Problem: Let $X$ be a compact Hausdorff space and $f : X \to X$ a continuous function. Show that there is some non-empty subset $A$ such that $f(A) = A$. Proof: Let $A_0 = X$, and let $A_n := f^n(X)$ for every $n \in \Bbb{N}$. Then $\{A_n\}_{n=0}^\infty$ is a sequence of non-empty, decreasing, compact sets. Hence $A := \bigcap_{n=0}^\infty A_n$ is non-empty....
I've been able to show that $f(A) \subseteq A$, but I'm having trouble showing $A \subseteq f(A)$. I could use some help.
> More generally, if f is monotonic, then the least fixpoint of f is the stationary limit of fα(0), taking α over the ordinals, where fα is defined by transfinite induction: fα+1 = f ( fα) and fγ for a limit ordinal γ is the least upper bound of the fβ for all β ordinals less than γ. The dual theorem holds for the greatest fixpoint.
In the mathematical areas of order and lattice theory, the Knaster–Tarski theorem, named after Bronisław Knaster and Alfred Tarski, states the following:
Let L be a complete lattice and let f : L → L be an order-preserving function. Then the set of fixed points of f in L is also a complete lattice.It was Tarski who stated the result in its most general form, and so the theorem is often known as Tarski's fixed point theorem. Some time earlier, Knaster and Tarski established the result for the special case where L is the lattice of subsets of a set, the power set lattice.The theorem has important...
It's complicated to explain what the transformation I'm dealing with is but basically it is a transformation that takes a 3D model and maps it to the surface of one polygon on another model. The issue is that I'd prefer to not have to recompute the normal vectors of the post-transformation model ...
Let $X=\{(x,y)\in \mathbb R^2:x=0 \text{or } x=\frac{1}{n}$ for some $n \in \mathbb N$, and $0\leq y\leq 1\}$. I know that equivalence class $[(0,0)]=\{(x,y)\in \mathbb R^2|(x,y)$ and $(0,0)$ in the same connected component$\}$. Why $[(0,0)]$ is not an open set? Under what topology it is not open? Seeing $(X,\tau)$ as a subspace topology of usual topology on $\mathbb R^2$? $[(0,0)]$ can not be written as the intersection of $X$ with any open ball in $\mathbb R^2$. am I correct?
Am I done a rigorous proof that $[0,0]$ is not open?
Hey!! Let $K$ be a normal extension of $F$ and $f\in F[x]$ be irreducible over $F$. Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$. Could you give me a hint how we could show that there exists $\sigma \in G(K/F)$ such that $g_2=\sigma (g_1)$ ?
Ok, I think the first step is since T is linear, it follows it will map a linear combination of vectors to a corresponding linear combination
so linear conbinations are preserved
In particular, linear combinations that gives the zero vector will also be preserved
You cannot have T(0)=/=0 as T(x)=T(x+0)=T(x)+T(0) thus T(0) must also be an additive identity, hence can only be a zero vector in the image. Hence subspace structure is preserved
Recall that to check whether something is a subspace, check whether it contains the zero vector and check that it is closed under addition and multiplication
$T^{-1}(0)=0$ by the same logic as $T(0)$ as described below because $T^{-1}$ is also a linear map
If, some elements in $W'$ are never hit/map by $T$, e.g.
no element $x\in W$ s.t. $T(x)=w'$ for some $w'\in W'$
why won't this affect that $T^{-1}(W')$ be a subspace?
this is what I can prove but never understand why
I randomly circle some elements in $W'$ to make them a subspace then something are never mapped back to domain and then I say: yeah, $T^{-1}$ keep subspace.
Hmm... if there is no $x$ such that $T(x)=w$ for some $w \in W'$, will that mean there is no $a,b,c$ such that $w=a+b+c$ such that $w = T(a+b+c) = T(a)+T(b)+T(c)$?
The domain of $T^{-1}$ is the range of $T$ be definition of being its inverse. That means, anything that is in $W' - \text{Ran}(T)$ lies outside the domain of $T^{-1}$. Thus the image of $T^{-1}$ is a subspace properly contained in $W'$
I'm really sorry about my misleading symbol, which is from my stopid book, but $T^{-1}$ I meant something like $R^{-1}$ where $R$ is a relation not necessarily function.
so for example $T$ not one-one then $T^{-1}$ never a function
The conclusion is the same, because points in $W'-\text{Ran}(T)$ were never reached from mapping by $T$, thus the inverse relation $T^{-1}$ that gives the set of points that map to it is empty. This holds even if $T$ is not injective
Since $T$ is not surjective, the preimage of $W'-\text{Ran(T)}$ is empty
@Alucard I'm taking a Discrete Mathematics course now, until then I need some reference guides and practice resources as combinatorics is taking a huge component
plaise if for a function $u:\Omega\subset \mathbb{R}^n\to \mathbb{R}$ we know that the positive part $u^+\neq 0$ why $|\Omega_+|>0$ where $\Omega_+=\{x\in \Omega, u(x)>0\}$
@Alucard Well I think the biggest takeaway for this is I can stop using Venn diagrams ... I'm too used to doing Boolean algebra and Venn diagrams and didn't really know about the probability tree.
Are there scenarios where a probability tree can't be used ?
Hi. I am not a native speaker of English. So, the bold part in the following sentence seems odd to me a bit. "$ \vec{a} $ satisfies $\varphi \Leftrightarrow \psi $ iff either $ \vec{a} $ satisfies both $\varphi $ and $ \psi $ or $\textbf{$ \vec{\textbf a} $ doesn't satisfy either $ \varphi $ or $ \psi $}$" It doesn't seem correct to me. What do you think?
I mean it doesn't seem to be equivalent to the phrase "$ \vec{a} $ satisfies neither $ \varphi $ nor $ \psi $ "
I do not see the difference between the last sentence and your sentence, but I understand your confusion. In any case, your last sentence is what is meant there.
(I imagine you parsed it as "a fails to satisfy at least one of phi, psi". This is not what they mean to say, but I also see how one can read ir that way.)
if for a function $u:\Omega\subset \mathbb{R}^n\to \mathbb{R}$ we know that the positive part $u^+\neq 0$ why $|\Omega_+|>0$ where $\Omega_+=\{x\in \Omega, u(x)>0\}$
@AlessandroCodenotti I can see that $\displaystyle \lim_{x\to b} f'(x) = k$ in your example, but I think $\displaystyle \lim_{x\to b^+} f'(x) $ is not defined.
What about if I have $f,g : \mathbb{R} \to \mathbb{R}$ with both continuous and with $g$ bounded, and $\phi : (-\infty, b) \to \mathbb{R}$. If I compute $\displaystyle \lim_{x\to b^+} f(x)g(\phi(x))$. My notes say $\displaystyle \lim_{x\to b^+} f(x)g(\phi(x)) = f(b)k $ since $g\leq k$ (bounded).
I think it shouldn't be defined.
Uhm does $\lim_{x\to +b}$ mean we approach $b$ from the left?
Assume there are two communities in this network: {A, B, C, D, G} and {E, F, H}. Which of the following statements is(are) True? Select all that apply. - The Common Neighbour Soundarajan-Hopcroft score of node C and node D is 2. - The Common Neighbour Soundarajan-Hopcroft score of node A and node G is 4. - The Resource Allocation Soundarajan-Hopcroft score of node E and node F is 0. - The Resource Allocation Soundarajan-Hopcroft score of node A and node G is 0.7.
Let $K$ be a normal extension of $F$ and $f\in F[x]$ be irreducible over $F$. Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$. Show that there exists $\sigma \in G(K/F)$ such that $g_2=\sigma (g_1)$.
I know that for finitely generated space surjectivity implies injectivity. Only problem is that we need to show $\hat b$ is surjective map. So can somebody explain me that tensor product calculation?
In the "polynomial" proof of the Cauchy-Schwarz Inequality, why does the discriminant of the polynomial have to be less than or equal to $0$? I never understood that...
I.e., the polynoimial $q(t) = \langle v - tu, v - tu \rangle = t^2 ||u||^2 - 2t \langle v,u \rangle + ||v||^2$.
Being like @TheSimpliFire and keep a question from last September open by commenting with valuable data in next year's October Now, that's dedication to a problem.
In mathematics, a field is a set on which addition, subtraction, multiplication, and division are defined, and behave as the corresponding operations on rational and real numbers do.
A field is thus a fundamental algebraic structure, which is widely used in algebra, number theory and many other areas of mathematics.
The best known fields are the field of rational numbers, the field of real numbers and the field of complex numbers. Many other fields, such as fields of rational functions, algebraic function fields, algebraic number fields, and p-adic fields are commonly used and studied in...
In mathematics, the p-adic number system for any prime number p extends the ordinary arithmetic of the rational numbers in a different way from the extension of the rational number system to the real and complex number systems. The extension is achieved by an alternative interpretation of the concept of "closeness" or absolute value. In particular, p-adic numbers are considered to be close when their difference is divisible by a high power of p: the higher the power, the closer they are. This property enables p-adic numbers to encode congruence information in a way that turns out to have powerful...