let U be a covering and construct a (possibly finite) sequence of elements of U: let U_1 be any element of U. Suppose we have gotten to U_k. If U_1, .., U_k cover X, then stop; if not, let U_{k+1} be an element in U which is not covered by them. If this procedure never stops, we would get an strictly increasing infinite chain, which is impossible. Therefore the construciton stops.
@Ben: Let $\mathscr{U}$ be an open cover of $X$. Pick $x_0\in X$ arbitrarily; there is some $U_0\in\mathscr{U}$ such that $x_0\in U_0$. Let $\mathscr{V}_0=\{U_0\}$. If $\mathscr{V}_0$ doesn’t cover $X$, pick $x_1\in X\setminus\bigcup\mathscr{V}_0$, choose $U_1\in\mathscr{U}$ such that $x_1\in U_1$, and let $\mathscr{V}_1=\mathscr{V}_0\cup\{U_1\}$.
Keep going: if $\mathscr{V}_k$ doesn’t cover $X$, pick $x_{k+1}\in X\setminus\bigcup\mathscr{V}_k$, choose $U_{k+1}\in\mathscr{U}$ such that $x_{k+1}\in U_{k+1}$, and let $\mathscr{V}_{k+1}=\mathscr{V}_k\cup\{U_{k+1}\}$.
Then $\bigcup\mathscr{V}_0\subsetneqq\bigcup\mathscr{V}_1\subsetneqq\bigcup\mathscr{V}_2\subsetneqq\cdots$, so by the ACC this construction must stop at some $n\in\Bbb N$, and at that point it must be the case that $\mathscr{V}_n$ covers $X$. This $\mathscr{V}_n$ is evidently a finite subcover of $\mathscr{U}$.
(I had to split it to keep it from being too long.)
@BrianMScott Perhaps it would be reasonable if @Ben posted this as a question and you as an answer on the main site...? (I mean, since you have already written it down and it might be useful to other users, too...)
@BenjaminLim Contradiction? It’s often an easier way to approach the problem. But if you then find that you didn’t actually need it $-$ that your proof is really a direct proof in disguise $-$ then why not write it as such?
there is a foot note in a book translated by jacobson (I don't recall which) in which he observes that the author he is translating is dragging a dead horse
last year I had a student who solved a problem in an exam like that: by midway along his solution he had gotten exactly what he wanted, and he went on and on carrying a dead horse
and managed to make such a mess that the whole thing was ruined :(
@Mariano : I have a question I am contemplating to ask on SE or MO, it sounds like this, there is a theorem to find this, hence there should also be a theorem to find this as well (by intuition and plausibility). (But i can't make any statement of the theorem, i don't know how it would sound) ? Would it be an agreeable question?
I posted a second answer to the question asking for a separable space with a basis such that there is an open set that cannot be written as a countable union of basis sets.
Take $X = \mathbb R$ and the basis consisting of all sets $\{x\} \cup (\mathbb Q \cap (x-\varepsilon, x + \varepsilon))$ for all reals $x$ and all $\varepsilon > 0$.
Then $\mathbb Q$ is a dense subspace and you can't write $\mathbb R$ as a countable union of basis sets.
@BrianMScott Is this correct? Or am I making a mistake?
Nate also posted another answer, after mine, so I assume he saw my post. And usually he posts a comment to my (wrong) answers, pointing out my mistake.
But now that I didn't get a comment I'm not sure what to think.
Given a matrix A and a B normal matrix, how can we prove that $BAB^{-1} = C$ is a diagonal matrix?
I've found that A is a Hermitian matrix, so this entails that it has n linearly independent eigenvectors, which gives us that A is a diagonalizable matrix.
So C is a diagonal matrix as A is a diagonalizable matrix.
There is a theorem in plane geometry which says that in a right triangle, if $c$ is the hypothenuse and it is divided into parts of lengths $c_a$, $c_b$, the legs and altitude fulfill $h^2=c_ac_b$, $b^2=cc_b$ and $a^2=cc_a$.
This theorem is called Euclid's theorem (or Euclid's theorem on altitude, Eudlid's theorem on leg) in Slovak/Czech, see Wikipedia.
To rephrase my question: I want to find out if there is a matrix $C \in R^{4x4} $, given a matrix $A \in R^{4x4} $ (A is symmetric, Hermitian) that this equation will be true: AC = CA + A
What I did was: AC = CA + A => A = AC - CA That's why I wanted to know if AB and BA matrices are similar then their subtraction would be 0.. But I've noticed that I am not sure if AC and CA matrices are similar @MartinSleziak
Yes I know, I've thought that and also thought of C as the zero matrix, but we are being asked if there is just one matrix C, for which the equation $AC = CA + A$ will be true. @MartinSleziak
When you say regular you mean that there is $C$ and $C^{-1}$ ?
:-) I'm reading the Stein & Shakarchi book, the section 3.1 of product measures and Fubini theorem. They start with $(X_1,M_1,\mu_1)$ and $(X_2,M_2,\mu_2)$ and then considering measurable rectangles $A\times B$ with $A\in M_1$ and $B\in M_2$. They claim that the collection $\mathcal A$ of measurable rectangles is an algebra
We assume that, let C be: $C = BAB^{-1}$ What can be said about the result of $ABAB^{-1}$ ? Yes I don't think that we have to sovle such a linear system..
@JonasTeuwen Sorry. My mistake. $\mathcal A$ is the collection of disjoint unions of measurable rectangles. They claim that the complement of a measurable rectangle is the disjoint union of 3 such rectangles while the union of two measurable rectangles is the disjoint union of at most six such rectangles
@leo @JonasTeuwen I don't get why the union of two measurable rectagles is the disjoint union of at most $6$ measurable rectangles. It seem to me like at most $3$ are needed: $(A_1\times B_1)\cup (A_2\times B_2)=(A_1\times B_1)\cup (A_2\backslash A_1\times B_2)\cup(A_1\cap A_2\times B_2\backslash B_1)$.
Of course, but Stein&Shakarchi actually write about at most six disjoint rectangles being needed. And these authors seem to have a very geometric viewpoint.
equation of a line of slope $m$ and passing through a point $(x_1,y_1)$ is $(y-y_1) = m(x-x_1)$. Now you have $(x_1,y_1)$ given. It has been given that the line is perpendicular to another line. Which means the slope of your line is the negative of the reciprocal of the slope of the other line. Hope this helps
Originally I thought "oh, someone always downvotes BillD, and Vadim got twice the upvotes as Mark even though their answers were the same and Mark was first, so that explains Vadim's downvote." But then I saw MT's comment and looked at all the other answers. I really don't see why they would downvote any of them, let alone more strangely all but mine.
I flagged the first portion of the downvotes and the posts were undownvoted. Do moderators do such things or must the downvoter do it?
(The flag was deemed helpful)
user19161
@ymar Downvoters cannot retract downvotes unless the post has been edited after the downvote. If there are many downvotes within a short period they may be detected and removed silently by the system. In other cases, human intervention is needed.
