Brian M. Scott
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 Jan 26 answered Histogram of duplication in n choose k Jan 26 awarded Nice Answer Jan 26 comment Infinity in “Extended Natural Numbers” @user208884: You’re welcome. Jan 26 comment Infinity in “Extended Natural Numbers” @user208884: It is, but you don’t need anything more formal than the definitions that I gave in the previous comment. The infinite series and the union of infinitely many sets are two fundamentally different kinds of animal. The former really is defined as a limit; the latter has a definition that does not rely on the sets being ordered in any way at all. Something is in the union of a bunch of sets if it’s in at least one of them; it’s in their intersection if it’s in all of them. That’s it: no first set, no ordering of the sets — just an unstructured collection of them. Jan 26 revised Infinity in “Extended Natural Numbers” added 353 characters in body Jan 26 comment Infinity in “Extended Natural Numbers” @user208884: No, the union does not require any notion of limit. And for unions and intersections there really is no notion of starting index. We don’t even need an index set. If $\mathscr{A}$ is any set of sets, then $\bigcup\mathscr{A}$ is by definition $$\left\{x:\exists A\in\mathscr{A}(x\in A)\right\}\;,$$ and $\bigcap\mathscr{A}$ is by definition $$\left\{x:\forall A\in\mathscr{A}(x\in A)\right\}\;.$$ Jan 26 comment What abstract structures allows us to describe “nets that converge toward each other”? You can certainly do it in uniform spaces, and all Tikhonov spaces are uniformizable. Jan 26 answered Direct sum of metrizable spaces. Jan 26 answered Infinity in “Extended Natural Numbers” Jan 26 comment How important is the own talent for research of your PhD supervisor? @Asaf: I agree: it varies enormously. I’m not at all sure that mine ever really internalized the definition of the property that I was studying, but I wouldn’t have traded her for anyone else. And I can think of two excellent graduate students in set theory while I was there who transferred to Madison from a very strong department elsewhere precisely because they wanted less rigid direction. Jan 26 comment How important is the own talent for research of your PhD supervisor? I can think of only one thing that you’ve missed: the OP should also think about how he works. There’s a very good reason that the list of things for which I thanked mine in my thesis ended ‘... but, above all else, for letting me emulate “the cat that walked by himself” in matters mathematical’! Jan 26 answered Show that the solution for the Diophantine equation $x^2 - y^2 = N$ is unique if and only if $|N|$ or $\frac{|N|}{4}$, respectively, is $1$ or prime. Jan 25 comment The Purpose of Master Thesis @GitGud: That’s true, so far as I can tell from my limited participation there, but I’m still not sure that the question would get better answers there than here. Mind you, I’m also not sure that it wouldn’t; I’ve seen several names there that I recognize from here. Jan 25 comment The Purpose of Master Thesis @GitGud: I’m not sure: the answers are somewhat discipline-dependent. Jan 25 answered Is every arrangement reachable by shuffling this way? Jan 25 answered how to prove $pr_i(\alpha \setminus \beta) \supseteq pr_i\alpha \setminus pr_i\beta$ Jan 25 comment How can we calculate the exponential form of a rotation matrix @Francis: Don’t think about the matrix; think about what is happening to the plane. You multiply by $A(\theta)$, and it rotates through an angle $\theta$. You multiply again by $A(\theta)$; what happens to the plane? And what’s the total effect of these two operations? Jan 25 comment What is the difference between necessary and sufficient conditions? @committedandroider: That’s right. Jan 25 comment Why do I generally see real solutions to recurrence relations? The problems are chosen to have real solutions. It is not too uncommon for students using these books to have little or no exposure to complex numbers (at least in the U.S.). Jan 25 comment What is the difference between necessary and sufficient conditions? @committedandroider: Yes, if you know that $\neg p\to\neg q$ and that $q$ is true, then $\neg p$ must be false, and therefore $p$ must be true. \\ If $p$ is not true, then $\neg p$ is true, and the implication $\neg p\to\neg q$ (which you are assuming) then automatically guarantees that $\neg q$ is true and hence that $q$ is false.