Tag Info

Hot answers tagged

17

The usual way is to use two different notations, one of which means that $A$ is finite, and the other means it's a subset of $B$.$$A\subset B,\quad|A|<\infty.$$


13

There are two options: You use the notation often. Then define it properly at the beginning (or when you first need it) and use whatever you think is reasonable. I'd suggest, as others: $$ A \subset_{\mathrm{fin}} B, \quad A \sqsubset B, \quad A \mathrel{\ddot{\subset}} B, \quad A \subset\!\!\!\!\!\cdot\!\!\cdot\, B \quad \ldots $$ You use it one or ...


10

In set theory there are two standard notations for the set of finite subsets of $X$: $[X]^{<\omega}$, $\mathcal P_\omega(X)$ or $\mathcal P_{\aleph_0}(X)$. In naive set theory, you can also find $\operatorname{Fin}(X)$ quite often. So if you'd want to write that $A$ is a finite subset of $B$, you could say that $A\in[B]^{<\omega}$ or $A\in\mathcal ...


7

Under the assumption that you wrote the term down correctly: $$ \begin{align} &\left((\{a,b\}\cup\{b,a\})\times(\{b,a\}\cap\{a,b\})\right)\setminus \left((\{b,a\}\setminus\{a,b\})\cup(\{a,b\}\times\{b,a\})\right)\\ &= \left((\{a,b\}\cup\{a,b\})\times(\{a,b\}\cap\{a,b\})\right)\setminus \left(\emptyset\cup(\{a,b\}\times\{a,b\})\right)\\ &= ...


5

$A\in[B]^{\lt\omega}$ where $[B]^{\lt\omega}$ denotes the set of all finite subsets of $B$.


5

If I had to invent a notation for this, the most suggestive I can think of is $$ A\in \mathcal P_{<\omega}(B). $$ One could define the RHS using another notation that in fact is more or less standard in some areas: $$ \mathcal P_{<\omega}(B) = \bigcup_{n \in \Bbb N}\binom Bn $$ which notation is derived from the that for the binomial coefficient ...


4

Probably use ordinal/cardinal numbers? $$|A| < \omega$$ $$|A| < \aleph_0$$


4

Hint: Show that there is a natural bijection between $P_{\text{fin}}(\mathbb{N})$ and $\mathbb{N}$; one particularly nice one is more apparent when you work in binary.


4

Let $X=\{a,\{a,b\}\}$. The set $X$ has two elements: nothing more, nothing less. Then, $$\mathcal P(X)=\big\{\emptyset,\{a\}, \{\{a,b\}\}, \{a,\{a,b\}\}\big\}$$ If things don't become clear, you can try "encapsulating" the element $\{a,b\}$ calling it $c$, so you can now write $X=\{a,c\}$.


4

-------------------------not that hard :)-------------------------


4

Choose $A=B$ then $A=C$ and use that $X\cap Y=Y\cap X$. These together say $$B=B\cap C,\quad C\cap B = C\qquad \implies B=(B\cap C)=(C\cap B)=C$$


4

The aleph numbers can certainly be multiplies, but not to give the results you mention. In fact, $2\aleph_0=\aleph_0$ and there is no such thing as fractional alephs so $0.5|A|$ is meaningless. For the sets $A$ and $B$ you mention it is very easy to establish a bijection and thus $|A|=|B|$.


3

You may think that is the "most natural" definition, but that's not how it is ever defined, for the reason that you gave. Instead, it can be recursively defined by the equations $$\begin{align} m + 0 & = m\\ m + s(n) & = s(m + n) \end{align}$$ which don't depend on any notion of "$n$ times". Note that for any sum of the form $m+n'$, either $n'$ ...


3

(I'm assuming that you are talking about infinite sets, since finite sets can easily produce many counterexamples.) Both these statements require the axiom of choice to be proved. The one about products even implies the axiom of choice itself (the one about unions is weaker than choice). If you already know that the axiom of choice implies that every set ...


3

You are on the right track, however you cannot use the phrase "the next one". The idea is:Let $S$ be an infinte set. Pick an element $x_1\in S$, then since $\{x_1\}$ is finite, $S\setminus\{x_1\}\ne\emptyset$, so pick an element $x_2\in S\setminus\{x_1\}$ and so on. Note the Axiom of Choice is involved here.


3

No, if $B$ is empty and $A$ is not empty, then there is no such function. To see this, note that $f\colon A\to B$ means that $f\subseteq A\times B$ and $\operatorname{dom}(f)=A$. But if $B=\varnothing$, then $A\times B=\varnothing$ and so $f=\varnothing$. Therefore $\operatorname{dom}(f)=A$ if and only if $A=\varnothing$. On the other hand, if $B$ is not ...


3

Select your favourite element $b \in B$. Then $A \times \{b\} \subseteq A \times B$ is a function $A \to B$; it is the constant function with value $b$. To derive this, we need just the axioms that are needed to meaningfully talk about functions. That is, Power Set, Separation, Union/Pairing, and Extensionality. NB. If $B = \varnothing$ then this argument ...


3

Let $\mathbb Q^n \subset \mathbb R^n$ the subset. We know that "boundary = closure - interior". We know the subset is dense, i.e. closure = everything. We see easily that interior is empty, for if we take a point in the rationals, every neighborhood will intersect the irrationals. Hence "closure - interior = all - nothing = boundary = $\mathbb R^n\subset ...


3

What you have encountered is the vacuous truthiness of an empty universe. A universal statement is falsified if there exists a counterexample within the domain of discourse.   However, there are no counterexamples within an empty set, so whatever is asserted about all its members is true. $$\forall y\in \emptyset ( P(y) ) \iff \neg \exists x\in ...


3

The statement $$x\in\bigcap F$$ means $$\hbox{for all $S\in F$ we have $x\in S$}.\tag{$*$}$$ Now, what is the truth value of a statement $$\hbox{for all $S\in F$,}\ldots\langle\hbox{whatever}\rangle$$ when $F$ is the empty set? The accepted answer is that such a statement is always true, regardless of the "whatever". To some extent this is a convention, ...


2

Let $I\in\wp(Y)$ be the image of $f$, i.e. $I:=\{f(x)\mid x\in X\}$. Then $F(I)=X=F(Y)$ and injectivity of $F$ tells us that $I=Y$ or equivalently: $f$ is surjective. Conversely let $f$ be surjective and $F\left(A\right)=F\left(B\right)$. Now let $y\in A$. Then $f\left(x\right)=y\in A$ for some $x$ (since $f$ is surjective) so that $x\in ...


2

Your answer for part c is correct. There are a number of more "formal" or "mathy" ways of writing it. For example, $A\cup B=\{n:n\in\Bbb{N}, 2|n\vee n\in\Bbb{P}\}$. This means, "the set of all $n$ where $n$ is a natural number, and $2$ divides $n$ or $n$ is prime." However, using the symbol $\Bbb{P}$ to represent the prime numbers is not universally ...


2

In the first case, consider $B = \emptyset$. $A \cup \emptyset = \emptyset$ implies $A = \emptyset$. Note that $\emptyset \cup B = B$ holds for all $B$, so $ A = \emptyset $ is a good solution. In the second case, consider $B = S$. $A \cap S = S$ implies $A = S$. Note that $S \cap B = B$ holds for all $B$, so $ A = S $ is a good solution.


2

Hint: To solve these kind of problems: $W= V$ you need to show that $W\subseteq V$ and $V\subseteq W$. In this case, you take an element $x$ that is in $W$ and show that it has to be in $V$, then you can conclude that $W\subseteq V$. Example: a) Let $x\in X\setminus(X\setminus A)$, then $x\in X \wedge x\notin (X\setminus A)$, then $x\in X \wedge (x\in A ...


2

It’s not in general true that $(A_1\times A_2)\times A_3=A_1\times(A_2\times A_3)$. Suppose, for instance, that $A_1=A_2=A_3=\{0\}$; then the only element of $(A_1\times A_2)\times A_3$ is $\big\langle\langle 0,0\rangle,0\big\rangle$, while the only element of $A_1\times(A_2\times A_3)$ is $\big\langle 0,\langle 0,0\rangle\big\rangle$, and ...


2

$(\implies)$ Suppose $A \cup B = B$. Take any $y \in A \cap B$, then $y \in A$ and $y \in B$, in particular $y \in A$. Then $A \cap B \subset A$. Now Let $y \in A$ then $y \in A \cup B$, as $A \cup B = B $ we have that $y \in B$. Try working the converse on your own using definition.


2

If the intended ordering is lexicographic, your argument is correct. In a general order-theoretic context, however, there is at least one other reasonable possibility: the question might be about the product partial order in which $\langle p_0,q_0\rangle\le\langle p_1,q_1\rangle$ iff $p_0\le p_1$ and $q_0\le q_1$. In that case they are not order-isomorphic: ...


2

What you wrote is completely indecipherable. You use $X$ twice, and it's not clear whether or not you make an appeal to the axiom of choice (or the well-ordering principle), which is something that you really shouldn't do. This is very similar to Hartogs theorem, only with surjections, and you can repeat the essence of the proof of Hartogs theorem, using ...


2

As you've sort of pointed out, $S$ contains $\Bbb Z^* = \{1,-1\}$. Now we prove that any number in $S$ has to be in $\Bbb Z^*$ (is $1$ or $-1$). Let $n\in S$. Then $$n = n^n$$ Now, we go up to $\Bbb Q$ for a bit and multiply by $n^{-1}$, assuming $n$ isn't $0$. Now we have $$1 = n^{n-1} = nn^{n-2}$$ Suppose that $|n| \geq 2$. In this case, if $n^{n-2}$ is ...


2

No. The equality must hold for every set $A$. If $A \cap B = A \cap C$ for every set $A$, then in particular $B = B \cap B = B \cap C$ so that $B \subset C$. Similarly $C = C \cap C = C \cap B$ so that $C \subset B$, implying $B = C$.



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