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21

Many authors take the existence of $1$ as part of the definition of a ring. In fact, I would disagree with Alessandro's comment and claim that most authors take the existence of $1$ to be part of the definition of a ring. There is another object, often called a rng (pronounced "rung"), which is defined by taking all the axioms that define a ring except you ...


16

How about simply "counterexamples related to the Banach fixed point theorem"? Or if you want to be more precise, "counterexamples to possible strengthenings of the Banach fixed point theorem".


10

I am also not a native English speaker, so not sure it this is a good suggestion, but I would use something like Necessity proofs or Necessity demonstrations or Necessity examples (meaning: examples that prove that every assumption of the theorem is necessary).


9

I'm currently teaching out of the 4th edition of Stewart's Galois Theory textbook. Stewart defines a ring to be what other authors might call a commutative ring with unity. The reason is simple: in this book, there is not much call for noncommutative rings, nor for rings without unity, and it gets old writing "commutative ring with unity" over and over, when ...


8

A possible English wording is "what happens if we drop the hypothesis of the Banach fixed-point theorem?"


7

I have looked at your website before, and I remember being a bit dissatisfied with the terminology "counterexamples"- from a technical point of view. For me, a counter-example is some constructible object that demonstrates the falsehood of some statement. Equivalently, a counter example demonstrates the truth of the negation of the original statement. So for ...


7

Probably the most standard term of the sort you're looking for is "equipotent", though it isn't used particularly often (more often, people will just say two sets "have the same cardinality" or "are in bijection"). People who think about categories a lot sometimes say "isomorphic as sets".


4

There's no name in common use for the set of expressions in exactly the form you quote. One problem with the concept is that they are not closed under multiplication, like polynomials are. For example we have $$ (a_2\sqrt x + a_1x)(b_2\sqrt x + b_1 x) = a_2b_2 x + (a_1b_2+a_2b_1) x^{3/2} + a_1b_1 x^2 $$ where the middle term looks neither like $ax^n$ nor ...


3

We say "$A$ injects into $B$", and may write $f: A \hookrightarrow B$.


3

In some contexts, when direct products are defined (e.g. vector spaces, groups, modules...) given two maps $$f: A \to B \\ g : C \to D$$ one defines $$f \times g : A \times C \to B \times D$$ as $(a,c) \mapsto (f(a), g(c))$. Sometimes (when you work with modules or vector spaces) products are denoted with the symbol $\oplus$ and are called direct sums. This ...


2

A mapping $f:X\to X$, for a set $X$, which satisfy $f\circ f=1$ is called involution.


2

First of all, I suspect there's a typo: I think it should be $$"{x^2-a^2\over x-a}=x+a."$$ (The "$x^2$" is an "$x$" in the OP.) E.g., set $x=2, a=1$ - in the OP this would yield 1=3. Second, as the comments have pointed out, even with that fix, it's false if $x=a$. There's an important conceptual issue lurking here: there are some functions, which are ...


2

You could call it a Puiseux polynomial.


2

I think the idea is that a particular player set contains all the nodes in which a given player makes a decision. There are $N+1$ of them because there are $N$ players and then we consider nature as an additional random player. In your diagram, if we call Nature player zero, then $g^0$ would include just the initial node. $g^1$ would include both of P1's ...


2

If $f$ is a nondecreasing function on $\mathbb R$, then its only discontinuities are jump discontinuities. That is, there is a discrete set $X$ such that $f$ is continuous on $\mathbb R\setminus X$ and for $x\in X$ the onesided limits $\lim_{t\to x^-}f(t)$ and $\lim_{t\to x^+}f(t)$ exist, but differ. For most purposes it does not really matter what $f(x)$ is ...


1

I would write this as $$ \nexists k: (2^k \equiv 12 \mod 14) $$


1

$\{k \in \mathbb Z \mid 2^k \equiv 12 \mod 14 \} = \emptyset$


1

Based on a non-decreasing function $f$ you can prescribe function $f^{-}$ by $x\mapsto\sup\left\{ f\left(y\right)\mid y<x\right\} $. This function is left-continuous and for each $y<x$ it satisfies:$$f(y)\leq f^{-}(x)\leq f(x)$$ Consequently we have $f^{-}(x)=f(x)$ if $f$ is left-continuous at $x$.


1

It is a polynomial in $y=x^{\frac 1{12}}$ namely $$p(y)=a_4y^3+a_3y^4+a_2y^6+a_1y^{12}+a_0$$and is thus an element of the polynomial ring obtained by adjoining a $12^{th}$ root of $x$ to the original polynomial ring.


1

If $M$ is a vector space, then we refer $A$ as an endomorphism. There are differences when a function takes an element from one space to an element in the same space. Take for example: $f:C[0,1]_{\|\cdot\|_1}\to C[0,1]_{\|\cdot\|_2}$, where $\|\cdot\|_1 \to$ integral norm $\|\cdot\|_2 \to$ maximum norm Note that $C[0,1]$ is complete under maximum ...


1

I would go for non-inclusive examples or exclusive examples, as in essence any violation to the hypotheses/preconditions of the theorem are supposed to produce results that are excluded from when every condition is fulfilled. For instance, on the level of definitions, when defining the term 'rectangle', you can state that a square and a triangle are ...


1

This is much more a mathematical question than an English language question. (Note that I didn't even look at the maths) On your website you say "We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled." Basically this theorem, like most theorems, takes the form "if conditions A, B and C are fulfilled, then ...


1

The term "identity" is vague without a specified domain. The above is not an identity for $x \in \mathbb{R}$ because it isn't true for $x = a$, but it is true for $x \in \mathbb{R} \setminus \{a\}$ because, by the definition, the statement is true for all $x$ in the domain. This answer takes into account the typo Emilio Novati noted.


1

I suppose that your question is about $$ \dfrac{x^2-a^2}{x-a}=x+a $$ and this is an idenity for $x\ne a$ but is not an identity for $x=a$ because the Left side is not defined.


1

When taking a calculus course a few years back some of the students were deaf. At every lecture there were 2 translators. It seemed that some concepts like "The mean value theorem" were easy to translate, but others consisting of words rarely used in everyday speech demanded more effort. I might be wrong, but I think that it is possible to "type out" words ...


1

While sample set could also be worthwhile terminology, it is worth recalling that any space in mathematics is always a set with some sort of operations defined. For example, in the theory of stochastic processes, we could let $\Omega = \mathcal{D}(\mathbb{R}_+, \mathbb{R})$, where $\mathcal{D}(\mathbb{R}_+, \mathbb{R})$ is the Skorokhod space of real-valued ...


1

The $L_p$ norm is more general, but you need to specify a measure space for it to make sense. You're integrating over $\mathbb X$ after all! The $l_p$ norm can be seen as a particular case of the above, as @Stephen Montgomery-Smith noted, with the counting measure on positive integers. So I don't think there really is any source of ambiguity: either I ...



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