New answers tagged

0

I have not seen any special symbol for this. I would write $S=\{\{a,b\}:a\in A\land b\in B\}.$ Of course if $x\in A\cap B$ then $\{x\}$ belongs $S.$ If you wish to exclude one-member sets you can write $T=\{\{a,b\}:a\in A\land b\in B\land a\ne b\}. $.... However there is are symbols $[A]^2$ for the set of all subsets of $A$ that have exactly $2$ members ...


1

Your statement is: $(*)$ Given non-empty sets $A$ and $B$, there is an injective map from $A$ to $B$ or from $B$ to $A$. The statement $(*)$ is equivalent to the axiom of choice, as follows. $AC \implies (*):$ Consider the set $P$ of all injective maps from a subset of $A$ to $B$. Partially order $P$ by set inclusion. (We're thinking of the functions ...


0

This is appropriate terminology, although one interpretation that may feel intuitive from a software perspective is to view each "component" of a tuple as the $i$th element in the tuple. Hence in $(a,b,c)$, $a$ is located as the 1st element of the tuple, $b$ the 2nd, and $c$ the third. This gives a very nice notational representation to a tuple as a vector. ...


0

It is easily shown for Von-Neumann ordinals. But each set is in bijection with its cardinal (a Von-Neumann ordinal) so it is sufficient to show for Von-Neumann ordinals.


-1

I will only answer in terms of discrete domains, where the number of elements is countable. It's very simple, if the domain contains fewer elements than the range then an injection is not possible. It's a simple pigeonhole principle problem where $n$ is the amount in the domain and $m$ is the amount in the range therefore any map which maps all elements of ...


1

It is not closed since there is a sequence of elements in this set which converge to $5$, yet $5$ is not in this set. Notice that it is also not open set. If $S \subset \mathbb{R}$ is open, then for every point in $x\in S$, there exists a ball $B_{\epsilon}(x)$ such that $B_\epsilon(x)\subset S$. This is clearly not true by taking $x = 4$.


2

In expansion of my comment, this is true assuming the axiom of choice. If $A$ and $B$ are sets, they are in bijection with cardinals $\kappa$, $\mu$ respectively. Then, $\kappa \in \mu$, $\mu \in \kappa$, or $\mu = \kappa$. In each case, have have an injection (i.e. inclusion\equality).


2

It depends on what $A$ is. In fact, (assuming the axiom of choice) the number of equivalence classes of $\mathcal{P}^*(A)$ for $A$ uncountable can be any infinite cardinality at all, by choosing $A$ appropriately. By definition, if $|A|=\aleph_{\alpha}$, the infinite cardinals that are less than or equal to $|A|$ exactly the cardinals $\aleph_\beta$ for $\...


2

If $|A|=\omega_\alpha$, there are $\omega+|\alpha|$ cardinal numbers less than or equal $|A|$ and hence $\omega+|\alpha|$ equivalence classes under $\sim$. Thus, the answer is $\max\{\omega,|\alpha|\}$. (I am assuming the axiom of choice here, as otherwise matters get really messy.)


0

It's unclear the question but i think i can give an answer: You can think "the set of all different sets consisting on exactly one element of each subset of a a set $S$" as the cartesian product: \begin{equation} \prod\mathcal{P}(X)\setminus\{\emptyset\}=\{f\mid f:I\to\bigcup_{i\in I}X_i, \text{ }f(i)\in X_i\} \end{equation} where I is an index set for $\...


0

I'm on a cell phone so let the kernel be $A$. Using the axiom of choice, construct a set $N'$ consisting of one element from each coset of $A$. There is an obvious bijection of $N'$ with $N$. Then every element of $M$ can be represented uniquely as the sum of an element of $N'$ and an element of $A$, hence there is a bijection between $N'\times A$ and $M$ ...


0

Intuitevely you are making the following infinite process: given an infinite set $X$ you choose any $a_0\in X$; then you choose $a_1\in X_1$, where $X_1=X\setminus\{a_0\}$. For any natural $n$ you choose $a_n\in X\setminus\{a_0,\dots,a_{n-1}\}$. This may seem to work without Choice, but the trick is that you need to do this process $\omega$ times. It's like ...


0

The axiom of the unordered pair guarantees existence for every set since it explicitly states that for any set, the union of that set with any other set exsits. So it stipulates that the union of any set with itself exists. The union of any set with itself is itself, therefore every set exists. Every subset of the Power Set is a set, therefore every subset ...


2

No, you cannot. For one thing, what should $f(0)$ be? Now it may be tempting to let $dom(f)$ be the nonzero ordinals to try and fix this. However, this still doesn't work: think about where $\omega$ goes. We must have $f(\omega)=n$ for some finite $n$, but then the infinitely-many ordinals $<\omega$ have to be squished into the finitely-many ordinals $&...


0

in the set theory we have $\{x_1,...,x_n\}=\{x_i\mid 1\leq i\leq n\}=\cup_{1\leq i\leq n}\{x_i\}$ so is the set of $n$ elements, and the set $\{x_i\}^{n}_{i=1}=\{x_1\}\{x_2\}\cdot\cdot\cdot\{x_n\}=\{(x_1,\cdot\cdot\cdot x_n)\}$ is a set of one element.


0

What you described is not a set, you probably mean $\{{x_1,x_2...,x_n}\}$ This is usually denoted as ${A=\{{x_i|i\in I}\}}$ Where $I$ is your index set $I=\{1,2,...n\}$


0

For the set $\{x_1,\ldots,x_n\}$ you can write $$\begin{cases}\{x_i\}_{i=1}^n \\ \\ \{x_i , 1\leq i\leq n\} \\ \\ \{x_i\}_{1\leq i\leq n}. \end{cases}$$


0

By distributive law: $$(A\cup B)\cap (A\cup C) =A\cup ( B\cap C)$$ That is done.


0

Hint: $( A \cup B ) \cap ( A \cup C ) = \{ x : ( x \in A \lor x \in B ) \land ( x \in A \lor x \in C ) \}$. You can use boolean algebra to simplify the latter, before converting it back into set-theoretic operations. The whole point is that $\cup,\cap$ correspond to $\lor,\land$ respectively.


5

Your "implicit usage" is actually in the very first line: $\aleph_0\le|S|$ is equivalent to the definition of a Dedekind-infinite set, which is strictly stronger than an infinite set, which only means that it is not finite (not in bijection with an element of $\omega$). A set is defined to be Dedekind-infinite if $|S|=|S|+1$, or equivalently there is a ...


2

If $f$ is injective, but not necessarily surjective, then the claim is true: Let $y \in f(A - X)$. There is $a \notin X$, such that $y = f(a)$. If $y \in f(X)$, then there is $x \in X$ such that $y = f(x)$. Thus, $f(x) = f(a)$, hence $x = a \in (A - X) \cap X$, which is impossible. Hence, $y \notin f(X)$, i.e. $y \in f(A) - f(X)$. Thus, $f(A - X) \subseteq ...


4

This is not true. Let $A = \mathbb{N} \times \mathbb{N}$, let $B = \mathbb{N}$ and let $f:A \to B$ via $f((a,b))=a$. This is clearly surjective. Now, let $X = \mathbb{N} \times \{1\} \subset A$. Then, $f(A - X) = B$, but $f(A) - f(X) = \emptyset$.


1

The set $X$ does have a smallest element, namely $(1, 1, 1, \dots )$. It's the set $A$ which has no smallest element, as follows: Let $a = (a_1, a_2, \dots )$ be any element of $A$. By the definition of $A$, there is some $n$ such that $a_n = 2$ and, for every $k \neq n$, $a_k=1$. Define $b = (b_1, b_2, \dots )$ by setting $b_{n+1} = 2$ and setting $b_k=...


1

Your confusion might lie in mis-parsing the definition: it's not 'finitely many'; it's the classic lexicographical order, in which we compare the two vectors element-wise until we find an index on which they disagree, and then use the natural ordering between the two values at that index. The reason that this fails in the case you've listed is that we have $...


1

Looking at $x \in X$ as a map $x \colon 1 \to X$, the operator $\tilde x$ is precisely the natural transformation given by co-Yoneda's embedding $$ \mathsf{Set}(X,-) \stackrel{\mathsf{Set}(x,-)} \longrightarrow \mathsf{Set}(1,-) \simeq \mathrm{id}_{\mathsf{Set}} $$ So you could call it the restriction along $x$, or precomposition by $x$, and denote it $x^\...


1

The first statement is ambiguous. It could mean either "There is an infinite number of sets $E_n$ for which $x\notin E_n$", or it could mean "There is only a finite number of sets $E_n$ for which $x\in E_n$". The second statement can only mean the latter. Addendum: For the first statement, suppose $\mathscr E$ is the collection of all $E_n$ under ...


2

I would say that the first sentence, as written, is ambiguous. It could either be interpreted as A) There is not an infinite number of $E_n$'s all of which contain $x$. B) There is an infinite number of $E_n$'s such that $x$ is contained in none of them. The sentence A) is equivalent to your second sentence "it is not the case that $x$ is in an ...


0

Think of the gaps between squares; how many pairs of squares can you think of with a difference of 3? There should be only 1: 1 and 4. To make sure that 6 and 9 are in $A$, k must be 5. This makes it an element of 4. but not any of the other possible answers.


3

You know that $6=x^2+k$ and $9=y^2+k$, for some integers $x$ and $y$ with $-3\le x<k$ and $-3\le y<k$. In particular $y^2-x^2=3$. There are only a few cases for this: $y+x=1$, $y-x=3$ $y+x=-1$, $y-x=-3$ $y+x=3$, $y-x=1$ $y+x=-3$, $y-x=-1$ Can you go on?


4

The ordinals have well-defined notions of successor, addition or multiplication, but not of predecessor, subtraction or division. So while it makes sense to talk about $\alpha + 1$ or $\alpha \cdot 2$ or so on, when $\alpha$ is an ordinal, it doesn't (always) make sense to talk about $\alpha - 1$ or $\frac{\alpha}{2}$, and so on. The ordinal $\omega$ is ...


2

Especially the first part of this answers your question. I read in the comments that there were more difficulties so decided to give a more complete answer. Let $S$ be a successor-set. If $\mathcal A:=\{B\in\wp(S)\mid B\text{ is a successor set}\}$ then $S\in\mathcal A$. This guarantees that $\cap\mathcal A$ is a well defined subset of $S$. Secondly ...


3

The proof almost works, except for here: [...] so that $x \notin C, y \in B$ If the set $B$ is empty, then it won't be possible to find such a $y\in B$, and so we won't be able to form the tuple $\langle x,y \rangle$ or say anything about it. In fact, if $A$ or $B$ is empty, then $A\times B = \varnothing$, in which case $A \times B$ is a subset of ...


2

You can just translate your statements in English and see if that helps. a For all x, both P(x) and Q(x) are true. b For all x, P(x) is true and for all x, Q(x) is true. c There exists x (or there is/are some value(s) of x) for which both P(x) and Q(x) are true. d There exists x for which P(x) is true and there exists x for which Q(x) ...


2

Your second expression doesn’t actually work: it doesn’t say that the sets $S_i$ belong to $\mathscr{S}$ or that you’re taking only some of the possible intersections of finite subsets of $\mathscr{C}$. It really is much easier to define $\mathscr{B}$ in terms of $\mathscr{S}$ and then define $\tau$ in terms of $\mathscr{B}$. If you really want to define $\...


2

In linear algebra the double-dual comes to mind, but the general term that's most relevant is "evaluation map". One somewhat common notation is $\mathrm{ev}_{x}(f)=f(x)$. Either ev (your ~) is the evaluation map, or each $\mathrm{ev}_{x}$ is the evaluation map at $x$.


2

In the context of functional analysis, the map $$ \Phi: A \to (\Bbb F)^{\Bbb F^A}\\ \Phi:x \mapsto (f \mapsto f(x)) $$ is called the "evaluation map".


5

It's called the canonical map into the double dual. It does not have another name that I know of. In a monoidal category with duals (I'm being vague), the uncurried version of it, $\operatorname{ev} : X \otimes X^* \to I$, is called evaluation.


0

One way to prove that the statements $p \rightarrow q$ and $p \vee q \iff q$ are logically equivalent is as follows. It suffices to show that their truth tables (which have $4$ rows each) are the same. Recall that the only assignment of truth values for $p$ and $q$ for which $p \rightarrow q$ evaluates to $F$ is when $p$ is $T$ and $q$ is $F$. For which ...


0

so we must prove: $(p\Rightarrow q)\Leftrightarrow [(p\vee q)\Leftrightarrow q)]$. suppose $(p\Rightarrow q)$ and prove the equivalence $(p\vee q)\Leftrightarrow q$ $\Rightarrow$): $ (p\Rightarrow q)\Rightarrow[ (p\Rightarrow q) \wedge (q\Rightarrow q)] \Rightarrow [(p\vee q)\Rightarrow q]$. $\Leftarrow$): $q\Rightarrow(p\vee q)$ obvious suppose $(p\vee ...


2

$(p\lor q)↔q$ $=((p\lor q)→q)\land((p\lor q)←q)$ $=(\overline{(p\lor q)}\lor q)\land((p\lor q)\lor \overline {q})$ $=((\overline{p}\land \overline{q})\lor q)\land(p\lor (q\lor \overline {q}))$ $=((\overline{p}\land \overline{q})\lor q)\land(p\lor T)$ $=((\overline{p}\land \overline{q})\lor q)\land T$ $=((\overline{p}\land \overline{q})\lor q)$ $=(\...


1

You got this far: $$p \vee q \equiv (\sim p \vee q) \wedge (p \vee q)$$ Use distributivity: $$p \vee q \equiv (\sim p \wedge p) \vee q$$ By Law of Contradiction, we know that the former part of this disjunction is always false, so using disjunctive syllogism we can conclude: $$p \vee q \equiv q$$


0

Let me give my take on it, set theory is the foundation of mathematics, we have sets we denote in various ways $\emptyset$ is the first from which we can build more, we have the infinite set $\omega$, from these we can make other sets with powersets, squaring, union, intersection etc. These contain elements which we can label anyway we want, $1,2,3,\ldots$, $...


0

What are you? You're a collection of atoms. But you're not just the atoms, you're also the structure they form. Your neurons and thought patterns. If I were to replicate this structure perfectly, on all its quantum information, which of your copies would admit not being the original one? And yet, at a given fixed instance of time, being frozen, you are ...


2

Apart from the fact that your proof could benefit from some better wording, it is perfectly correct. I would still suggest you try to reword it more clearly as that will help you later on. Make it more clear what you want to do. Something like: We wish to prove that $R$ is reflexive i.e. that for every $x$, $xRx$ is true: $$\forall x: xRx$$ Let $x_0\in ...


1

Maybe I'm wrong, but I think this is rather a deep question. Or at least, I ponder it frequently, perhaps once a month. I don't have an answer, but let me just speculate out loud. Lets just pretend, for a moment, that we kind of accept finitism, just a little bit. We accept finitism a little bit. Now $\mathbb{N}$ is the initial pointed monounary algebra. ...


0

The following is mostly just my personal opinion, please keep this in mind. It is really pointless to think everything in mathematics is a set. It is true (in classical set theory), but this view does not help you at all, unless you are studying set theory specifically. By that logic it is also somewhat pointless to think of the set of real numbers $\{1,1/...


0

I'm not sure if I've understood the point you're getting at, but if we have a set $X$ and a bijection $f:X \to \mathbb R$, $f$ will induce a structure $(X, +, \cdot, \leq)$ on $X$, just by using $f$ to identify $x$ with $f(x)$. Different bijections will give rise to different structures on $X$. So the structure certainly isn't intrinsic to $X$ (or $\mathbb ...


1

You're right about $T$ and mostly right about the equivalence classes, with only one definite shortcoming: You write "If $ -\infty < x \leq 0 \text{ and } 3 \leq x < \infty$" -- this condition is not satisfied by any $x$. What you mean is, "If $ -\infty < x \leq 0$ or $3 \leq x < \infty$". Generally, though, I would note that you're not being ...


0

I usually pronounce $\varnothing$ as "null set" in the same way that I pronounce $<$ as "less than." Alternatively, I also read $A=\varnothing$ as "$A$ is null" as well as "$A$ is the null set." I also use empty, but others have mentioned that, and I thought I should add null as an alternative pronunciation. I would call the symbol itself either "null set ...


2

It’s basically correct. There’s a typo at the very beginning, where you meant to write ‘For each $n$ there exists’ (instead of $j$). And you need to pass to a convergent subsequence only once: the tail sequence $\langle x_n:n\ge k(j)\rangle$ already converges to $\hat x$, since it’s a subsequence of a sequence converging to $\hat x$. For a proof in case $X$ ...



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