddings of $S^1$ into $\bf R^2$. Jan 5 comment Periodic solution of differential equation y′=f(y) A small gap in this answer: the map $\pi_C \circ y$ can be surjective. But the half-ray argument still works, it's just little harder to locate the set $X$ which will exist whenever the winding number is $0$. I'll update the answer later, but here's the sketch: every half-ray intersects $y(S^1)$ in even number of points (counted with multiplicity). It must sometimes happen that some of those points merge into a double-point or an interval like $X$, otherwise there would have to some self-crossings or the curve would have a disconnected image. Jan 5 answered Periodic solution of differential equation y′=f(y) Jan 5 comment Periodic solution of differential equation y′=f(y) Let $f(x) = |x| x$. Then $y(x) = |x|^3/3$ is a periodic solution. But perhaps I misunderstand your condition on $f$. Is it supposed to be $\forall x \in {\bf R}^2 \forall k \in {\bf R} f(kx) = k^2f(x)$? Jan 5 comment Periodic solution of differential equation y′=f(y) Can you write down explicitly what you mean by periodic? Jan 5 revised Mirror point with respect to Riemann circle (Möbius transformation) Fixed the mapping argument Jan 5 comment Mirror point with respect to Riemann circle (Möbius transformation) @Daniel: indeed, thanks for pointing this out, I was too hasty. Jan 5 answered Mirror point with respect to Riemann circle (Möbius transformation) Jan 5 comment How a group represents the passage of time? @wonghang: precisely. Another often occurring group is integers that occurs in discrete dynamics. You obtain it by simply taking any map $f$ on a given space and iterating it and its inverse. Although to be fair, most dynamics are not reversible, maps not invertible, so one uses natural numbers instead and moves from groups to monoids or semigroups. But dynamics coming from physics usually are reversible (with the notable exception of heat equation). Jan 4 comment Uniform continuity of $f(x)=x+\frac{\sin x}{x}$ in $(0,\infty)$ @GinKin, your argument is wrong. You can't calculate a limit of $\infty - \infty$ by simply saying it's $\infty$. It's undefined and you need to use some other way. Taylor series for $\sin(x)$ is $x - x^3 / 6 + \cdots$ so that $\sin(x)/x$ = $1 - x^2 / 6 + \cdots$ has $0$ derivative at $0$. Jan 4 comment Open/Closed Subsets of Metric Space I'll add that $U$ is not closed, since $f \equiv 5$ is a limit of functions in $U$. For $V$, consider the function $f \equiv 0$ and argue that in all of its neighborhoods there are functions with negative measure. Jan 4 comment Doubt on differential Glad to hear that, you're welcome. Jan 4 comment Doubt on differential I think your confusion is only notational. The expression $df(v)$ defines a function on $U$ by $[df(v)](p) =[df(p)]v(p)$. All of these objects (vector fields, form fields, functions) are local, so by applying one object to the other what we really mean is that we do it at every point $p \in U$. Jan 4 comment Doubt on differential What do you mean by "vector field doesn't stay in $U$"? Where does it go? I am also not clear about your confusion in the last paragraph. You know how to apply $v$ pointwise, so what's the problem with applying it globally? Same way $v(p)$ corresponds to derivation at $p$, $v$ globally corresponds to a derivation $X \in Der(U)$ in the algebra of derivations on $U$ that take smooth functions to smooth functions and obey Leibniz's rule. Jan 4 answered How a group represents the passage of time?