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17

A monotonically non-increasing sequence of natural numbers is eventually constant, so it is completely identified by specifying the finite initial subsequence up to the point at which it becomes constant. There are only countably many finite sequences of natural numbers.


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.


5

"$S \subseteq T$" is short-form for "$\forall x \in S\ ( x \in T )$". "$S \nsubseteq T$" is short-form for "$\neg( S \subseteq T )$", which would be equivalent to "$\neg \forall x \in S\ ( x \in T )$", which is equivalent to "$\exists x \in S\ ( \neg( x \in T ) )$", which is not equivalent to "$\forall x \in S\ ( \neg( x \in T ) )$". Notice that the latter ...


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 ...


5

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 $\...


5

You could just note that the set $\{\emptyset\}$ has one element (namely $\emptyset$) while $\emptyset$ does not contain any elements. So, they are not equal.


5

What the author must have had in mind is something like By definition the circle with center $O$ and one point $A$ consists of the endpoints of all line segments starting at $O$ whose length is $OA$. However, if we believe in set theory, then we can already form the set $$ \{ X \mid \overline{OX} = \overline{OA} \} $$ which is exactly the circle! So ...


4

As you're asking for a hint, I suggest trying to find intervals $A$ and $B$ as counter examples. More hints:


4

Obviously, $1\notin A$. And $Y\setminus\{1\}$ does not meet the condition. $A=Y\setminus\{1,13\}$. The sum of the elements of $A$ ($p=2,029,091$) is prime. Let $x\in A$. Since $x$ is not $1$ and certainly is not $p$, we have that $x$ does not divide $p$. Since $x$ does not divide $p$, $x$ does not divide $p-x$. But $$p-x=\sum_{y\in A\setminus\{x\}} y$$


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 ...


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 $f(A) - f(X) = \emptyset$.


4

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.)


4

$S$ is a union of open balls. $S$ is a countable union of open balls. .. or of Cartesian products of open intervals. $S$ is $\{x\in\mathbb R^n\mid f(x)>0 \}$ for some continuous function $f:\mathbb R^n\to \mathbb R$. $S$ is $\{x\in\mathbb R^n\mid f(x)>0 \}$ for some $\mathcal C^\infty$ function $f:\mathbb R^n\to \mathbb R$. $S$ is $\{x\in\mathbb R^n\...


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 ...


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?


3

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).


3

As others have stated, $\emptyset$ is a set with no elements and $\{\emptyset\}$ is a set with one element. Therefore these two sets can't be equal. To address your specific question: As for b. the empty set is empty therefore it can not include elements or be a subset (else from the empty set itself), is this claim concludes that $\{\emptyset\}\...


2

As Asaf points out, Konig's Theorem applies here. It states (Jech, Set Theory, Millenium Edition, Theorem 5.10): If $\kappa_i < \lambda_i$ for all $i \in I$, then $$ \sum_{i\in I} \kappa_i < \prod_{i\in I} \lambda_i. \tag{Konig} $$ Take $I = \Bbb N$, and for all $n\in I$, $\kappa_n = \lvert A_n\rvert$, $\lambda_n = \lvert A \rvert$. Then by (...


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 ...


2

Vacuous truth "all elements of A are not elements of B" is NOT the same as saying "there is an element of A that is not an element of B". $\forall x \in A; x \not \in B$ is NOT enough to show $A \not \subset B$. To show $A \not \subset B$ you must show $\exists x \in A; x \not \in B$ and you can not do that if $A = \emptyset$. To pound it home. $A \...


2

"$X$ is a subset of $A$", in symbols $X\subset A,$ means $\forall x(x\in X\implies x\in A).$ If $X$ has no elements, this is true vacuously; hence $\emptyset\subset A,$ i.e., the empty set is a subset of $A.$ "$X$ is disjoint from $A$", in symbols $X\cap A=\emptyset,$ means $\forall x(x\in X\implies x\notin A).$ If $X$ has no elements, this is true ...


2

The boundary is the full set; there is no interior point. Note that there is no open interval that contains $2$ that is conatained in your set. Same for the other points in the set (or in fact any real). Thus the interior is empty. And every open interval containing one of the points, will also contain a point not in the set. So every point in the 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 $X$ ...


2

In practice you just check it by brute force. You can make it a bit faster to conduct by drawing a diagram with nodes indicating the elements 0, 1, 2, 3 and arrows between elements that are related (and such a diagram is good to do anyway in order to train your intuition). Then, instead of checking every combination of pairs for transitivity, you can just go ...


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".


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 ...


2

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


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 $\...



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