Aloizio Macedo
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 20h comment What's exactly the deal with differentials? (Confessions of a desperate calculus student) (inuing...) For example, $dx$ can mean a one-form, a "symbol of the variable you are integrating" (although I dislike this use), $\frac{d}{dx}$ can mean an operator or a tangent vector (although a tangent vector can be defined as an operator, it can also be defined as different ways, so this observation is still relevant). And, like $\infty$, you see people arguing things like $du/dx=du/dy.dy/dx$ just due to notation and without proper reasoning or rigorous background. 20h comment What's exactly the deal with differentials? (Confessions of a desperate calculus student) @MattSamuel Not at all! Like I said, it is like $\infty$: for instance, you can talk about the one-point compactification, and $\infty$ is really a point in the space. Or you can talk about limit at infinity, which can be either symbolic, or some real limit when talking about some compactification. Or you can say that $\#A=\infty$ when you mean it is not finite. But you do see people arguing things like $\infty/\infty=1$ when talking about limits. I almost agree with your teacher... but I think there is a not something a differential really is. (cont... 1d comment What's exactly the deal with differentials? (Confessions of a desperate calculus student) "Differentials" are like $\infty$... it can be given different meaning in different contexts, sometimes only "symbolic" meaning, sometimes notation-abused by people who actually know what they are doing, and sometimes simply abused by people who don't know what they are doing. 1d accepted Why is the sequence exact? 2d comment Compactness in a topological space This might be useful: math.stackexchange.com/questions/371928/… 2d comment Why is the sequence exact? @EricWofsey Could you please elaborate in an answer? 2d comment Why is the sequence exact? @NajibIdrissi Yes, I think so. I would like such proof for terms of generality, but I think that in this special case (with these special maps) there are things which can be simplified. For instance, Alex says that I would need the splitting for the left term, but the left term is an inclusion, hence the exactness is clear (or isn't it?). 2d comment Why is the sequence exact? @NajibIdrissi Yes, Bredon mentions that the fact that $\Delta_{*}$ is free abelian implies the sequence is split, and hence that the splitting map I mentioned exists. But your answer in the question you linked seems circular to me... you say that the result follows due to the fact that a free abelian group is flat... but in the linked wikipedia article, a flat module is something that preserves exactness through tensoring... which is what I'm trying to prove, essentially (it's actually more general). 2d asked Why is the sequence exact? Nov 23 comment Is every closure of a metric space a completion? What is $1$ in $(0,1)$? Nov 23 answered Is the writing of the proof ok? Nov 23 comment We can define the derivative of a function whose domain is a subset of rational numbers? Since noone mentioned this yet... $\mathbb{Q}$ is not a Banach space. Differentiation in non-Banach spaces lack a lot of fruitful theorems (for instance, the Inverse Function Theorem). Also, as Eric said, lacking the mean value theorem is a bad thing. Nov 23 comment Circular argument in proof? @learner Responding your question on the comment, Rudin defines $e$ as the series and proves $e$ equals the limit of this question. Nov 23 comment Why $\pi(x) \approx \frac{x}{\ln(x)}$? en.wikipedia.org/wiki/Prime_number_theorem Nov 22 comment Why do we care about differential forms? (Confused over construction) I always thought that the widely-used phrase of "differential forms are things which we integrate" is an understatement. There are many uses for differential forms and its many "incarnations", which is reflected by the plethora of equivalent definitions of a differential form. Nov 22 comment Is $f:\Bbb Q\rightarrow \Bbb R$ continuous? 1 - $x \mapsto x^2$ is continuous because is the restriction of a continuous function (I'm assuming you know $x \mapsto x^2$ is continuous in the real numbers... if you don't know, you surely know the identity is and the multiplication of continuous function is continuous, so there it is.) 2 - $A,B$ are open because they are the preimage of open sets (namely, $(2, \infty)$ and $(-\infty, 2)$ 3 - For a function to be ("globally") continuous it suffices to prove that the preimages of open sets are open. This is a fundamental fact of continuity, and surely you have that on your textbook. Nov 22 answered Is $f:\Bbb Q\rightarrow \Bbb R$ continuous? Nov 21 reviewed Close Taylor's expansion notation problem. Nov 21 reviewed Leave Open Compute the topological K-group of Klein bottles Nov 21 reviewed Close connectedness in metric space