# Tag Info

16

Yes, though this has nothing to do with set theory really and is much more about propositional logic. If we replace "$A \in B$" with the propositional variable $\textrm{P}$, then your statement is simply an instance of $$\textrm{P} \lor \lnot \textrm{P}$$ which is of course always true.

11

Yes. This is a trivial consequence of a theorem by Steinhaus: Suppose that $X$ has a positive measure, then $X-X=\{x-y\mid x,y\in X\}$ contains an interval around $0$. It is not hard to prove that if $X$ is infinite, then $X$ and $X-X$ are equipotent (there is a surjection from $X^2$ onto $X-X$, and there is an obvious injection from $X$ into $X-X$). ...

10

Consider the function $f: \mathbb{N} \times \mathbb{N} \longrightarrow \mathbb{N}$ defined by $f(a,b)=2^a \, 3^b$. Then this function is one-one. Consequently cardinality of $\mathbb{N} \times \mathbb{N}$ is "no more than" (if I may use this phrase) that of $\mathbb{N}$, hence countable.

8

The first, and perhaps deepest problem is on the first line. $X/{\sim}$ is not the union of the equivalence classes. It is the set of equivalence classes. More specifically, $\bigcup_{x\in X}[x]=X$, whereas $X/{\sim}=\{[x]\mid x\in X\}$. It is a set whose elements are subsets of $X$. So a function from $X$ to $X/{\sim}$ is a function mapping points of $X$ ...

8

HINT: Show that $\Bbb R$ and $\Bbb{R^2}$ have the same cardinality. Now show that $\Bbb R^2$ is the union of $\frak c$ sets of size $\frak c$. (Note that the proof itself generalizes to any infinite set $A$ such that $A$ and $A^2$ have the same cardinality; which is all of them if you assume the axiom of choice.)

6

In fact , $$\overline{A\times B}=\bar {A}\times \bar{B}$$for any two subsets $A,B\in \mathbb R$. Here, $$\overline{\mathbb Q\times \mathbb Q}=\bar {\mathbb Q}\times \bar{ \mathbb Q}=\mathbb R\times {\mathbb R}$$

6

Consider the statement $x \in \bigcap_{m=1}^\infty \bigcup_{n=m}^\infty A_n$. This means exactly that $x \in \bigcup_{n=m}^\infty A_n$ for every $m$. (Here I have expanded the definition of the intersection.) This means exactly that for every $m$, $x \in A_n$ for some $n \geq m$. (Here I have expanded the definition of the union.) This means exactly that $x$ ...

5

Yes it does. The statement that the closure of $\Bbb Q$ is $\Bbb R$ says that given an element $x$ of $\Bbb R$ you can find a sequence of elements of $\Bbb Q$ that converge to $x$. If you know that, if I give you an element $(x,y)\in \Bbb {R \times R}$ can you find a sequence in $\Bbb {Q \times Q}$ that converges to it?

4

Whenever you have $$x\in f^{-1}(A)$$ you can say $f(x)\in A$ and conversely. Thus $$x\in f^{-1}(A)\implies f(x)\in A\implies f(x)\in B\implies x\in f^{-1}(B)$$

4

Take an arbitrary $x \in f^{-1}(A)$. By definition, $f(x) \in A$. As $A \subset B$, $f(x) \in B$. As $f(x) \in B$, it must also lie in ....

4

Well. Just off the bat, $\omega_1$ is an ordinal, and it cannot be expressed as a "polynomial" in $\omega$ in any nontrivial way. Moreover, since for infinite ordinals, $\alpha,\beta$ we have that $|\alpha+\beta|=|\alpha\cdot\beta|=|\alpha^\beta|=\max\{|\alpha|,|\beta|\}$ (where the arithmetic is ordinal arithmetic, of course), it follows that any finite ...

4

$S^T$ is the set of all functions with domain $T$ and codomain $S$. The reason for the notation is $$\left \vert {S^T}\right \vert = |S|^{|T|}$$

4

It is entirely valid to have a relation that relates every element of $A$ to every element of $B$. It can even have fancy properties -- for example if $B$ is a singleton set, the relation $A\times B$ will be a function!

4

Notice that $\Bbb R \times \{0\} \subset \Bbb R \times \Bbb R$ has the same quantity of elements that $\Bbb R$¹. So if you prove that $\Bbb R$ is uncountable, you're done. ¹ Actually $|\Bbb R| = |\Bbb R^n|$ for every $n$.

4

Observe that $$C\le C.\aleph_0\le C.C=C$$ Now use Cantor-Bernstein theorem to deduce the statement. For proving that $C.C=C$ you can use decimal expansion of real numbers and the function $$f:[0,1]\times[0,1]\to [0,1]\\ f(.r_1r_2r_3\dots,r'_1r'_2r'_3\dots)=.r_1r'_1r_2r'_2r_3r'_3\dots$$

4

You denote (not notate!) the set of rational numbers by $\mathbb Q$ and that of the real numbers by $\mathbb R$ ; therefore the set of irrational numbers can be written as $\mathbb R \backslash \mathbb Q$ or $\mathbb R - \mathbb Q$, depending on your taste. Your proof could simply go as follows : since $\sqrt 2 \in \mathbb R \backslash \mathbb Q$ but $(\sqrt ... 3 Usually, the set of irrational numbers is written simply as$\mathbb R\setminus \mathbb Q$. As for the symbolic proof, my advice is "Avoid symbolic proofs." A good proof in mathematics is not one that is fully written with symbols alone. A good proof is written in words, but is still mathematically rigorous, such that there is no doubt that any ... 3 Depends on which topology you are talking about. If it is the topology on the real number system$\mathbb{R}$, then$\mathbb{R}$is the closure of itself. If it is the topology on the extended real number system$\bar{\mathbb{R}}$, then$\bar{\mathbb{R}}$is the closure of$\mathbb{R}$. 3 This might depend on the form of logic you are using. It is certainly true if you use "standard" logic. And as has already been pointed out it just follows from basic logical axioms. Surprisingly though there are (fairly often) used logical systems where the statement$P\vee \neg P$(called the law of excluded middle) is not provable. An example of this ... 3 First responding to your claim "science and specially mathematics are based on the set theory": I agree with Zev. Today we could pick an entirely different foundation for mathematics and I'm rather certain that physicists, biologists, and chemists won't change their attitude with how they apply mathematics to their work. You don't need to think of simple ... 3 First, list the pairs in which the sum is$0$: $$(0,0)$$ Then those in which the sum is$1$: $$(1,0),(0,1)$$ Then those in which the sum is$2$: $$(2,0),(1,1),(0,2)$$ Then those in which the sum is$3$: $$(3,0),(2,1),(1,2),(0,3)$$ And so on. 3 Separate the set of primes$P$into two sets$A_p,B_p$by considering the partial products of$(1-1/2)(1-1/3)(1-1/5)\cdots$with each factor of the form$1-1/p$where the primes$p$run through the sequence$2,3,5,\cdots$of primes. We alternately go for partial products each at most$1/2,$so that$A_p$begins with$2$[since$1-1/2\le 1/2$] and then$B_p$... 3 To a question in title No,$2^A \not\subseteq A$because a power set of any$A$has a cardinality strictly greater than$A$itself. To a question in text No,$f(a)\subseteq 2^A$does not hold in general, as$f(a)$is by definition an element of$2^A$, and an element of a set is not (in general) a subset of the same set. Althoug in some special cases it ... 2 If$X\stackrel f\leftarrow A\stackrel g\to Y$is a cotriad, the pushout is $$X\stackrel{\bar g}\to Z\stackrel{\bar f}\leftarrow Y$$ where$Z$is the quotient space of$X\sqcup Y$by the relation generated by$f(a)\sim g(a)$, and$\bar f$is the composite$Y\to X\sqcup Y\to Z$. As often with relations, one only says what generates it, and the whole relation ... 2 That is occasionally used as an alternative for$\prod$, meaning an indexed Cartesian product. From the Wikipedia article on Cartesian products: 2 Yes, you do the diagonalization like with rational numbers which is basically the same thing. 2 Another easy way :take a pair, write down the numbers in any numeration basis. For example (decimal) 1234 and 987. Add extra zeros where needed so they are the same length, and shuffle them 1 2 3 4 0 9 8 7 -------- 10293847 Reverse operation : take digits with even (resp. odd) positions. EDIT: and of course there are explicit formulas, here given as 3 ... 2 Two sets$A$and$B$are equal, if and only if for all$x$we have$x\in A \Leftrightarrow x\in B$. Alternatively you can show$A\subseteq B$and$B\subseteq A$. We have: $$x\in A \cup \emptyset \Leftrightarrow x\in A \vee x\in \emptyset \Leftrightarrow x\in A$$ so$A\cup \emptyset = A$. (Note, that$x\in A\Rightarrow x\in A\vee \phi$is true for any ... 2 Hint: Try writing out what each of$A,B,C$look like explicitly.$A = \{x~:~x~\text{is a positive integer divisible by 2}\} = \{2,4,6,8,10,12,\dots\}B = \{x~:~x~\text{is a positive integer divisible by 4}\} = \{4,8,12,16,\dots\}C = \{x~:~x~\text{is a positive integer divisible by 6}\} = \{6,12,18,24,\dots\}$Remember also the definition of each of ... 2 Let$R=T\times\{S\}.$By folding in$\{S\}$, we've made each element of$R$"too big" to also be an element of$S$. (One way to make precise this notion of "too big" is the following: the transitive closure of any element of$R$is strictly larger than the transitive closure of any element of$S\$.) Note that we've used implicitly a non-trivial fact about ...

Only top voted, non community-wiki answers of a minimum length are eligible