New answers tagged

1

Yes, the limit may well exist if $D_f=[0,\infty)\times[0,\infty)$. For instance, if $f\colon[0,\infty)\times[0,\infty)\longrightarrow\Bbb R$ is defined by $f(x,y)=x$, then $\lim_{(x,y)\to(0,0)}f(x,y)=0$. What this means is that$$(\forall\varepsilon>0)(\exists\delta>0):\|(x,y)\|<\delta\implies\bigl|f(x,y)\bigr|<\varepsilon,$$which is true: for ...


0

Definitions are just ways to synthetize complex concepts in words that are nearer to our natural language. So a definition should always be an "if and only if". If it's not is simply poor writing.


1

What that article says is just that the images $f(A_0)$ and $f(A_1)$ form a partition of $A$, which means $$ A=f(A_0)\cup f(A_1) $$ and $f(A_0)\cup f(A_1)=\emptyset$. The phrase "the union of the images of these functions" should be understood as the collection of the images of these functions.


1

Yes, the union of these images is $A$, so that is maybe poorly phrased. What they mean is that the images of these injections partition $A$. That is $$ P = \{ f_i(A_i) : 0 \leq i < n \} $$ partitions $A = \biguplus_{0 \leq i < n} A_i$, where $f_i: A_i \to A$ is the injection corresponding to $0 \leq i < n$. So in your example we have $n = 2$ and so $...


1

What should be the correct formula for the nth hyperoperation? Well, there is no answer because both of them are currently called Hyperoperations. Your definition gives what are called lower-hyperoperations. The recursive relation for lower h.os is the one you give: $$h_{n+1}(x,y+1)=h_n(h_{n+1}(x,y),x)$$ As you can read on Wikipedia, this is only one of ...


0

What exactly a function $f:X\to Y$ is? Typically we define it as a subset of $X\times Y$ such that for any $x\in X$ there is precisely one $y\in Y$ such that $(x,y)\in f$. And so "well defined function" means: "a subset of $X\times Y$ we just defined is actually a function" which boils down to showing that (a) for any $x\in X$ there is $y\...


1

As $\mathbb R$ is a subset of itself and is a set of real numbers both those definitions are the same. If $f(x) = \sin x$ has $[-1,1]\subset \mathbb R$ as it's range. $g(x) = x^2$ has $[0,\infty)\subset \mathbb R$ as its range and $h(x) =x^3$ has $\mathbb R\subset \mathbb R$ as its range. And $j(x) = 7$ has $\{7\} \subset \mathbb R$ as its range. All are ...


3

A real function is a function whose codomain is $\Bbb R$. So, its range is necessarily a subset of $\Bbb R$, but it doesn't have to be the whole $\Bbb R$.


5

(Disclaimer: This is not at all a historical account - it is a path of intuition that leads to vector spaces, not the only path) Vector spaces! (a.k.a. stuff you can add and scale, but not multiply) Vector spaces are intentionally capturing the notion of quantities that can be sensibly added together or scaled, but not necessarily anything else. This ...


2

When authors claim that something needs to be proven to be well defined that is because the definition given is under suspicion of having some flaw: it might not define anything at all. For example: (Fake) Definition. Let $x$ be the smallest strictly positive real number. Since there is no smallest positive real number, the definition above doesn't define ...


0

Your "narratives" 1–3 are okay, but 4 and 5 are not quite right and can not define the real and complex number fields: You can also solve the equation $x^2=2$ in the field $\mathbf{Q}(\sqrt{2})$, not only in $\mathbf{R}$. Similarly, you can solve the equation $z^2=-1$ in $\mathbf{Q}(i)$, not only in $\mathbf{C}$. Instead, you might want to say ...


4

Here is one possible thought process. It seems natural to attempt to generalize calculus to functions that take a list of numbers as input and return a list of numbers as output. This leads us to introduce $\mathbb R^n$. Since the fundamental strategy of calculus is to approximate a nonlinear function locally by a linear function, we must define what it ...


6

For a function to be well-defined, its output needs to be unambiguous. So if $a$ = $b$, then we need $f(a) = f(b)$. Let's look at something that's not well-defined. Let's try to define addition on the rational numbers as $$ \frac{a}{b} + \frac{c}{d} = \frac{a + c}{b + d}. $$ Let's look at $\frac{1}{2} + \frac{1}{3}$. Under this proposed definition of ...


5

Ultimately, well-defined means simply: defined, however in a special kind of defining functions. Namely, we want to define a map $f\colon A\to B$ in terms of given maps $ g\colon C\to A$ and $h\colon C\to B$. We attempt to do so by saying that for given $x\in A$, we pick $z\in C$ with $g(z)=x$ and then set $f(x):=h(z)$. But is that really a definition of a ...


1

A function $f: A \to B$ is well defined if for every $x \in A$, $f(x)$ is equal to a single value in $B$. So if $f(x)$ could equal two different values in $B$, then it is not well defined. Or if $f(x)$ is not equal to a value in the set $B$, then it is not well defined.


3

I'll describe well-definedness in terms of a simpler function. Let's say that I wanted to describe a function on non-zero rational numbers by $$f(p/q)=q/p$$ In this case, it falls on me to show that the value of $f$ at a certain rational number is independent of the way that the ratio is formed, for instance that $f(1/2)=f(2/4)$. If I can show that, then ...


5

Introduction Here is the formal definition of a function. You need two sets $A,B$ (domain, codomain) and a subset $f$ of $A\times B$, that is, a collection of pairs of the form $(a,b)$. A pair $(a,b)$ in this set means that "$a$ is in relation to $b$". To obtain a function you need two additional constraints: every element of $A$ is in relation to ...


0

Saying "given by", "verifying", "satisfying", etc. is more suitable for defining functions using words. Otherwise, remember that you can also use the notation: \begin{align*} f: & A \longrightarrow B \\ & a \longmapsto f(a)=b \end{align*} And we'll say it's well defined if any point $a\in A$ has an image through $f$ ...


0

You should say "with $f(a)=b$" or "satisfying $f(a)=b$". Some people might say "where $f(a)=b$", but I recommend "where" Immediately before definitions.


1

That's a definition. You don't prove a definition. That's how we define a differentiable function from $\Bbb R^2$ into $\Bbb R$. We could alse define a differentiable function from $\Bbb R$ into $\Bbb R$ as a function $f$ such that, for each $x_0\in\Bbb R$, there is some $a\in\Bbb R$ such that$$\lim_{x\to x_0}\frac{f(x)-f(x_0)-a(x-x_0)}{x-x_0}=0.$$


1

In the second definition if you take $h=(h_1,0,..,0)$ and let $h_1 \to 0$ you see immediately that $v_1=\frac {\partial f} {\partial x_1}$. Similarly, $v_i=\frac {\partial f} {\partial x_i}$ for all $i$. However, the existence of partial derivatives does not guarantee the existence of Frechet derivative. For counter-examples see Is diffirentiability in ...


0

A partition of a set $S$ is a collection of subsets $S_i\subset S$ such that the union $\bigcup S_i=S,$ and each distinct $S_i$ have empty intersection. In the case of $S=[a,b]\subset \Bbb R^1$, this gives you non-overlapping subintervals of $[a,b]$ which together forms the entirety of $[a,b]$.


5

Here is a formal definition of partitions of intervals: In mathematics, a partition of an interval $[a, b]$ on the real line is a finite sequence $x_0, x_1, x_2, ..., x_n$ of real numbers such that $$ a = x_0 < x_1 < x_2 < ... < x_n = b. $$ Such partition gives you $n$ "subintervals" of $[a,b]$, $$ [x_{k-1},x_{k}],\quad k=1,\cdots,n\...


4

A partition of an interval is a division of the interval into subintervals; the subintervals are the ‘pieces’ of the partition, so to speak, the subintervals of which the partition is composed. The norm of a partition is not the sidest subinterval of the partition: it is the width of the widest subinterval. Consider, for instance, the partition of $[0,1]$ ...


2

I think that you are looking for the phrase ``ideal determined variety''. Might want to start with this paper: http://www.mathematik.uni-marburg.de/~gumm/Papers/Ideals%20in%20universal%20algebras.pdf


4

No, it's also a set. A function with domain $A$ and codomain $B$ is a subset $f$ of $A\times B$ such that, for each $a\in A$, there is one and only one $b\in B$ such that $(a,b)\in f$.


2

In your question you define an atlas on a set $M$ as a collection of bijections $x_\lambda : U_\lambda \to V_\lambda \subset \mathbb R^n$ with open $V_\lambda$ such that suitable conditions are satisfied. This induces a unique topology on $M$ such that all $U_\lambda$ are open in $M$ all all $x_\lambda$ are homeomorphisms. Thus an alternative approach is to ...


1

As I see the problem is that, after you extend the square root function from $[0,\infty )$ to $\{-1\}\cup [0,\infty )$ adding a symbol $i:=\sqrt{-1}$ to $\mathbb{R}$, you still need to construct a minimal field (or an algebra) that contains $\mathbb{R}\cup \{i\}$, and you still would need to define what is the value of $i^2$. One approach could be to set a ...


4

Here is an answer which addresses the underlying logical issues that your question raises. I could define a jackalope to be be a rabbit with antelope horns, but then the trouble arises whether any rabbit with antelope horns exists, because if not then my definition is vacuous nonsense. Similarly, you could define $i$ to be $\sqrt{-1}$, but then you would ...


7

Strictly speaking, you can't just make definitions and hope they work. Practically speaking though, mathematicians have a lot of machinery built precisely to make sense of definitions that basically go, "Hm, I wish I had a element that had this property" - and these sorts of definition are ubiquitous in many branches of math. This extends well ...


2

You should stick to the definition of an indefinite integral. Given a function $f:I\to\mathbb{R}$ defined on an open interval $I$, if $F:I\to\mathbb{R}$ is a function such that $F'(x)=f(x)$ for every $x\in I$, then we call $F$ an antiderivative of the function $f$. It is easy to check the two following two facts: If $F$ is an antiderivative $f$ on the ...


0

Looking back at the question, I began to wonder why I defined the curvature of a circle as $\frac1r$ in the first place. I never really thought deep about it when thinking about curvature, and I wondered if the answer would lie there. Intuitively the amount of "time" it would take for a circle to curve is inversely proportional to its radius. ...


2

$$ \lim_{\Delta x\to\ 0} \frac{\arctan(x+\Delta x)-\arctan x}{\Delta x} = \lim_{u\,\to\,v} \frac{u-v}{\tan u - \tan v} $$ Here we have $u = \arctan(x+\Delta x)$ so $\tan u = x+ \Delta x$ and so on. Why is it that $u\to v$ as $\Delta x\to 0\text{?}$ In effect this says the arctangent function is continuous. That can only follow from properties of the tangent ...


2

Let $y=\arctan x,\,y+\Delta y=\arctan(x+\Delta x)$ so$$\begin{align}\arctan^\prime x&=\lim_{\Delta x\to0}\frac{\arctan (x+\Delta x)-\arctan x}{\Delta x}\\&=\lim_{\Delta x\to0}\frac{\Delta y}{\tan(y+\Delta y)-\tan y}\\&\stackrel{\star}{=}\lim_{\Delta y\to0}\frac{\Delta y}{\tan(y+\Delta y)-\tan y}\\&\stackrel{\dagger}{=}\frac{1}{1+\tan^2y}\\&...


4

$$\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x(x+\Delta x)}\right)}{\Delta x}=\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x^2}\right)}{\Delta x}$$ Now let's introduce the variable $u$ by $$\tan u=\dfrac{\Delta x}{1+x^2}$$ Then $$\Delta x=(1+x^2)\tan u$$and your limit is $$\lim_{u\to\ 0}\frac{\arctan(\tan u)}{(1+x^2)\tan u}=\...


0

Switch to the straight function, find derivative and notice that $$\dfrac{dy}{dx}=\dfrac{1}{dx/dy}=$$ So $$x= \tan(y)$$ In the way you have obtained d.c. you can also find: $$ \dfrac{dx}{dy} = \sec^2 y ={1+x^2}$$ $$ \dfrac{dx}{dy} = \dfrac{1}{\sec^2} =\dfrac{1}{1+x^2}.$$


3

Notation of a preceding exclamation point can represent the subfactorial (a.k.a. derangement number). In this usage, \begin{align*} !0 &= 1 \\ !1 &= 0 \\ !n &= (n-1)(!(n-1) + !(n-2)), n > 1 \text{.} \end{align*}


4

Actually it's the opposite: the non-symmetric part of the bilinear form is redundant. Any bilinear form $B(u, v)$ has a canonical decomposition (over a field of characteristic $\neq 2$) into a symmetric part and a skew-symmetric part $$B(u, v) = \frac{B(u, v) + B(v, u)}{2} + \frac{B(u, v) - B(v, u)}{2}$$ and the skew-symmetric part of $B$ does not contribute ...


1

Both are correct and exactly identical. In particular $f(a)=a$ and $f(b)=b$ for both your proposed definitions.


2

The ruleset given by the book is not incomplete. The example derivation you give holds up to scrutiny as well. You get (seemingly) paradoxical conclusions because restriction (iv) does not actually hold in any of your examples. In your first example, the formula $S$ denotes the following: "$v_2 \text{ is a function } \wedge \langle v_1,w \rangle \in ...


0

The most common set theory, ZFC, has no prohibition on anything being a member of a set. It does not require them, but you can have a model with things that are not sets being members of sets. Those things might as well be sentences. This will not allow you to consider the structure of the sentences themselves. Each one is an atom. You could add some ...


2

Perhaps an example will help. Consider the standard $\mathbb{R}$. You may be aware that intervals of the form $(a,b)$ are open. But intervals of the form $[a,b]$ are not, these are closed. In particular singletons $\{a\}$ are not open. So consider now open intervals $U_n=(-1/n, 1/n)$ for any $n\in\mathbb{N}$. You can then check that $\bigcap_{n=1}^\infty U_n=...


1

The condition that $\bigcup F_n=G$ is unusual, for instance $(\{1,\dots,n\})_{n\ge 0}$ is a reasonable Følner sequence in $\mathbf{Z}$. If $(F_n)$ is Følner, however, one can modify it to get a covering sequence (i.e. $\bigcup F_n=G$). Just fix an enumeration $(g_n)$ of $G$ and set $F'_n=F_n\cup\{g_n\}$. It's still Følner. Indeed if $G$ is finite, ...


0

For the question title: A topological space is a pair $(X, \mathcal{T})$ of a set and a topology $\mathcal{T}$ is a collection of subsets of $X$ (i.e. $\mathcal{T} \subset \wp(X)$) satisfying a particular set of axioms. For the question in the main body: A topology on a set can be defined in many different ways. You can find some of these ways listed here....


1

A subset of a topological space $X$ can be both open and closed. The properties are not mutually exclusive. $\varnothing$ and $X$ are open subsets of $X$ by definition. The closed sets are those whose complement is open. Therefore, $\varnothing$ and $X$ are also closed. Comments) You should distinguish $X$ from $\tau$. We call a set $X$ together with its ...


0

Since $X=\emptyset^\complement\;$ and $\;\emptyset=X^\complement\;$ so in any topological space they are both closed and open sets(clopens). In a topological space these are the only clopens $\iff$ the space is connected


2

You are wrong when you claim that “a topology on $X$ is a collection of open subsets of $X$ which satisfy three properties”. A topology on $X$ is a colletion $\tau$ of sets which satisfy three properties, and then we say that a subset $A$ of $X$ is open when (and only when) $A\in\tau$. And we say that $F\subset X$ is closed when $F^\complement$ is open. So, ...


1

By picking an origin we can think of $A$ as a vector space $V$ WLOG, and then by replacing $f$ with $f - f(0)$ (which preserves the desired condition) we can assume that $f(0) = 0$ WLOG. Then the condition is that if $p_1 - q_1 = p_2 - q_2$ then $f(p_1) - f(q_1) = f(p_2) - f(q_2)$. Rearranging and setting $q_2 = 0$, the condition is that $$f(a + b) = f(a) + ...


0

The following is a published definition for a symmetric $n$-ary relation, with the qualifier of strongly to denote that all permutations must hold. [An $n$-ary (finitary) relation $\rho$ is said to be] strongly symmetric if $(x_1, \cdots, x_n) \in \rho$ implies $(x_{\sigma(1)}, \cdots, x_{\sigma(n)}) \in \rho$ for any permutation $\sigma$ of the set $\{ 1, \...


1

Because of point 2.: $$\left< x , y \right> = \overline{\left< y , x \right>}$$ applied for $x= y$, you get $$\left< x , x \right> = \overline{\left< x , x \right>}.$$ Therefore $\left< x , x \right>$ is a real number.


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