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revised A maximal inequality on distance to median, so called Lévy's inequality?
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Apr
23
comment What resources are there for learning Russian math terminology?
@AlexB. Similar phenomenon for Chinese. And I wonder whether there is canonical translations for these terminologies, even those appear in advanced undergraduate courses or graduate courses? For Chinese, I don't know much. I don't know how to translate some terms from relatively elementary mathematics, say, generic point, fibration, quasi-coherent sheaf, lying over, going up, going down (Cohen-Seidenberg theorems), etc. No need to mention more modern mathematics.
Apr
19
comment Quotient of measurable functions
Hint: if $f\colon Y\to Z$ and $g\colon X\to Y$ are measurable maps between measurable spaces, then so is $f\circ g$. A continuous function between topological spaces is a fortiori measurable with respect to Borel $\sigma$-algebras (namely, Borel measurable).
Apr
19
comment Graduate school self-doubt (currently an undergraduate)
And the example of Galois isn't a good one from another aspect: finally Galois succeeded in concours and was admitted into École Préparatoire, which would been renamed as École Normale Superieure later.
Apr
19
comment Graduate school self-doubt (currently an undergraduate)
@EricNaslund Sorry for my bad French. I visited the wikipedia, and there is a footnote which seems to me second what Graphth said: Plusieurs récits circulent sur cet examen. Selon le mathématicien Joseph Bertrand, l'examinateur à l'oral est Dinet, qui pose des questions classiques pour lesquelles il exige des réponses détaillées, type d'examen peu adapté à Galois, que les exercices trop scolaires impatientent : une légende raconte que Dinet lui ayant posé une question trop simple sur les logarithmes, Galois lui aurait jeté une éponge à la figure (Verdier 2003, p. 13-14).
Apr
18
comment $g(x)\le f_0(x_0)$ for all $g \in B$ and $x \in A$.
It suffices to show that $A\times C(A,\mathbb R^n)\to\mathbb R^n,(x,f)\mapsto f(x)$ is continuous. Note that $A$ is locally compact.
Apr
18
comment How to determine the ideal defining $\overline{f(Z)}$ where $f$ is a regular map of affine variety.
Atiyah & Macdonald, Ex 1.21: Let $\phi\colon A\to B$ be a ring homomorphism. Let $X=\operatorname{Spec}(A)$ and $Y=\operatorname{Spec}(B)$. $\phi$ induces a mapping $\phi^*\colon Y\to X$. If $\mathfrak b$ is an ideal of $B$, then $\overline{\phi^*(V(\mathfrak b))}=V(\mathfrak b^c)$ where $\mathfrak b^c=\phi^{-1}(\mathfrak b)$ is the contraction of $\mathfrak b$.
Apr
9
comment $K[[X]]$ is not a finitely generated $K[X]$-module.
@FanZheng And note that $K[[X]]$ is a priori an integral domain. We can assume that $K$ is algebraically closed (seen from a scalar extension to its algebraic closure), where classical Nullstellensatz works.
Apr
9
comment How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$?
@fleablood On what you said about the unboundedness of the preimage. The boundedness is meaningless.
Apr
9
comment $K[[X]]$ is not a finitely generated $K[X]$-module.
@FanZheng For me, an affine variety should be irreducible and reduced. (anyway, in most of classical algebraic geometry we should assume that $K$ is algebraically closed)
Apr
8
comment How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$?
@fleablood No. Show that, for all metric space $(X,d)$, there is a metric $d'$ on $X$ such that $d,d'$ induces the same topology (the same collection of open sets) and $d'$ is bounded, namely $\exists M,\forall x,y\in X\colon d'(x,y)\le M$. Hint: Consider $d'=min(d,1)$.
Apr
8
comment How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$?
@ahorn There is nothing stronger: show that $\overline{f(\overline A)}=\overline{f(A)}$.
Apr
8
comment How to know if one problem is more difficult than another one?
Personally I think it's just a guesswork and some kind of intuition. Mathematics isn't very different from other works. People measures difficulties from experience and/or feelings, but it could be just false. Recently, Google's AI AlphaGo beats Lee Sedol in the go game. It's astounding to professional go players that, AlphaGo is very good at estimating the state, which was considered an esoteric ability of human beings.
Apr
8
comment Why is a convolution well-defined?
$y\mapsto f(x-y)g(y)$ isn't necessarily of $L^1$ for all $x$, but it's $L^1$ for almost all $x$ by an application of Fubini's theorem. That's what Young's inequality (or Minkowski's integral inequality) tells you.
Apr
8
comment How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$?
Hint: Show that $f(\overline A)=\overline{f(A)},\forall A\subseteq X$ if and only if $f(F)$ is closed for all closed $F\subseteq X$.
Apr
8
comment Are there any problems in probability theory where it cannot be solved or defined with respect to Borel sets but only with Lebesgue measurable sets?
In probability theory, I think one should work with Borel measurable functions but not Lebesgue measurable ones, because probability measures are Borel measures but not necessarily absolutely continuous with respect to Lebesgue measure. And usually in probability theory one should work with not only a probability measure, but a sequence of probability measures. Yes, as @DaveL.Renfro pointed out, one can work with universally measurable things.
Apr
8
comment $K[[X]]$ is not a finitely generated $K[X]$-module.
I have no simple proof, but if $K[[X]]$ is of finite type over $K$, then it should be an affine variety over $K$, but $K[[X]]$ is local.
Apr
8
comment $K[[X]]$ is not a finitely generated $K[X]$-module.
I think it's also true that $K[[X]]$ isn't of finite type over $K[X]$, and that the transcendental degree of $\operatorname{Frac}(K[[X]])$ over $K(X)$ is infinite.
Apr
8
comment Uniform integrability and stopping times
Is $(X_n)_n$ a $L^1$-martingale?
Apr
8
revised Compact operators are orthogonally equivalent to a diagonal matrix?
added 88 characters in body