5

The result is false as stated: if $I=\Bbb N$, $B_n=\Bbb N$ for each $n\in\Bbb N$, and $A_n=\{n\}$ for each $n\in\Bbb N$, then $\bigcap_{n\in\Bbb N}B_n=\Bbb N$, which is infinite, and $\bigcap_{n\in\Bbb N}A_n=\varnothing$, which is finite, but $$\bigcap_{n\in\Bbb N}(B_n\setminus A_n)=\bigcap_{n\in\Bbb N}(\Bbb N\setminus\{n\})=\varnothing$$ is finite.


4

If they're disjoint, take $f : [0,1) \to A$ and $g : [0,1) \to B$ bijections, then you can basically concatenate these two bijections as $h : [0,2) \to A \cup B,h(x)=\begin{cases} f(x) & x \in [0,1) \\ g(x-1) & x \in [1,2) \end{cases}$. If you don't like the use of $[0,2)$ for some reason then you can use $x \mapsto 2x$ to make the domain of $h$ be $[...


3

Welcome to MSE. The answer to your question is NO You can write it as $$S = \bigcup_{i = 1}^{n}S_i$$ where $S_i \cap S_m = \varnothing$ for all $1 \leqslant m < i \leqslant n$. Note here: $i \not = m$. Hope this helps. Note: You can also write it as $$S = \bigcup_{S_i \cap S_m = \varnothing}^{}S_i$$


3

It is not uncommon to use the symbols $\uplus$ or $\sqcup$ to denote disjoint unions, but there are two different interpretations of that: First, $A\uplus B$ could mean the union of two sets that are isomorphic (in whichever way appropriate) to $A$ and $B$ but with their elements renamed to guarantee they will always be disjiont. Alternatively, $A\uplus B$ ...


3

Yes, you can. Simply map every (ordered) pair of subsets $(A, B)\in \mathcal P(\mathbb N)^2$ to $(2A) \cup (2B+1)$. To find this idea, try to think of the cardinals. The cardinal of natural integer sets is $2^{\aleph_0}$ and the cardinal of ordrered pairs of natural integer sets is $(2^{\aleph_0})^2=2^{2\aleph_0}=2^{\aleph_0}$. How do you fit $2\aleph_0$ in $...


3

Hint: there are many ways. One way is to identify $c$ with the set $S$ of positive irrational numbers, which can be uniquely represented as infinite sequences of positive integers (using continued fractions). Now you have $S = A \cup B$ where $A$ comprises the sequences that start with an even number and $B$ comprises the sequences that start with an odd ...


3

Suppose $B \subseteq f(B)$. Take any $a \in A$. Then $f(a) \in B \subseteq f(B)$ so we can write $f(a)=f(b)$ for some $b \in B$. Since $f$ is injective this gives $a =b$ so $a \in B$. We have proved that $A \subseteq B$ which is a contradiction. [ It is given that $B \subseteq A$ but $B \neq A$].


2

Here is how the construction of that bijection mirrors the freeing of the first two rooms in Hilbert's Hotel. The function $h$ matches the rational numbers $$ 0, 1/1, 1/2, 1/3, 1/4, 1/5, \ldots $$ in order to the rational numbers $$ 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, \ldots $$ That says the inverse function $h^{-1}$ creates "rooms" for the extra guests ...


2

You haven't quite shown that $h$ is one-to-one or onto, but you're very close! You have shown that two restrictions of $h$ are one-to-one, that no two elements map to $\frac12,$ and that infinitely-many elements of $(0,1)$ are in the range of $h,$ but that's not enough. You've hinted around the rest, but it should be made explicit. I would proceed in a ...


2

Saying that two sets $X, Y$ have the same cardinality says that there exists some function $f: X \longrightarrow Y$ which is one to one and onto. It says nothing about what other functions can exist. For instance, if $X = Y = \mathbb R$ we can consider $g: \mathbb R \longrightarrow \mathbb R$ such that $g(x) = 0$. Even though $\mathbb R$ and $\mathbb R$ ...


2

For the case of necklaces we have the cycle index $$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$ We get with $k$ colors $Q_j$ (this is PET) $$[Q_1^{n/k} \cdots Q_k^{n/k}] Z\left(C_n; \sum_{j=1}^k Q_j\right) = \frac{1}{n} \sum_{d|n} \varphi(d) [Q_1^{n/k} \cdots Q_k^{n/k}] \left(\sum_{j=1}^k Q_j^d\right)^{n/d}.$$ We see that $d$ must divide $n/k:$ $$...


2

STEP 1. Prove that there is a bijection $f$ between $\mathbb{R}$ and $[0,1)$. STEP 2. Divide $\mathbb{R}$ into countable pieces of copies of $[0,1)$.


2

One easy way to get to an answer is to start by taking any set $X$, and simply looking at the simplest (but not trivial) collection of sets which guarantee that every finite subset of $X$ will be a subset of some element in the collection: simply take all finite subsets of $X$. This is not linearly ordered set, and it's easy to see that if $X$ is uncountable,...


2

No. $A\setminus B\ne\emptyset\ne B\setminus A$ is sufficiently short and clear.


1

The definition of "antiset" given on the Wolfram MathWorld page you linked to (and also on Wiktionary!) is: A set which transforms via converse functions. This "definition" seems meaningless to me without more context. What is a "converse function"? What does it mean for a set to "transform via converse functions"? ...


1

Suppose that $A \setminus B$ were countable (i.e. not uncountable). Then $A = A\setminus B \cup B$ would be countable too as a (disjoint) union of two countable sets. But $A$ is given to be uncountable. Contradiction. So $A\setminus B$ is uncountable.


1

Hint: Consider the case where $[w,x] = [0,1]$ and $f\colon [0,1] \to [y,z]$ is defined by $$f(t) = tz + (1-t)y$$


1

Yes, $\forall X \subseteq A (X \notin \mathcal F \!\implies\! A \setminus X \in \mathcal F)$ is a property only for ultrafilters: Suppose that there exists a subset $X$ of $A$ such that $X \notin \mathcal F$ and $A \setminus X \notin \mathcal F$. Then $\mathcal F \cup \{X\}$ has the finite intersection property: if $S_1,\dots,S_n \in \mathcal F$ then $(S_1 \...


1

Yes, sets can contain arbitrary elements as long as they're distinguishable.


1

Yes, just because $A=B$, so both products are just $A\times A = \{(1,1), (1,2), (2,1), (2,2)\}$.


1

For example, let $$ C=\{\,\{x,A\}\mid x\in B\,\}.$$


1

The only way this can be satisfied with the graphs having disjoint edge sets is if $G_1$ and $G_2$ are both empty graphs (and hence equal, since they have the same vertex set). This is because any edge of $G_1$ is an edge of $G_1\cup(G_2\cap G_3)$, but not of $G_2$, contradicting the relationship. So $G_1$ has no edges, and $G_1\cup(G_2\cap G_3)=G_2\cap G_3$....


1

Defining $G_0$ as the graph with the common vertex set but no edges, we can write $$G_2 = G_1\cup (G_2\cap G_3) = G_1\cup G_0 = G_1$$ Done! Note: We can go on to observe that the edge sets of $G_1$ and $G_2$ must also be empty, but this isn't required to establish the equality.


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