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7

The intersection of two sets is a set. $1$ is an element of the set $\{1\}$. That is, $X\cap Y=\{x:x\in X\:\text{and}\: x\in Y\}$. In our case, $X=\{1,2\}, Y=\{1\}$, so $X\cap Y=\{1\}$, the set of elements in both sets $X,Y$.


5

Yes. For instance $$ A = \{1\}, B = \{2\}\\ A = \text{All the negative integers},B = \text{All the positive integers}\\ A = \{2n \mid n \in \Bbb Z\}, B = \{2n+1\mid n \in \Bbb Z\}\\ A = \Bbb Z \setminus \{1\}, B = \Bbb Z \setminus \{2\}\\ A = \{5n \mid n \in \Bbb Z\}, B = \{3n \mid n \in \Bbb Z\} $$


5

If you can argue that $\Pr(A \cap B) \le \Pr(A)$, then you can also argue that $\Pr(A \cap B) \le \Pr(B)$ by symmetry, giving the desired result. If you need a formal argument to show that $\Pr(A \cup B) \ge \Pr(A)$, consider writing $A \cup B$ as a union of the disjoint events $A$ and $B \setminus A$.


5

Remember that the difference between a partial order and a total order is that in a partial order, you can have incomparable elements. Can you think of sets $A$ and $B$ such that $A\not\subseteq B$ and $B\not\subseteq A$? To clarify: such an example will merely show that $(\mathcal{P}(\mathbb{Z}), \subseteq)$ is not a total order. You still need to show ...


5

The power set of the empty set has 1 element, the empty set. $2^0=1$ EDIT: Credit as well to Tunococ, Andreas Caranti, and Mufasa for their comments. I guess I jumped the gun since this is my first answer!


4

You are going to have a bad time proving that. Consider A={0}, B={0}, C={1}, D={1}. Then $A\times B=\{(0,0)\}$, $C\times D = \{(1,1)\}$, $\big(A\times B\big)\bigcup \big(C\times D\big)=\{(0,0),(1,1)\}\ne \big(A\bigcap C\big)\times \big(B\bigcap D\big)=\emptyset$


4

$y^2$ is positive, so $y^2 \leq 3y^2$. Thus, if $x,y$ are such that $x^2 + 3y^2 = 1$ then $x^2+y^2 \leq 1$.


3

For a category $C$, and two objects $X,Y$ in $C$, the hom-set (for homomorphism set) is given by $$C(X,Y) = \{f \mid f \text{ is an arrow } f: X \rightarrow Y \text{ in } C\}.$$ Now, despite the name, there is no a priori reason why this should be a set, rather than a class. Indeed, it is not difficult to give an example where a hom-set is a proper class, ...


3

No such function exists. To see that, suppose you had such an $f$. Let's compute a few examples. Case I: $$A=\{1,2,3\},\,B=\{3,4,5\},C=\{1,4,6\}$$ Then it is easy to compute everything in your expression. We get $$0=f(3,3,3,1,1,1)$$ Case II: $$A=\{1,2,3\},\,B=\{1,4,5\},C=\{1,6,7\}$$ Then we get $$1=f(3,3,3,1,1,1)$$


2

Um... yes? If $a \in (0,1]$ then $a \in (0,2)$ so $(0,1] \subset (0,2)$. If $a \in (1, 2)$ then $a \in (0,2)$ so $(1,2) \subset (0,2)$. So $(0,1] \cup (1,2) \subset (0,2)$. If $b \in (0,2)$ then $b > 0$ and $b < 2$ and either 1) $b > 1$ or 2) $b \le 1$. If 1) $b \in (1,2)$ and $b \in (0,1] \cup (1,2)$ If 2) $b \in (0,1]$ and $b \in (0,1] \cup ...


2

The way you worded that is a little strange. I have an alternative approach. Let $I$ be the event that a person is infected, $D$ be the event that a person develops the disease, and $\bar I, \bar D$, not those events. Then $$P(I\bar D)= P(\bar D|I)P(I) = (1-P(D|I))P(I) = (1-.30)(.5) = 0.35 = 35\%$$ where in the second step I used the product rule.


2

This doesn't make any sense. You're treating the variables as if they were sets, but they aren't. For instance, if $X$ and $Y$ are random variables, what does $X\cap Y$ mean? And the result you're getting is not true: for instance, if $M=2X$, then $Y=X$, so $E(XY)=E(X^2)$, which is usually not $0$.


2

Power set is the set containing all subsets of S. {a,b} is a subset of S. So, it is an element of P. {{a}} is also a subset of S. So, it is an element of P. similarly, the rest follow.


2

You can definitely have an ordered pair (or ordered tuple) with the same element, and you should definitely write that element twice (or more). Tuples are not sets in the sense that order and repetition matter. So you can't just ignore these like you do with sets.


2

It appears that the set $b$ is the set of subsets of elements of $a,$ i.e.: the union of the set of all power sets of the elements of $a.$ Notationally, $$b=\bigcup\bigl\{\mathcal P(d):d\in a\bigr\}.$$ Your interpretation turns out to be correct precisely when $a$ has exactly one element. Your friends' interpretation is close, but off by a level. The ...


2

As pointed out by @RobertIsrael in the comments, in this particular problem, it's easier to do the induction on $m$ because it's easy to assign an extra element in $E$ to an element in $F$ while it's not that straightforward the other way around. I must admit though, this problem can be easily solved without the help of induction. Note that you can assign ...


2

You prove the first statement is false with a counter-example. Consider the diagonal line in $\Bbb R\times\Bbb R$. It is not of the form $A'\times B'$ because every point is of the form $(x,x)$. The second is also false. Consider $A=[0,1]$, $B=[1,2]$ and $C=B$ and $D=A$. Then the LHS is a union of two squares of side length $1$ while the RHS is a square ...


2

To understand lattices first you need to understand partially ordered sets. A partially ordered set is a set with an ordering operation $ \leq $ that sometimes works. For example the set of all people ordered by ancestry is a partially ordered set (poset). If your mother is your ancestor and your mother's mother is your ancestor then your grandmother is ...


2

It is only a preorder. If $u$ is any non-trivial unit, and $x$ is any non-zero element, then $x|ux|x$, but $x \neq ux$.


2

By De Morgan's Law: $(B^c \cap (B \cap A)^c)^c = B \cup (B \cap A) = B$


2

Usually this is an axiom (the Powerset Axiom). Since apparently Tao adopts a different axiom instead — existence of $Y^X$, the set of all functions $X\to Y$ for any $X,Y$ — you'll have to use that to prove the usual Powerset axiom. You can use existence of empty set and the axiom of pairing to define a set $T = \{\emptyset, \{\emptyset\}\}$, and prove that ...


2

If $A=\bigcup_{a\in D}F_a$, then $A^c=\bigcup_{a\in \{0,1\}^n\setminus D}F_a$.


2

Consider any set $X$ with $|X| \geq 2$. Then if $\mathcal{T}$ is the indiscrete topology on $X$ we have: $id_X: (X,\mathcal{P}(X)) \rightarrow (X,\mathcal{T})$ is continuous and the pre-image of every subset is open in $(X,\mathcal{P}(X))$. I'm not sure what is meant by 'more than one pre-image value'? If $f:X\rightarrow Y$ is a function, then for any ...


2

Sure: let $\mathbf{Y}$ be any space where $y_0\in \mathbf{Y}$ is not open, and consider the constant function $f\colon x\mapsto y_0\colon \mathbf{X}\to\mathbf{Y}$; then $f^{-1}(\{y_0\}) = \mathbf{X}$.


2

I would like to make a few points about this question. As aptly pointed in Hanul's answer, there is no consistent theory which proves any contradictory sentence, and in particular, any sentence of the form $\phi \iff \neg A$. Furthermore, every inconsistent theory proves all contradictions, since inconsistent theories prove every sentence. But, what does ...


2

Using that $P(A\cup B) = P(A)+P(B)-P(A\cap B)$, you get that $P(A\cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$ (simply because $P(A \cup B)\geq P(A)$). So you can say that $P(A\cap B) \leq \operatorname{min}\{P(A),P(B)\}$. Now, if $A=B$ then you get the equality, so that is the largest possible value.


2

$A \cap B\subseteq A$; therefore $\Pr(A\cap B)\le\Pr(A)$. In the same way, prove $\Pr(A\cap B)\le\Pr(B)$.


2

I can see how this may be confusing to someone not familiar with ordinals, or more precisely $\omega_1$. Let me try to explain: Let $\prec$ be the well-order on $S_\Omega$. First let us prove that every element in $S_\Omega$ actually has an immediate successor: Fix $a \in S_\Omega$. Then $a_\downarrow :=\{x \in S_\Omega \mid x \prec a\}$ is well-ordered by ...


2

To avoid considerations of $\frac{1}{\infty}$ or other a-bit-handwavy arguments that could end in confusion or mistakes, an option is to go back to the definition, or at least elementary methods: Fix any $x \in(0,1)$. Will $x$ belong to $\bigcap_{n=1}^\infty (0,\frac{1}{n})$? In particular, will it belong to $(0,\frac{1}{n})$ for $n > ...



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