Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

On some occasions I have seen the following situation: We want find out whether a set of a given cardinality $\varkappa$ has some property P. If this property is invariant under bijective maps, then it doesn't matter which set of cardinality $\varkappa$ we choose. But sometimes choosing a set which has some additional structure might make proving/disproving whether the set has this property.

I am posting some examples of this method as answers below - I hope they help to clarify what I have in mind.

Do you know of some other interesting examples?

share|improve this question

8 Answers 8

This is a very trivial example which also has easy solutions not invoking the idea of transfer via bijection, but I will include it anyway.

We want to show that the set $\mathbb N$ can be decomposed into infinitely many infinite sets. We take any bijection $f\colon \mathbb N \times \mathbb N \to \mathbb N$ and then the sets $f[\mathbb N\times\{i\}]$ for $i\in\mathbb N$ yield the required decomposition.

There have been several questions at MSE about this problem:

share|improve this answer

This is perhaps not as nice as other answers given here, but it still might be a nice exercise for students in the first course where they learn about cardinalities. I have seen this exercise in some introductory textbook on set theory. (It was the book T. Šalát a J. Smítal. Teória množín, which is written in Slovak language. But it is very probable that the same exercise appears in many different books, too.)

EDIT: Now I found the same exercise as Problem 6.34 in Schaum's Outline of Set Theory and Related Topics by Seymour Lipschutz (Google Books link).


Question: Show that the set $\mathbb Q\times\mathbb Q$ can be written as $V\cup H$, where the set $V$ has the property that all vertical section $V\cap\{x\}\times\mathbb Q$ are finite, and the set $H$ has the property that all horizontal sections $H\cap \mathbb Q\times\{y\}$ are finite.

Solution. For any bijection $f \colon \mathbb Q\to\mathbb N$ we have bijection $f\times f \colon \mathbb Q\times\mathbb Q \to \mathbb N\times\mathbb N$ given by $f(q_1,q_2)=(f(q_1),f(q_2))$. This bijection preserves vertical and horizontal lines. So in fact this bijection transforms the problem to the set $\mathbb N\times\mathbb N$.

Solving the same problem for $\mathbb N\times\mathbb N$ is easy; just take $V=\{(x,y)\in\mathbb N\times\mathbb N; y\le x\}$ and $H=\{(x,y)\in\mathbb N\times\mathbb N, y>x\}$. $\mathbb \qquad\square$

(If we look back to the original problem, which asks about $\mathbb Q\times\mathbb Q$, the fact that this set can be decomposed in such way might seem a little surprising at first.)

share|improve this answer

Here is another example I like. It requires a mild understanding of forcing: One can prove abstractly that if a partial order is countable, and any condition can be extended in two incompatible ways, then the partial order is forcing isomorphic to Cohen's forcing. (This means that there is an isomorphism between the Boolean completions of the two posets.)

Conditions in this poset are finite functions $f$ whose domain is a (finite) subset of $\omega$ and whose range is $\{0,1\}$. A condition $p$ extends a condition $q$ iff, as functions, $p$ extends $q$, that is, $p\supseteq q$.

This forcing adds a subset of $\omega$ (a Cohen real). One of its many nice properties is that forcing with this poset makes the ground model reals have measure $0$.

One nice way of proving this consists of considering, for each $\epsilon$, the poset $C_\epsilon$ whose conditions are finite unions of rational intervals, whose total length is less than $\epsilon$. A condition $p$ extends a condition $q$ iff it extends it as a set of reals. Note that $C_\epsilon$ is countable, and any condition can be extended in two incompatible ways: Given $p$, we can find extensions $p_1$ and $p_2$ such that the total length of the intervals making up $p_1\cup p_2$ is $\epsilon$ or larger. This means, because of the abstract result mentioned above, that $C_\epsilon$ is Cohen forcing in disguise. (This is the "transfer".)

Now, forcing with $C_\epsilon$ gives us an open set $I_\epsilon$ whose measure is at most $\epsilon$, and such that any ground model real is in $I_\epsilon$. This shows that forcing with Cohen forcing makes the ground model reals have outer measure at most $\epsilon$. Since $\epsilon$ was arbitrary, we are done.

share|improve this answer
    
I mentioned this example here: mathoverflow.net/questions/46757/… –  Andres Caicedo Apr 14 '13 at 20:40
    
For a stronger result: mathoverflow.net/questions/63497/… –  Andres Caicedo Apr 14 '13 at 20:41

A particularly nice use of this "transfer" idea can be seen in the following proof of the $\Delta$-system lemma. I first saw this in Shelah's book on Proper forcing. Assaf Rinot recently posted the argument on his (highly recommended) blog.

A version of the lemma states that if $\kappa$ is a regular uncountable cardinal, and $\mathcal A$ is a family of finite sets, with $|\mathcal A|=\kappa$, then we can find a subfamily $\mathcal B\subseteq\mathcal A$, again with $|\mathcal B|=\kappa$, that forms a $\Delta$-system, that is there is a fixed finite set $r$ such that whenever $a\ne b$ are in $\mathcal B$, then $a\cap b=r$.

To prove this, we note that we may as well assume that $\mathcal A\subseteq[\kappa]^{<\omega}$ (this is the "transfer" part). Since $\kappa$ has uncountable cofinality, we may in fact assume that $\mathcal A\subseteq[\kappa]^n$ for some fixed $n$. We then proceed by induction on $n$. The result is clear if $n=1$. Assuming the result for $n$, and that $\mathcal A\subseteq[\kappa]^{n+1}$, we ask whether for some $\gamma$ there are $\kappa$ elements $a\in\mathcal A$ such that $\gamma=\min(a)$.

If yes, we can replace $\mathcal A$ with $$ \mathcal A'=\{a\setminus\{\gamma\}\mid a\in\mathcal A\mbox{ and }\min(a)=\gamma\}, $$ and use the inductive assumption to conclude.

If not, then we proceed recursively (using that $\kappa$ is regular): We enumerate $\mathcal A=\{a_\alpha\mid \alpha<\kappa\}$. Given any $\gamma$, there are fewer than $\kappa$ values of $\alpha$ such that $\gamma\in a_\alpha$: Else, there are $\kappa$ members of $\mathcal A$ whose minimum is at most $\gamma$, and by regularity we can fix $\kappa$ with the same minimum. But then we can start with $\alpha_0=0$, find $\alpha_1$ such that $a_0\cap a_\beta=\emptyset$ for $\beta\ge\alpha_1$, then find $\alpha_2>\alpha_1$ such that $a_{\alpha_1}\cap a_\beta=\emptyset$ for $\beta\ge\alpha_2$, etc. The sets $\{a_{\alpha_i}\mid i<\kappa\}$ are pairwise disjoint.

share|improve this answer
    
The $\Delta$-system lemma is also called the sunflower lemma, especially when considering its finite versions. I'll use this answer as an excuse to draw attention to this MO question: mathoverflow.net/questions/115069/… –  Andres Caicedo Apr 14 '13 at 18:43

The partition calculus is a generalization of Ramsey theory to the transfinite. In its basic setting, we ask, given ordinals $\alpha,\beta,\gamma$, whether $$ \gamma\to(\alpha,\beta)^2, $$ meaning that if the edges of the complete graph on $\gamma$ vertices, $K_\gamma$, are colored red and blue, then either there is a red copy of $K_\alpha$ (that is, a subset of $\gamma$ of order type $\alpha$, any edge between two of its vertices being red), or a blue copy of $K_\beta$.

To investigate partition relations satisfied by ordinals of the form $\omega^\alpha$ for $\alpha$ countable (where the exponentiation is in the ordinal sense), in many cases it is convenient to work not with $\omega^\alpha$ directly, but rather with some other set that can be ("naturally") ordered in type $\omega^\alpha$. For example, for $n<\omega$, we typically work with $$ W(n)=\{(m_0,m_1,\dots,m_{n-1})\in\mathbb N^n\mid m_0<m_1<\dots<m_{n-1}\}, $$ ordered lexicographically. Of course, we could replace $W(n)$ with the set of subsets of $\mathbb N$ of size $n$. Notationally, $W(n)$ is more convenient on occasion.

To see how this is used, I indicate here rather briefly how Haddad and Sabbagh proved that $\omega^2\to(\omega^2,n)^2$ for all $n<\omega$, a result first due to Specker.

We consider a partition of $[W(2)]^2$ into two pieces, $A_0$ and $A_1$ ($A_0$ corresponds to red, and $A_1$ to blue). Use this to partition $[\omega]^4$ into 16 pieces $B_{ijkl}$ where $i,j,k,l\in\{0,1\}$: Given $\{a,b,c,d\}\in[\omega]^4$, with $a<b<c<d$, we put $\{a,b,c,d\}\in B_{ijkl}$ iff

  • $\{(a,b),(c,d)\}\in A_i$,
  • $\{(a,c),(b,d)\}\in A_j$,
  • $\{(a,d),(b,c)\}\in A_k$, and
  • $\{(a,b),(a,c)\}\in A_l$.

We now invoke Ramsey's theorem to conclude that there is an infinite $H\subset\omega$ such that all tuples in $[H]^4$ are in the same $B_{ijkl}$. Quickly checking the possibilities for $i,j,k,l$ gives us a monochromatic $I\subset W(2)$ as required: If one of $i,j,k,l$ is $1$, one easily gets $I$ of any finite size we want with $[I]^2$ all blue. If $i,j,k,l$ are all $0$, then we can write $H$ as $\bigcup_n H_n$ where the $H_n$ are infinite and pairwise disjoint, and let $I=\{(x,y)\mid x\in H_0$ and $y\in H_{x+1}$ and $x<y\}$. Then $I$ has type $\omega^2$ as $[I]^2$ is all red.

More sophisticated applications of this idea of conveniently representing $\omega^n$ or $\omega^\omega$ or other indecomposable ordinals can be seen in the work of Jean Larson. See for example Part I of these slides.

share|improve this answer
    
Section 7.2 of "Combinatorial set theory", by Williams, contains additional details of this argument. –  Andres Caicedo Apr 14 '13 at 19:52
    
In the late 60s, Haddad and Sabbagh announced a series of results on the computation of Ramsey numbers $R(\alpha,n)$ for finite $n$ and $\alpha$ countable. None of the details of their arguments were ever published (and most remain unpublished today), with the exception of their short new proof of Specker's theorem. –  Andres Caicedo Apr 14 '13 at 19:54

This is another classical example: A nonempty family $\mathcal I$ of subsets of $\omega$ is independent iff for any pairwise distinct $A_1,\dots,A_k,B_1,\dots,B_l$ in $\mathcal I$, we have that $$ \bigcap_{i=1}^k A_i\cap\bigcap_{j=1}^l(\omega\setminus B_j) $$ is infinite. The question of whether there is an independent family of size continuum is clearly invariant under replacement of $\omega$ with any other infinite countable set $X$.

For example, we can take $X=\{(a,A)\mid a\subseteq{\mathbb N}$ is finite, and $A\subseteq{\mathcal P}(a)\}$.

Given $S\subseteq{\mathbb N}$, let $t_S=\{(a,A)\in X\mid a\cap S\in A\}$. The claim is that ${\mathcal I}=\{t_S\mid S\subseteq{\mathbb N}\}$ is as wanted.

For this, first note that $S\mapsto t_S$ is injective. Now, suppose $X_1,\dots,X_n,Y_1,\dots,Y_m$ are infinite, pairwise distinct, subsets of ${\mathbb N}$. Write $A=t_{X_1}\cap\dots\cap t_{X_n}\cap(X\setminus t_{Y_1})\cap\dots\cap(X\setminus t_{Y_m})$. For each $i,j$ with $1\le i\le n$ and $1\le j\le m$, let $\alpha_{ij}\in X_i\triangle Y_j$. One easily shows that for any finite $F\supseteq\{\alpha_{ij}\mid 1\le i\le n,1\le j\le m\}$ there is an ${ \mathcal F}$ such that $(F,{\mathcal F})\in A.$ This proves the result.

share|improve this answer
1  
Several proofs of existence of such independent families are collected in Stefan Geschke: Almost disjoint and independent families. He says there that this proof is due to Felix Hausdoff. –  Martin Sleziak Apr 14 '13 at 18:52
    
Thanks for the reference, Martin! –  Andres Caicedo Apr 14 '13 at 18:56

Suppose that $|A|=\aleph_0$. We are asking whether there exists a system of almost disjoint system of subsets of the set $A$, which has cardinality $\mathfrak c$. Recall that a system $\mathcal S$ is called almost disjoint if $A\cap B$ is finite for any two sets $A,B\in\mathcal S$ such that $A\ne B$.

Clearly this property is invariant under bijections.

If we take $A=\mathbb Q$ and we take for every real number $r$ a sequence $(q_n)_{n\in\mathbb N}$ of rationals, which converges to $r$, then the system of sets of the form $$A_r=\{q_n; n\in\mathbb N\}$$ for every $r\in\mathbb R$ give as an almost disjoint family of cardinality $|\mathbb R|=\mathfrak c$.

Such almost disjoint systems appeared in several other questions at MSE, for example here and here.

share|improve this answer

Suppose that $|A|=\aleph_0$. We are asking whether there exists a chain in $(\mathcal P(A),\subseteq)$ that has cardinality $\mathfrak c$.

Clearly this property is invariant under bijections.

If we take $A=\mathbb Q$ and $$A_r=(r,\infty)\cap\mathbb Q$$ for any real number $r\in\mathbb R$, then $\{A_r; r\in\mathbb R\}$ is a chain of cardinality $|\mathbb R|=\mathfrak c$.

This proof have appeared in an answer here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.