# Tag Info

10

Yes, this proof uses countable choice. In an essential way, too. It is consistent (without choice) that there is a dense set if reals without a countably infinite subset. In particular every convergent sequence from that set must be eventually constant. But density means that every real is in the closure. Other proofs that use the axiom of choice include: ...

8

Let $y\in\mathbb{R}$. As $f\circ g\circ h$ is surjective, there exists $x\in\mathbb{R}$ such that $(f\circ g\circ h)(x) = y$. So, if we let $x' = (g\circ h)(x)$, we have $f(x') = y$, which proves that $f$ is surjective.

6

If the sets are not disjoint, in the right-hand side $\lvert A\rvert+\lvert B\rvert$, formula, the elements of $A\cap B$ are counted twice.

5

$\mathbb{R}=(f\circ g\circ h) (\mathbb{R})\subseteq{f(\mathbb{R}})\subseteq \mathbb{R}$, so $f(\mathbb{R})=\mathbb{R}$. even $f\colon E \to F$, $g\colon D \to E$ and $h: C \to D$, then since $\mathbb{F}=(f\circ g\circ h) (\mathbb{C})\subseteq{f(E})\subseteq \mathbb{F}$, so $f(\mathbb{E})=F$.

5

Think of a box. Empty set is like an empty box. But a box with another box inside it is no longer empty.

4

$\{\emptyset \}$ has one element, while $\emptyset$ has $0$ elements.

4

Let $X$ be a metric space; let $A \subset X$; let $B \subset A$; then $B$ is called open (resp. closed) in $A$ if and only if there is some $C$ open (resp. closed) in $X$ such that $B = C \cap A$. From this definition it follows that the Cantor set is clopen in itself. However, the Cantor set is not clopen in $\mathbb{R}$, as you have remarked.

4

A first pass might be to write $$\{ \sqrt{n} \mid a \le n \le b \}$$ which can be written equivalently as $$\{ m \mid m = \sqrt{n} \text{ for some } a \le n \le b \}$$

4

You might have a look at older related questions, like: every non-principal ultrafilter contains a cofinite filter. Is an ultrafilter free if and only if it contains the cofinite filter? Ultrafilters containing a principal filter at MathOverflow It is shown there that an ultrafilter is free if and only if it contains the filter consisting of all cofinite ...

4

I don't know how you got to Therefore f(a) = b , g(a) = b , h(a) =b , so f is surjective. by the way you don't need A,B,C and D because you know by definition of f,g,h that they go from R to R..... so A,B,C,D would simply be R As you said we know per definition $\forall y \in R\; \exists x\in R : f(g(h(x)))=y$ Then simply substitude $g(h(x))$ by $x'$ ...

4

The question is asking you to prove that a relation, $\sim$, is an equivalence relation. A binary relation is technically a subset of the cartesian product of a set on itself. Although the words may seem foreign, you've been using relations your whole life. Think of it as a way of "comparing" things. For example, $5<7$, we say that $5$ is "less than" ...

4

Better to consider it as $B_f$ - we take a function $f:A\to P(A)$ and define $B$ in terms of it. In particular, we are stating exactly when $x\in B_f$. That is: $$x\in B_f\iff x\in A \text{ and } x\notin f(x)$$ That entirely determines $B_f$, given an $f$. And what the argument shows is that no matter what $f$ you give men, $B_f$ is not in the range of ...

4

Based on your remark that you allready proved existence I preassume that you have proved that $A:=\cup\mathcal F$ satisfies the conditions (a) and b). Let it be that the set $A'$ also satisfies these conditions. So: (a') $\mathcal{F}\subseteq\wp\left(A'\right)$ (b') $\forall B\left[\mathcal{F}\subseteq\wp\left(B\right)\implies A'\subseteq B\right]$ Then ...

4

The claim that any surjective function has a right inverse is indeed (against standard background assumptions) equivalent to Axiom of Choice. There are whole books on the Axiom of Choice, and on the costs and benefits of accepting it (it has surprising consequences). The Wikipedia entry is not a bad place to start finding out more about AC. And for a more ...

4

It's not a big deal or a problem, but it does require the Axiom of Choice. Here's the point. We want to base our math on set theory. In elementary situations sets are just things that have elements, and the way they work is just the way things with elements obviously work. That's the level at which you'd say what's the big deal, we just select one of these ...

3

Probably the clustering of empty set symbols and nested set braces is the most confusing here. The empty set has only one subset: $\emptyset$ itself, hence $P(\emptyset)$ is a one-element set with $\emptyset$ as only element. For any one-element set $X=\{a\}$ we have $P(X)=\{\emptyset,X\}$ as the empty set and the full set itself are the only subsets of ...

3

Write the disjoint unions and use your original result. That is: $$\begin{cases}A\cup B=A\setminus B\cup B\setminus A \cup A\cap B \\ A = A\setminus B \cup A\cap B\\ B=B\setminus A\cup A\cap B\end{cases}.$$ Since you know that these are all disjoint, you can use the original result to write your proof as $$|A\cup B|=|A\setminus B|+|B\setminus ... 3 |A|+|B| contains twice those elements that are contained in both sets. So, if you want to calculate the true number of elements of A\cup B, |A\cup B| then you have to subtract the number of elements that are taken into account twice, that is, you have to subtract |A\cap B| form |A|+|B|. As a result$$|A\cup B|=|A|+|B|-|A\cap B|.$$3 While writing down this answer, I found that @Martin Sleziak gave an answer first with the same idea as in here (with even some details identical). I leave my answer for my personal record, but if people think I should erase it, I will. Let \mathcal{D} be the filter generated by E. Case 1. If \cap E \neq \varnothing, then \mathcal{D} is a ... 3 (Work in progress.) Say a subset A of \mathbb{R} is difference-unique if for all z \not = 0, z \in \mathbb{R} there is at most one pair a_z, b_z \in A with a_z - b_z \in A. Say A is nice if additionally for all z \not = 0, z \in \mathbb{R} there is exactly one a_z, b_z \in A with a_z - b_z = z. (That is, your property.) It is immediately ... 3 It is clear that such relation is a function. Let k be an arbitrary positive integer and let k=2^{a_1}\cdot3^{a_2}\cdots p_t^{a_t}be its prime factorization, where p_i is the i-th prime and a_i is the largest integer e for which p^e\mid k. Then 3^{a_2}\cdots p_t^{a_t} is odd. You should be able to continue from here to show that the function ... 3 Yes, what you’ve done is fine. You could simplify it by taking Z=\{a\}, so that ZZ=\{aa\}; clearly neither is a subset of the other. 2 You can order by size every finite subset of positive rationals. If you could conclude from every finite subset on the infinite set, then ordering of all positive rationals by size would be possible. Of course this is not possible. For instance there is no smallest positive rational. But this shows that you can not conclude from every finite subset on the ... 2 As hardmath has already hinted, A∩(B∪C)^c and B∪C are disjoint sets, so ,$$ P((A∩(B∪C)^c)∪(B∪C)) = P(A∩(B∪C)^c) + P(B∪C)$$by addition rule But$$(A∩(B∪C)^c)∪(B∪C) = A∪B∪C$$Therefore,$$P(A∪B∪C) = P(A∩(B∪C)^c) + P(B∪C)  which is the desired result.

2

What you are describing makes me think about formalism, which was one of the most prominent or popular schools of mathematical philosophy with many famous mathematicians in favour of it, for instance David Hilbert. The school (or philosophy) stated that a mathematical theory should be possible to express as systematic string manipulations according to ...

2

Prove that for every $k\in\mathbb N$ there are $m,n\in\mathbb N$ such that $k=2^{m-1}(2n-1)$ (surjectivity), and secondly that this combination $\langle m,n\rangle$ is unique (injectivity). Example: $k=672=2^5\times 21$ (first write all factors $2$; then the second factor is odd). Here $m=6$ and $n=11$.

2

That is not true. $2 \in \{2,4,6\}$ and $\{2\}\subset \{2,4,6\}$. However the way natural numbers may be defined is $0=\{\}$, $1=\{0\}$, $2=\{1,0\}$, ... This is a very common definition, and it shows that $1\in 2$ as counter intuitive as that sounds. This is how natural numbers are defined in set theory. ...

2

You could write $A\backslash M\ne\emptyset$. Meaning that when you take all the men out of the alphas, there are alphas remaining.

2

Yes, De Morgan laws hold for arbitrary unions and intersections, as can easily be verified directly. The proof in the finite case is essentially the same as for the infinite case.

2

You are correct: $T$ is one-to-one. In fact, since $\max T(n)=n$, we see that $\max$ is a left inverse of $T$.

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