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9

[2016-07-25]: Section Differential Geometry added. Although OP narrowed down the post, there are still many more important historical facts which should be addressed to adequately answer the question, than I can give in this answer. Nevertheless here are some aspects, which might be interesting. At least we will see, OP is right when he thinks that many ...


7

A great mathematician once said to me : making mistakes is not a big deal. What's important is to make the right mistakes. You can interpret this in whatever way you want, but I think the point is this : getting a 96% because you computed something fast in Calculus and disregarded a + or - sign is not relevant. You will still get A+ and there is a reason ...


6

In the bigger scheme of things (beyond grades assigned in undergraduate courses), mathematics does not self-select for perfectionism, nor does it award perfectionism to any realistic extent. Yes, you need to be able to calculate accurately. However, there's a lot more to mathematics than calculation. To be an accomplished mathematical professional, you have ...


5

The mostly commonly used AC has 3 forms: choice function: suppose you have a bag of inhabited sets $A$, then you can just say: oh, let $f$ be a function on $A$, with the action $f(x)\in x$ for each $x\in A$. zorn's lemma: suppose you have a partial order $A$ with the property that every linearly ordered subset (chain) is bounded above, then you can say: ...


4

Let me add to Lee Mosher's remarks about math and CS: First, I've done both, and made plenty of mistakes in both places. But in CS, because there are ways of testing things relatively easily, the mistakes are often easier to find. A consequence of this is that you end up developing habits that keep you from making mistakes in the first place. Here's a (...


4

It's overloaded notation, for sure. But I have never seen it mean anything other than the following: Let $f : X \to Y$. Then, for $A \subseteq X$, $f(A) := \{ f(x) : x \in A \}$. The reason there isn't much notice is probably because it's a very common convention. It may even be explicit at some point in some books on Real Analysis. It turns out that ...


3

The definition of $f$ isn't changing. $f(U)$ is simply standard notation for the image of $U$ under $f$, and similarly for the inverse image $f^{-1}(U)$. Of course $f:X \to Y$ is not the same function as its direct image function, which you could denote $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$ if you really wanted. If category theory interests you,...


3

I'm a perfectionist, but I unintentionally make lots of careless mistakes all the time. Like you, I feel a sharp twinge of disappointment or embarrassment, depending on the situation, but it goes away almost immediately. The reason is that I consider intention and not outcome to be the deciding factor, and so as long as I had put in what I deem sufficient ...


2

Not sure if it answers your question, but you can define them as follows: $\color\red{\neg{x}}$ as $\color\green{2-x}$ $\color\red{{x}\vee{y}}$ as $\color\green{({x}+{y})\bmod3}$ $\color\red{{x}\wedge{y}}$ as $\color\green{({x}\times{y})\bmod3}$ $\color\red{{x}\implies{y}}$ as $\color\green{\neg{x}\vee{y}}$ $\color\red{{x}\oplus{y}}$ as $\color\green{[\neg{...


2

As mentioned by skyking, your definition does not work. Take any statement $P$ that is not provable (Your example of "$x \in S$" is one if you allow non-sentences). Then $( P \lor \neg P )$ is unconditionally true, but you would have assigned it the 'truth-value' $2$. So your definition is self contradictory. However, what you seem to want is Kleene's 3-...


2

Well because you only have exams a few times a semester I can see where the perfectionism might come in. A couple of computational errors and you already are down to an A- (or worse!) depending on the grader. Also if the courses are not challenging you might feel like you are obligated to get a perfect score because of your aptitude. Also because math is ...


2

The whole point of "indexed family" is that it is a function. It is a function mapping the index to some object. So if you consider the first statement, it just says in a more complicated language that every family of non-empty sets has a choice function. Why more complicated? Because it requires the understanding that an indexed family is a function, ...


2

Some authors use different notation, i.e., for $f\colon A\to B$, the associated direct image function is denoted $f_\to\colon \mathcal P(A)\to \mathcal P(B)$. However, since this association is very common it is standrd to simply write $f$ for the extended set valued function, without warning as you say. Some care must be exercised, since for instance if $...


2

Algebraic geometers in fact study an object that is just called $\mathbb{P}^n$. It is a scheme, and it contains the data of $K \mathbb{P}^n$ for any field $K$ (but is more general than this, e.g. $K$ can be replaced by an arbitrary commutative ring; at this level of generality the definition becomes more subtle). As mentioned in the comments, what makes $\...


2

Unfortunately, you are not free to morph the two halves in the very restricted fashion described in your question. You simply do not have that much control over the three given sets. Those three sets can each be much wilder and off kilter than you are imagining. If your proof were correct, then the given plane $P$ that you found, passing through the centers ...


1

When I first learned that equations of degree five or higher may not be solved using an equation similar to the infamous quadratic formula, I felt a great amount of despair.


1

This is going to be a "spherical cow shooting milk in all directions in a frictionless vacuum" approach. The basic idea is fine, but variations may make this model inaccurate. You can make every point where a person can choose to change his path a node in a directed graph. Each street at an intersection would be a node, an exit would be a node, an onramp ...


1

You study topology generally if you are interested in understanding stuff like connectivity, compactness and continuity. Topology actually aids analysis quite a bit - for instance in differential geometry we have the Atiyah-Singer index theorem. In fact, you cannot really properly understand why or how calculus varies depending on the space without studying ...


1

As a child, I learned logic with the book "The Game of Logic", by Lewis Carroll. This does not use the modern symbols, but I think that it is a great book to introduce the basic concepts.


1

A possible sub-base would be to take balls of only rational radius. Any ball of irrational radius can be realized as the infinite union of balls of rational radius. E.g. $ \ \ B_\pi(x) = B_3(x) \cup B_{3.1}(x) \cup B_{3.14}(x) \cup \cdots$ I hope I've interpreted the definitions correctly. I'm taking "sub-base" to mean a subset $S \subset B$ that serves ...


1

Even if you can obtain a "100%" grade, don't be complacent about it: a stricter examiner might have given you only 98%! You can still be a perfectionist; but what you are presently trying to perfect is the wrong thing. You are trying to perfect the output, whereas what you should be trying to perfect is something like the ratio of output to input (according ...


1

It is not rare for math papers to have minor issues. Referees do the best job they can, but their job is not to find every minor issue - the author should ideally do that. Papers are written by humans, in any case, so perfection is rarely achieved. This doesn't have as much impact as you might think because people don't blindly cite proofs from other ...


1

@Peter Smith wrote: So it is worth noting that e.g. John Corcoran can write "Three Logical Theories" as late as 1969 (Philosophy of Science, Vol. 36, No. 2 (Jun., 1969), pp. 153-177), finding it still novel and necessary to stress the distinctions between different types of logical theory. Here it is https://www.academia.edu/9855795/Three_logical_theories


1

If $R$ is a finite commutative ring with maximal ideals $I_1,\dots,I_k$ then by the Chinese Remainder Theorem there is a surjective homomorphism $$R\to R/I_1\times\dots\times R/I_k,$$ and so $\vert R\vert\geq\prod_{i=1}^k\vert R/I_i\vert$. Since each $R/I_i$ has at least two elements, this gives $\log_2(\vert R\vert)$ as an upper bound for the number of ...



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