5

As you've noted if $g^5=e$ for some $g\in G$ then $x^5=e$ for any $x\in R_g=\{g,g^2,g^3,g^4\}$ and these sets are either equal or disjoint, and $|R_g|=4$ if $g\neq e$. And so if $k$ is the number of nontrivial solutions to $x^5=e$ and $k<\infty$ then indeed $4$ divides $k$. Note that I didn't assume that $G$ is finite, only that $k$ is. Also note that ...


4

Let me start by making a couple observations: As soon as you have an uncountable set of things, you can also get a countable set of the same type of thing by just throwing out a bunch of stuff. So in one sense your question is trivial. On the other hand, things get more interesting when we demand concreteness: are there "reasonably-definable" examples of ...


4

Although it is not easy to generate the bijective function $f\colon\mathbb{N}\times\mathbb{R}\to\mathbb{R} $, we can instead use Cantor-Bernstein theorem which states that, if there are injections $\mathbb{N}\times\mathbb{R}\to\mathbb{R} $ and $\mathbb{R}\to\mathbb{N}\times\mathbb{R} $, then they have the same cardinal number. Therefore it suffices for us to ...


4

Envision the famous Hilbert Hotel, where there are an infinite number of rooms numbered $1, 2, \ldots$. All the rooms are occupied. To create infinitely many more vacancies, the hotel manager reassigns each guest from room $n$ (which they currently occupy) to room $2n$; that is, all the odd-numbered rooms will become vacant and all the even-numbered rooms ...


4

I am not sure if Cantor did it this way, but this argument works: any number $x$ in $[0,1]$ has an expansion to base $2$: $x=\sum \frac {a_k} {2^{k}}$ where $a_k =0$ or $a_k=1$ for each $k$. This expansion is not unique but it can be made unique by avoiding expansions with $a_k=1$ for all but finitely many $k$ (except when $x=1$). Now let $r \in \{0,1,2,...,...


4

If both $n$ and $a$ are fixed, then the set is , the singleton $\{na\}$ If $a$ is fixed and $n \in \Bbb N$ is a varying quantity, then the set is $\{na: n \in \Bbb N\}=\{a,2a,3a,\cdots\}$ In both cases, $A$ is one set with cardinality $1$ and infinite(of course, $\aleph_0)$ respectively!


3

Your claim is not true if $X$ is an infinite set. Consider the function $f(x)=2x$ mapping the natural numbers to itself. It is injective but not surjective.


3

The proof is correct, there's no need to improve anything unless you want to be very formal. You can omit the quantifier $\forall$, because if something is true for all $x$, then it's particularily true for the one $x$ that you're given.


3

Yes. Let $f: X \to Y$ be a bijection. Then show that $\hat{f}: \mathscr{P}(X) \to \mathscr{P}(Y)$ given by $$\hat{f}(A) := f[A] (= \{f(x): x \in A\})$$ is a bijection between their power sets.


3

You want to prove that, for any set $A$, if $|\mathbb N\times\mathbb R|\ge|A|$, then $|\mathbb R|\ge|A|$. No bijections are needed. You just have to show that $|\mathbb R|\ge|\mathbb N\times\mathbb R|$; then you will have $|\mathbb R|\ge|\mathbb N\times\mathbb R|\ge|A|$. In order to show that $|\mathbb R|\ge|\mathbb N\times\mathbb R|$, you need an ...


3

No, the statement is correct: the minimum of $x$ and $y$ in a linearly ordered set is also the infimum: the maximal lower bound for $\{x,y\}$ (which could exist in any partially ordered set with incomparable $x$ and $y$, see lattices etc.). Leinster is really saying that a linearly ordered set is a lattice. It might be (I don't know the book) that soon ...


3

No, those are two different concept: countable comes from set basic theory; the main property, one might argue, of a set, is that it has cardinality: it has a number of elements. This number can be 0 (if the set is empty), infinite or any number in between. The way we compare the size of two sets is by trying to construct isomorphisms (bijections) between ...


3

We want to prove, that $A\times (B-C)=(A\times B)-(A\times C)$ So let $(x,y)\in A\times (B-C)$. Then $x\in A$ and $y\in B$ and $y\notin C$. Hence $(x,y)\in A\times B$ and $(x,y)\notin A\times C$ and we conclude $(x,y)\in (A\times B)-(A\times C)$. Go through this proof one by one. If you struggle with one point, go and look at the definition of the symbols. ...


3

As Dave points in a comment, you can conclude that $\inf A \geq \varepsilon$, yielding $\inf A \geq \varepsilon > 0,$ whence $\inf A > 0$. Why can't you conclude that $\inf A = \varepsilon$? Well, suppose, for example that $A=\{2\}$ and $\varepsilon=1$. Then $\inf A = 2$. Of course you still have that $\inf A > 0$, which is what you wanted to ...


2

In Cantor's cardinal arithmetic, $2\cdot\aleph_0=\aleph_0 $. When dealing with levels of infinity, the rules are different (than in the finite case).


2

First of all, you try to work with limits and want to use that an expression at the limit point equals the limit of said expression as we approach the limit point. But for that you first of all need to know that the function you consider is defined at the limit point. So, how do you define division at infinity? And even if defined, you'd need continuity for ...


2

Perhaps $\infty=\frac12\cdot\infty$ seems wrong, but $0=\frac12\cdot0$ should illustrate that there are numbers that remain unchanged when divided by $2$. It is possible that $2\cdot\infty=\infty$. What you describe above means that the "density" of the set of even natural numbers is $\frac12$, but the concept of density is different from that of cardinality....


2

Both are correct, because $\aleph_0 = 2\aleph_0$ For example, $|\mathbb{N}|=\aleph_0$ and $|\mathbb{Z}| = \aleph_0$ as well. I think this illustrates why that is true.


2

The paper that Cantor published his result in is Ein Betrag zur Mannichfaltigkeitslehre from 1878. (the link is to the original German version). In this he reduces to the irrationals in $[0,1]$, which have a unique representation in terms of continued fractions of positive integers, and where he can easily associate $n$ irrationals to a single new one in a ...


2

A typical example of a partially ordered, non-linearly ordered set is $\mathcal{P}(X)$, all subsets of $X$ (a set with at least $2$ elements) ordered by inclusion. If $A$ and $B$ are non-comparable sets, they still have a lower bound $\emptyset$ and upper bound $X$. Consider what $\sup \{A,B\}$ is, we cannot have maximum as the sets are incomparable, but $...


2

It's clearly in bijective correspondence with $\Bbb Q^4$: if $f$ is such a function, map it to $(f(\sqrt{2}),f(\sqrt{3}),f(\sqrt{5}),f(\sqrt{7})) \in \Bbb Q^4$. The function $f$ is completely determined by these 4 rational values, and all tuples define such a function. The square $A^2$ of a countable set $A$ is countable, and we can apply this twice to the ...


2

Note that $$ A=(A\cap B) \cup (A-B)$$ The sets on the RHS are disjoint so $$|A|=|A\cap B)|+|A-B|$$


2

Let $m$ be the minimum of $x,y$. If $a \le m$, then $a\le m\le x,y$, so $a\le x,y$. Conversely, suppose $a\le x,y$. If $m < a$, then $m < a \le x,y$, so $m < x,y$, contradiction, since $m$ must be equal to one of $x,y$. Thus, since we can't have $m < a$, it must be the case that $a\le m$. Let's take a numerical example . . . The minimum of $...


2

You can also write it as $a\Bbb N$.


2

When showing that a relation does not have a property, usually the fastest and most obvious way is to give a concrete counterexample. For instance, $1\simeq0$ and $0\simeq1$, or $0\not\simeq\frac12, \frac12\not\simeq0$ and $0\neq\frac12$, either pair proving that the relation doesn't have trichotomy. Of course, some times general arguments can show ...


2

The power set is a subset of itself, and it is a sigma algebra, but there may be others. In practice, for example in measure theory on $\mathbb{R}$, we don't take the power-set because it is too complicated. Instead, we work with a smaller sigma-algebra, like the Borel sigma-algebra on $\mathbb{R}$, which is generated by open intervals. This ...


2

I guess you're asking about why they said $a \in X, b \in Y$ in the second definition but not in the first definition. In fact, the definitions are equivalent and the choice is purely a matter of style. In particular, every object belongs to some set, so you could always in the first definition have said "let $A$ be a set containing $a$, and $B$ containing $...


2

Let's first think about the definition of the cartesian product you gave. I claim that it is just a formalization of the intuitive notion that you have in mind. Namely, an element of $\prod_{\alpha \in \mathcal{A}} A_{\alpha}$ should be an "$\mathcal{A}$-indexed tuple" $(x_{\alpha})$, with each $x_{\alpha} \in A_{\alpha}$. But what is the precise definition ...


2

After doing some more research, I saw that the most natural notion to use here is that of an ideal in the set-theoretical sense. In fact, if we define $\mathcal{A}':=\{A'\subseteq\mathbb{N}\ |\ \exists A\in\mathcal{A},F\subseteq\mathbb{N}\text{ finite}:\ A'\subseteq A\cup F\}$, then $\mathcal{A}'$ is a set-theoretical ideal on $\mathbb{N}$. Furthermore, my ...


1

Reading these through: Set $1$ contains "any integer which is greater than $5$" Set $2$ contains "any number which is an integer and greater than $5$" It's clear to see they are the same thing;.


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