is it possible there's a proof out there that we haven't figured out yet that doesnt use choice, that the countable union of countable sets is countable?
I didn't find proofs that the latter implies choice, only that choice implies the latter
Let us denote by ACC the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by UCC the assertion that a countable union of countable sets is countable.
UCC is a simple theorem of ZF+ACC.
Proof Suppose for every $i\in\omega$...
I know that this question has been asked a lot here, some of them are duplicate of each other. I’ve read every single of them, but my problem has not been resolved. I can just memorize that Axiom of Choice (AC) is needed, but I want to be clear of this, logically.
Note : “Countable” here means i...
Munkres brings a proof that it's possible to construct the first uncountable ordinal without choice. then he says the result is less exciting than it sounds at first glance because you need the theorem about countable unions of countable sets to prove that every countable subset has a supremum, and without that aspect it's not a useful set
In mathematics, specifically in axiomatic set theory, a Hartogs number is a particular kind of cardinal number. It was shown by Friedrich Hartogs in 1915, from ZF alone (that is, without using the axiom of choice), that there is a least well-ordered cardinal greater than a given well-ordered cardinal.
To define the Hartogs number of a set it is not necessary that the set be well-orderable: If X is any set, then the Hartogs number of X is the least ordinal α such that there is no injection from α into X. If X cannot be well-ordered, then we can no longer say that this α is the least well-ordered...
Yup, you find the first ordinal that does not inject into some given set
$\omega_1$ is constructed by taking any set $X$, consider the class of all ordinals that can be injected into $X$, use the axiom schema of replacement to map these ordinals into the order type of $X$ (this is the impredicative step, because it requires a definable class function which is not present beforehand) and the resulting set of maps is itself an ordinal, $\omega_1$
It might seems like we can't prove the existence of an uncountable ordinal without the axiom of choice. For example, here's the following incorrect argument:
"Let s be the order type of the set of all countable ordinal numbers, then s must exceed all countable ordinal numbers so it's uncountable...
We know that min(x)+min(y)!=min(x+y) then how can we find min(x+y) if we only know the values min(x) and min(y), i.e arrays x y are not available. If not possible then what is the closest approximate of min(x+y)
We know that min(x)+min(y)!=min(x+y) then how can we find min(x+y) if we only know the values min(x) and min(y), i.e arrays x y are not available. If not possible then what is the closest approximate of min(x+y)? can min(x+y) be approximated in any way?
@Islam If the array x and y does not have some notion of continuity (e.g. can be fit into some nice functions such as exp, sine, polynomials etc.) it is in general impossible to get min(x+y) if you don't know the full details of x,y
@Islam If x and y are linear and monotonous, then in principle the partial data will allow you to extrapolate to the unknown region to get the best linear regression of x,y and thus allow min(x+y) to be computed closely
the brute force way to tell if it's diagonalizable is to check if the characteristic polynomial splits into $k$ linear factors over the underlying field
when the matrix is $k \times k$
there are sufficient conditions that don't involve actually looking at the characteristic polynomial
but with a 2x2, brute force isn't that hard
it has $k$ linear factors if it has $k$ distinct eigenvalues over the field you're considering
The first isomorphism theorem for sets is the statement that if $f:X \to Y$ is a map, then the induced map $X/\sim \to \operatorname{Im}(f)$, where $a \cong b \Leftrightarrow f(a)=f(b)$, given by $[a] \mapsto f(a)$ is a bijection
if $L/K$ is transcendental, then this is clear. If we suppose that $L/K$ is minimal, then for any transcendental element $x \in L$ by minimality we get $L=K(x)$, but becaue $L$ is transcental, $K(x^2)$ is a proper subextensioin
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Deduce that if the characteristic polynomial of $T$ splits, then any
non-trivial $T-$invariant subspace of $V$ contains an eigenvector of
$T.$
Let $W$ be a $T-$Invariant subspace. $W\neq\{0\}$($\because$ Given that ...
one of my professors once told me that, as an immature mathematician, he believed that statements were more mathematically correct if they used more symbols and fewer words
Secret the upper bound of correctness for mathematical statements assumes that the levels of correctness are well orderable which requires choice. assuming choice, the upper bound is $\text{true}$
my professor told me about a professor of his that used $\mathbb Z_2$ as the ring to compute homology groups because then you don't have to worry about sign errors, as $-1 = 1$
or more precisely
he always made sign errors and his students encouraged him to use $\mathbb Z_2$
reminds me of the joke about how a mathematician wakes up to find himself in a burning room of his apartment. He looks outside, grabs a piece of paper, and calculates the velocity and trajectory needed to land on the ground with minimal impact upon hitting the ground
unfortunately, because of a sign error, he jumped out and fell up into the sky
Fluxional spacetime: The metric and its derivative has significant time dependence, suggesting raging gravitational fields in the vicinity
Unstable spacetime: Requires a local metric in the spacetime manifold to describe, as the topology of spacetime started to vary erratically in both spatial and temporal directions, which means neighbouring patches may have completely different metrices
Highly unstable spacetime: The patches become infinitesimally small, as the topological variations from one point to the next become arbitrarily large. Spacetime is on the brink of becoming chaotic
The Bulk: Spacetime becomes chaotic and random. Pockets of stable spacetimes can pop in and out of existence
Approaching timelessness: Causal ordering started to become ambiguous and then lost. Only pure spatial structures remain
Approaching spacelessness: Topology started to relax into pretopology, become increasingly chaotic, random, indeterministic, indeterminable, blurred, and finally lost completely. No human being nor any physical matter can survive in this realm
Approaching ambigurity: Metaphysical entities started to blend and blur with each other, as the law of identity become increasingly fragmented and ultimately fail
The Ambigurity: Everything is completely homogenised, no matter it is laws, identity, god, space, time, metaphysics, magic, spiritual, ineffables, nothing etc., and become one with this amorphous homogeneous "dead"
Let $R$ be a ring and $G$ a group. How can I explicitly describe the adjunction between $RG$-modules and $G$-sets using the structure arrows? Given a group arrow $\rho :G\to \mathsf{Set}(X,X)$ for instance I don't see how to formally get a ring arrow $RG\to R$-$\mathsf{Mod}(RX,RX)$ and vice versa.
The group algebra $\dashv $ group of units seems to move between algebra arrows $\mathsf{Mod}(RX,RX)$ and group arrows $G\to R$-$\mathsf{Mod}(RX,RX)^\times $, but I don't recognize the latter as the group of bijections of some set.
hi. suppose $B \subset A \times A $ is countably infinite, i need to show that $A$ has a countably infinite subset, i thought taking $\Pi_1(B) \cup \Pi_2(B) $ , but im not sure how to show that this set is countably infinite, someone can help?
@Abr001am lots of series can be nicely reduced to telescoping sums. It usually takes some time to develop the ability to see the telescoping sums for harder problems.
If I have $\kappa$ countable sets, when their union is not countable? only if $\kappa$ is uncountable? Using AC make differences?
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Deduce that if the characteristic polynomial of $T$ splits, then any
non-trivial $T-$invariant subspace of $V$ contains an eigenvector of
$T.$
Let $W$ be a $T-$Invariant subspace. $W\neq\{0\}$($\because$ Given that ...
