%LaTeX 2.09 \documentstyle[12pt]{article} \def\ep{\epsilon} \def\concat{{\;\hat{}\;}} \def\om{\omega} \def\ka{\kappa} \def\ga{\gamma} \def\Ga{\Gamma} \def\al{\alpha} \def\be{\beta} \def\de{\delta} \def\rh{\rho} \def\De{\Delta} \def\Si{\Sigma} \def\forces{\mid\vdash} \def\su{\subseteq} \def\ra{\rangle} \def\la{\langle} \def\se{\setminus} \def\supp{{\rm supp}} \def\dom{{\rm dom}} \def\cof{{\rm cof}} \def\rmiff{{\mbox{ iff }}} \def\rmand{{\mbox{ and }}} \def\rmor{{\mbox{ or }}} \def\pf{\par\noindent{\bf Proof:}\par} \def\qed{\par\noindent$\Box$\par\medskip} \def\pfof#1{{\bigskip \noindent{\bf Proof of Theorem \ref{#1}.}\medskip }} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \def\res{\upharpoonright} \def\rr{{\Bbb R}} \def\rat{{\Bbb Q}} \def\po{{\Bbb P}} \def\cc{{\goth c}} \input amssym.def \input amssym %% if amssym.def and amssym.tex are not available uncomment lines below %\def\Bbb#1{{\bf #1}} %\def\goth#1{{\bf #1}} %\def\upharpoonright{|_} \begin{document} \def\note{\footnote{This appears in Journal of Symbolic Logic, 63(1998), 29-49.}} \begin{center} {\Large Orthogonal Families of Real Sequences\note} \end{center} \begin{center} Arnold W. Miller\\ and \\ Juris Steprans\\ \end{center} \bigskip For $x,y\in\rr^\om$ define the inner product $$(x,y)=\Sigma_{n\in\om}x(n)y(n)$$ which may not be finite or even exist. We say that $x$ and $y$ are orthogonal if $(x,y)$ converges and equals $0$. Define $l_p$ to be the set of all $x\in\rr^\om$ such that $$\sum_{n\in\om}|x(n)|^p<\infty.$$ For Hilbert space, $l_2$, any family of pairwise orthogonal sequences must be countable. For a good introduction to Hilbert space, see Retherford \cite{R}. \begin{theorem} \label{thm1} There exists a pairwise orthogonal family $F$ of size continuum such that $F$ is a subset of $l_p$ for every $p>2$. \end{theorem} It was already known\footnote{Probably Kunen was the first. His example is given in the proof of Theorem \ref{thm1-5}. Earlier work was done by Abian and examples were also constructed by Keisler and Zapletal independently. At any rate, we know definitely that we didn't do it first.} that there exists a family of continuum many pairwise orthogonal elements of $\rr^\om$. A family $F\subseteq\rr^\om\se {\bf 0}$ of pairwise orthogonal sequences is orthogonally complete or a maximal orthogonal family iff the only element of $\rr^\om$ orthogonal to every element of $F$ is ${\bf 0}$, the constant 0 sequence. It is somewhat surprising that Kunen's perfect set of orthogonal elements is maximal (a fact first asserted by Abian). MAD families, nonprincipal ultrafilters, and many other such maximal objects cannot be even Borel. \begin{theorem} \label{thm1-5} There exists a perfect maximal orthogonal family of elements of $\rr^\om$. \end{theorem} Abian raised the question of what are the possible cardinalities of maximal orthogonal families. \begin{theorem} \label{thm2} In the Cohen real model there is a maximal orthogonal set in $\rr^\om$ of cardinality $\om_1$, but there is no maximal orthogonal set of cardinality $\ka$ with $\om_1<\ka<\cc$. \end{theorem} By the Cohen real model we mean any model obtained by forcing with finite partial functions from $\ga$ to $2$, where the ground model satisfies GCH and $\ga^\om=\ga$. \begin{theorem} \label{thm3} For any countable standard model $M$ of ZFC and $\ka$ in $M$ such that $M\models \ka^\om=\ka$, there exists a ccc generic extension $M[G]$ such that the continuum of $M[G]$ is $\ka$ and in $M[G]$ for every infinite cardinal $\al\leq\ka$ there is a maximal orthogonal family of cardinality $\al$. \end{theorem} \begin{theorem} \label{ma} (MA$_\ka$($\sigma$-centered)) Suppose $X\su \rr^\om$, $||X||\leq\ka$, $X\cap l_2$ is finite, and for every distinct pair $x,y\in X$ the inner product $(x,y)$ converges. Then there exists a $z\in \rr^\om\se l_2$ such that $z$ is orthogonal to every element of $X$. \end{theorem} The question arises of whether uncountable families of pairwise orthogonal elements of $\rr^\om$ must somehow determine an almost disjoint family of subsets of $\om$. In the following result we give a perfect family of orthogonal elements of $\rr^\om$ each of which has full support. It is possible to modify Kunen's example, Theorem \ref{thm1-5}, or the method of Theorem \ref{thm1} to produce such a perfect set by replacing the zeros by a very small sequence of positive weights. If this\footnote{ Briefly, use the weights $a_n={1\over{2^{4n}}}$ off of the comb on level $n$. Use weights $+-b_n$ on the comb with $b_0$ on the root node, $-b_n$ on the teeth, and $+b_n$ on the branch. By choosing $b_{n+1}$ so that $$b_0^2 + 2b_1^2 + ... + 2b_n^2 - 2b_{n+1}^2 + \sum_{i\leq n+1} a_i^2(2^i-2) + \sum_{i>n+1} a_i^2 (2^i -4) =0,$$ the inner products will be zero.} is done, then the resulting elements will be $l_2$ almost disjoint'' in the following sense. Given $x$ and $y$ there will be $X$ and $Y$ almost disjoint subsets of $\om$ such that $$\sum_{n\notin X}x(n)^2<\infty \rmand \sum_{n\notin Y}y(n)^2<\infty.$$ Equivalently, $$\sum_{n\in\om}\min\{|x(n)|,|y(n)|\}^2<\infty.$$ Note that the supports of $x$ and $y$ are almost disjoint iff the function $$n\mapsto\min\{|x(n)|,|y(n)|\}$$ is eventually zero. Consequently, the minimum function is one measure of the almost disjointedness of $x$ and $y$. Note, however that if the inner product of $x$ and $y$ converges, then $\min\{|x(n)|,|y(n)|\}\to 0$ as $n\to\infty$. \begin{theorem}\label{support} There exists a perfect set $P\su\rr^\om$ such that every pair of elements of $P$ are orthogonal, $\supp(x)=\om$ for every $x\in P$, and if we define $$h(n)=\min\{|x(n)|:x\in P\}$$ then for every $p$ $$\sum_{n<\om}h(n)^p=\infty.$$ \end{theorem} K.P.Hart raised the question of whether there could be a maximal orthogonal family in $l_2$ which was not maximal in $\rr^\om$. This was answered by Kunen and Steprans independently. \begin{theorem} \label{l2} (a) There exists $X$ which is a maximal orthogonal family in $l_2$ such that for all $n$ with $1\leq n \leq\om$ there exists $Y\su \rr^\om\se l_2$ with $||Y||=n$ and $X\cup Y$ a maximal orthogonal family in $\rr^\om$. Furthermore, every maximal orthogonal family containing $X$ is countable. (b) There exists a perfect maximal orthogonal family $P\su\rr^\om$ such that $P\cap l_2$ is a maximal orthogonal family in $l_2$. \end{theorem} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%% Proofs %%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} The Proofs \end{center} \pfof{thm1} Here is the basic idea. Take the full binary tree and attach pairwise disjoint finite sets $F_s$ to each node, see figure \ref{comb}. Instead of taking branches, take "combs", which are a branch together with nodes which are just off the branch, e.g. for the rightmost branch the comb would be as in figure \ref{comb}. The elements of the comb will be the support of the sequence. Attach to the branch nodes $F_{111..1}$ positive weights and to the off branch nodes $F_{111...10}$ negative weights. Then when two combs eventually disagree the lowest level pair of nodes gives a negative value. By choosing the sizes of the $F_s$'s and attached weights correctly this negative value will cancel out all the positive values above. \begin{figure} \unitlength=1.00mm \special{em:linewidth 0.4pt} \linethickness{0.4pt} \begin{picture}(130.00,50.00) \put(40.00,50.00){\makebox(0,0)[cc]{$F_{\la\ra}$}} \put(20.00,30.00){\makebox(0,0)[cc]{$F_{0}$}} \put(60.00,30.00){\makebox(0,0)[cc]{$F_{1}$}} \put(10.00,10.00){\makebox(0,0)[cc]{$F_{00}$}} \put(30.00,10.00){\makebox(0,0)[cc]{$F_{01}$}} \put(50.00,10.00){\makebox(0,0)[cc]{$F_{10}$}} \put(70.00,10.00){\makebox(0,0)[cc]{$F_{11}$}} \put(25.00,35.00){\line(1,1){10.00}} \put(55.00,35.00){\line(-1,1){10.00}} \put(10.00,15.00){\line(1,1){10.00}} \put(20.00,25.00){\line(1,-1){10.00}} \put(50.00,15.00){\line(1,1){10.00}} \put(60.00,25.00){\line(1,-1){10.00}} \put(90.00,50.00){\makebox(0,0)[cc]{$F_{\la\ra}$}} \put(80.00,40.00){\makebox(0,0)[cc]{$F_0$}} \put(100.00,40.00){\makebox(0,0)[cc]{$F_1$}} \put(90.00,30.00){\makebox(0,0)[cc]{$F_{10}$}} \put(110.00,30.00){\makebox(0,0)[cc]{$F_{11}$}} \put(100.00,20.00){\makebox(0,0)[cc]{$F_{110}$}} \put(120.00,20.00){\makebox(0,0)[cc]{$F_{111}$}} \put(110.00,10.00){\makebox(0,0)[cc]{$F_{1110}$}} \put(130.00,10.00){\makebox(0,0)[cc]{$F_{1111}$}} \put(83.00,43.00){\line(1,1){4.00}} \put(97.00,43.00){\line(-1,1){4.00}} \put(93.00,33.00){\line(1,1){4.00}} \put(103.00,37.00){\line(1,-1){4.00}} \put(103.00,23.00){\line(1,1){4.00}} \put(113.00,27.00){\line(1,-1){4.00}} \put(113.00,13.00){\line(1,1){4.00}} \put(124.00,17.00){\line(1,-1){4.00}} \end{picture} \caption{Combs \label{comb}} \end{figure} Define the sequence $r_n$ by $r_0=1$ and $$r_{n+1}=\sum_{i \leq n} r_i.$$ (This makes $r_n=2^{n-1}$ for $n>0$, but it is irrelevant.) Let $p_n>2$ be a sequence decreasing to 2. Next construct integers $k_n>r_n$ and reals $\ep_n>0$ such that $$\ep_n^2 \cdot k_n =r_n$$ and $$\ep_n^{p_n}\cdot k_n \leq {1\over n^2}.$$ To do this first pick $k_n$ so that $$k_n^{{p_n\over 2}-1}\geq n^2\cdot r_n^{{p_n}\over 2}$$ then let $$\ep_n^2={r_n\over k_n}.$$ Thus $$\ep_n^{p_n}\cdot k_n =({r_n\over k_n})^{p_n\over 2}\cdot k_n= {{r_n^{p_n/2}}\over{k_n^{p_n/2-1}}}\leq {1\over n^2}.$$ Let $2^\om$ be the set of infinite sequences of 0 and 1's and let $2^{<\om}$ be the set of finite sequences of 0 and 1's. Let $\{F_s: s\in 2^{<\om}\}$ be pairwise disjoint subsets of $\om$ such that $||F_s||=k_n$ when $s\in 2^n$. For $x\in 2^\om$ and $n\in\om$ define $$s^x_{n+1}=x |_{n+1}$$ and $$t^x_{n+1}=x |_n \concat (1-x(n)).$$ Thus $s^x_{n+1}$ is the first $n+1$ bits of $x$ while $t^x_{n+1}$ is the first $n$ bits of $x$ followed by the opposite bit $1-x(n)$. Define $y_x\in\rr^\om$ by $$y_x(m)=\left\{\begin{array}{rl} \ep_{n+1} & \mbox{if m\in F_{s^x_{n+1}} for some n}\\ -\ep_{n+1} & \mbox{if m\in F_{t^x_{n+1}} for some n}\\ \sqrt{2}\cdot\ep_0 & \mbox{ if m\in F_{\la\ra} } \\ 0 & \mbox{ otherwise } \end{array}\right.$$ First note that $y_x\in l_p$ for any $p>2$: $$\sum |y_x(m)|^p=(\sqrt{2}\ep_0)^p\cdot k_0 +\sum (\ep_{n+1}^p ||F_{s_{n+1}}||+ \ep_{n+1}^p ||F_{t_{n+1}}||)$$ and $$\sum (\ep_{n+1}^p ||F_{s_{n+1}}||+ \ep_{n+1}^p ||F_{t_{n+1}}||)= \sum 2\ep_{n+1}^p k_{n+1}.$$ But for all but finitely many $n$ we have that $p_n0$ (if its negative a similar argument works). Note that for any $t\in T$ either \begin{itemize} \item $t\concat 0\in T$ or \item there exists $w>0$ such that $t\concat w\in T$ and $t\concat -w\in T$. \end{itemize} In other words if a node doesn't split it continues with 0 and if it does split, then one way is positive and other negative. Using this it is easy to construct a sequence $$s=s_0\su s_1\su\cdots\su s_m\su\cdots$$ with $s_m\in T_{n+m}$ such that for $m>0$ $$s_m(n+m-1)\cdot x(n+m-1)\geq 0.$$ It follows that if $b=\cup\{s_m:m\in\om\}$ and $(x,b)$ converges, then $$(x,b)\geq (x\res n,s)>0,$$ and so, in any case, $x$ is not orthogonal to $b$. \qed \begin{figure} \unitlength=.9mm \special{em:linewidth 0.4pt} \linethickness{0.4pt} \begin{picture}(150.00,80.17)(6,0) \put(80.00,80.00){\makebox(0,0)[cc]{$1$}} \put(40.00,70.00){\makebox(0,0)[cc]{$1$}} \put(120.00,70.00){\makebox(0,0)[cc]{$-1$}} \put(20.00,60.00){\makebox(0,0)[cc]{$\sqrt{2}$}} \put(60.00,60.00){\makebox(0,0)[cc]{$-\sqrt{2}$}} \put(120.00,60.00){\makebox(0,0)[cc]{$0$}} \put(20.00,50.00){\makebox(0,0)[cc]{$0$}} \put(60.00,50.00){\makebox(0,0)[cc]{$0$}} \put(100.00,50.00){\makebox(0,0)[cc]{$\sqrt{2}$}} \put(140.00,50.00){\makebox(0,0)[cc]{$-\sqrt{2}$}} \put(10.00,40.00){\makebox(0,0)[cc]{$2$}} \put(30.00,40.00){\makebox(0,0)[cc]{$-2$}} \put(60.00,40.00){\makebox(0,0)[cc]{$0$}} \put(100.00,40.00){\makebox(0,0)[cc]{$0$}} \put(140.00,40.00){\makebox(0,0)[cc]{$0$}} \put(10.00,30.00){\makebox(0,0)[cc]{$0$}} \put(30.00,30.00){\makebox(0,0)[cc]{$0$}} \put(50.00,30.00){\makebox(0,0)[cc]{$2$}} \put(70.00,30.00){\makebox(0,0)[cc]{$-2$}} \put(100.00,30.00){\makebox(0,0)[cc]{$0$}} \put(140.00,30.00){\makebox(0,0)[cc]{$0$}} \put(10.00,20.00){\makebox(0,0)[cc]{$0$}} \put(30.00,20.00){\makebox(0,0)[cc]{$0$}} \put(50.00,20.00){\makebox(0,0)[cc]{$0$}} \put(70.00,20.00){\makebox(0,0)[cc]{$0$}} \put(90.00,20.00){\makebox(0,0)[cc]{$2$}} \put(110.00,20.00){\makebox(0,0)[cc]{$-2$}} \put(140.00,20.00){\makebox(0,0)[cc]{$0$}} \put(10.00,10.00){\makebox(0,0)[cc]{$0$}} \put(30.00,10.00){\makebox(0,0)[cc]{$0$}} \put(50.00,10.00){\makebox(0,0)[cc]{$0$}} \put(70.00,10.00){\makebox(0,0)[cc]{$0$}} \put(90.00,10.00){\makebox(0,0)[cc]{$0$}} \put(110.00,10.00){\makebox(0,0)[cc]{$0$}} \put(130.00,10.00){\makebox(0,0)[cc]{$2$}} \put(150.00,10.00){\makebox(0,0)[cc]{$-2$}} \put(45.00,75.00){\line(6,1){30.00}} \put(115.00,75.00){\line(-6,1){31.00}} \put(23.00,63.00){\line(2,1){13.00}} \put(57.00,63.00){\line(-2,1){14.00}} \put(120.00,68.00){\line(0,-1){5.00}} \put(20.00,58.00){\line(0,-1){5.00}} \put(60.00,58.00){\line(0,-1){5.00}} \put(117.00,57.00){\line(-4,-1){14.00}} \put(123.00,57.00){\line(4,-1){14.00}} \put(13.00,43.00){\line(4,5){4.00}} \put(27.00,43.00){\line(-4,5){4.00}} \put(60.00,48.00){\line(0,-1){5.00}} \put(100.00,43.00){\line(0,1){5.00}} \put(140.00,48.00){\line(0,-1){5.00}} \put(10.00,38.00){\line(0,-1){5.00}} \put(30.00,33.00){\line(0,1){5.00}} \put(52.00,34.00){\line(5,4){5.00}} \put(63.00,38.00){\line(1,-1){4.00}} \put(100.00,33.00){\line(0,1){5.00}} \put(140.00,38.00){\line(0,-1){5.00}} \put(10.00,23.00){\line(0,1){4.00}} \put(30.00,27.00){\line(0,-1){4.00}} \put(50.00,23.00){\line(0,1){4.00}} \put(70.00,27.00){\line(0,-1){4.00}} \put(92.00,23.00){\line(5,4){5.00}} \put(104.00,27.00){\line(1,-1){4.00}} \put(140.00,23.00){\line(0,1){4.00}} \put(10.00,13.00){\line(0,1){4.00}} \put(30.00,17.00){\line(0,-1){4.00}} \put(50.00,13.00){\line(0,1){4.00}} \put(70.00,17.00){\line(0,-1){4.00}} \put(90.00,13.00){\line(0,1){4.00}} \put(110.00,17.00){\line(0,-1){4.00}} \put(134.00,13.00){\line(4,5){4.00}} \put(143.00,18.00){\line(4,-5){4.00}} \put(10.00,7.00){\line(0,-1){2.00}} \put(30.00,5.00){\line(0,1){2.00}} \put(50.00,7.00){\line(0,-1){2.00}} \put(70.00,5.00){\line(0,1){2.00}} \put(90.00,7.00){\line(0,-1){2.00}} \put(110.00,5.00){\line(0,1){2.00}} \put(130.00,7.00){\line(0,-1){2.00}} \put(150.00,5.00){\line(0,1){2.00}} \end{picture} \caption{Kunen's perfect tree \label{kunentree}} \end{figure} \pfof{thm2} For $x\in \rr^\om$ define the support of $x$: $$\supp(x)=\{n\in\om: x(n)\not=0\}.$$ Let us say that $x$ and $y$ are strongly orthogonal iff $\supp(x)$ and $\supp(y)$ are almost disjoint (i.e., intersection is finite) and $(x,y)=0$. Let $S$ be all the elements of $\rr^\om$ with infinite support. Let $S^*$ be all elements $x$ of $S$ such that if $n\in\supp(x)$ then $|x(n)|>1$. Let $\po$ be any countable partially ordered set. \begin{lemma} Suppose $\{x_n:n\in\om\}\subseteq S^*$ are pairwise strongly orthogonal. Let $p\in\po$ and $\tau$ be a $\po$-name such that $$p\forces \tau\in S \mbox{ and }\forall n \;(x_n,\tau)=0.$$ Then there exists $q\leq p$ and $y\in S^*$ such that $$q\forces (y,\tau)\not= 0$$ and $y$ is strongly orthogonal to $x_n$ for every $n\in\om$. \end{lemma} \pf Case 1. $p\forces$$\supp(\tau)\setminus\cup_{n1$ such that $$q\forces |u_n\cdot\tau(k_n)|>1.$$ Now construct $y\in S^*$ so that $$\supp(y)=\{k_n:n\in\om\}\cup\{l_n:n\in\om\}\cup \{r_n:n\in\om\}$$ and $y(k_n)=u_n$ and $(y,x_n)=0$ for all $n$. [$(y,x_n)=0$ is accomplished by picking the values of $y(l_n)$ and $y(r_n)$ inductively. Use that given $x,u,v\in\rr$ with $u\not=0$ and $v\not=0$ we can pick $a,b\in\rr$ with $|a|>1$ and $|b|>1$ and $au+bv=x$. This is possible because if we let $b={x-au \over v}$, then as $a\to\infty$ we have $|b|\to\infty$.] But note that $$p\forces ||\{n: |\tau(k_n)y(k_n)|>1\}||=\om$$ hence we are done with case 1. Case 2. There exists $r\leq p$, $N<\om$, and $F$ finite such that $$r\forces \supp(\tau)\subseteq \cup_{n1 for each n\in\supp(x_{n_0}), we must be able to find distinct k_1,k_2\in Q such that the vectors \la\tau^G(k_1),\tau^G(k_2)\ra and \la x_{n_0}(k_1),x_{n_0}(k_2)\ra are not parallel. [Else \la\tau^G(k):k\in Q\ra would be parallel to \la x_{n_0}(k):k\in Q\ra and this would give that (x_{n_0},\tau)\not=0.] Thus we may find a pair of real numbers u,v (take \la u,v\ra=\la x_{n_0}(k_2),-x_{n_0}(k_1)\ra) such that |u|>1 and |v|>1 with ux_{n_0}(k_1)+vx_{n_0}(k_2)=0 but u\tau^G(k_1)+v\tau^G(k_2)\not=0. Now build (similarly to case 1) y\in S^* strongly orthogonal to each x_n and such that y(k_1)=u, y(k_2)=v and$$\{k_1,k_2\}=\supp(y)\cap (\cup_{n N^r$. For$\al\in F^r\se \be$define$\hat{s}_\al: N^q\to \rat$by$\hat{s}_\al\res{N^r}=s_\al^r$and$\hat{s}_\al(n)=0$for all$n$with$N^r\leq n < N^q$. Consider the precondition$t$defined by $$F^t=F^r\cup F^q$$ and $$\la s_\al^t,P_\al^t\ra =\left\{ \begin{array}{ll} \la \hat{s}_\al,P_\al^r\ra & \mbox{ if \al\in F^r\se\be } \\ \la s_\al^q,P_\al^q\ra & \mbox{ if \al\in F^q } \end{array} \right.$$ But then$t$extends to a condition by Lemma \ref{precond}, showing that$p$and$r$are compatible and proving the Claim. It follows from the Claim that if$A\su\po_\be$is a maximal antichain of$\po_\be$, then$A$is a maximal antichain of$\po_\ga$, and hence the lemma is proved. \qed \begin{lemma} Let$\ga$be a limit ordinal and \label{finsup} suppose$\tau$is a$\po_\ga$-name for an element of$\rr^\om$,$F\in[\ga]^{<\om}$,$H\in\om$, and$p\in \po_\ga$have the property that $$p\forces ||\supp(\tau)||=\om\rmand\supp(\tau)\su \bigcup_{\be\in F}\supp(x_\be)\cup H.$$ Then there exists$\al_1\in\ga$and$q\leq p$such that $$q\forces (\tau,x_{\al_1})\not=0.$$ \end{lemma} \pf Without loss we may assume for every$\al\in F$$$p\forces (\tau,x_\al)=0.$$ Find$r\leq p$and$\al_0\in F$such that $$r\forces \supp(\tau)\cap \supp(x_{\al_0}) \mbox{ infinite.}$$ Let$G$be$\po_\ga$-generic with$r\in G$and let$Q$be the infinite set defined by $$Q=(\supp(\tau^G)\cap \supp(x_{\al_0}))\se (H\cup \bigcup\{ \supp(x_\be): {\be\in F, \be\not=\al_0}\}).$$ Then there must be$k_0 \max(F^t)$and define the precondition$q$as follows: $$F^q=F^t\cup\{\al_1\},$$ $$(\la s^q_\al,P_\al^q\ra : \al\in F^t)= (\la s^t_\al,P_\al^t\ra : \al\in F^t),$$ and define$s_{\al_1}:N^t\to\rat$by $$s_{\al_1}(l)=\left\{ \begin{array}{rl} v & \mbox{ if l=k_0}\\ -u & \mbox{ if l=k_1}\\ 0 & \mbox{ otherwise} \end{array}\right.$$ Define $$P^q_{\al_1}=\{(0,\be):\be\in F\se \{\al_0\}\}\cup\{(N^t-1,\al_0)\}.$$ The precondition$q$satisfies all requirements to be an element of$\po_\ga$except possibly the orthogonality condition. But note that for$\be\in F$$$\sum_{nN\; |\tau(n)\cdot x_\ga(n)|<1.$$ Let$G$be$\po_{\ga+1}$-generic over$M$with$q\in G$. Choose$n>N,N^q$with $$n\in\supp(\tau^G)\se\cup \{\supp(x_\al):\al\in F^q\cap\ga\}.$$ Since$\tau$is a$\po_\ga$-name and$G\cap\po_\ga$is$\po_\ga$-generic over$M$there exists$r\in G\cap\po_\ga$and$m\in\om$such that $$r\forces |\tau(n)|> {1 \over {m+1}}.$$ We may assume without loss of generality that$N^r>n$and$r\leq q\res\ga$. Note that$s_\al^r(n)=0$for all$\al\in F^q\cap \ga$. Now define$\hat{s}_\ga:N^r\to\rat$as follows. $$\hat{s}_\ga(k)=\left\{ \begin{array}{ll} s_\ga^q(k) & \mbox{ if k1 which would be a contradiction. \qed \begin{lemma} Suppose G is \po_\ga-generic over M (\cof(\ga)>\om) and \{x_\al:\al<\ga\} are the generic family of mutually orthogonal elements of \rat^\om, then in M[G] for every y\in \rr^\om there exists \al<\ga such that y and x_\al are orthogonal. \end{lemma} \pf First note that by any easy density argument for any m0 be such that for any (y,k,\ga)\in P$$|\sum_{k\leq n N_0$ and $y\in F\se \{x\}$ $$|\sum_{N_0\leq n N_0 be minimal such that$$ \sum_{N_0\leq n b.$$(This exists since x is not in l_2.) Choose \rh with |\rh|\leq 1 and so that$$\rh\cdot\sum_{N_0l.$$\end{lemma} \pf Given p just let x\in X\se F^p. As in the proof of Lemma \ref{ma1} we can extend p to equal x as much as we like. Since x is not in l_2 and p has no requirements mentioning x we can get the norm of s^q greater than l. \qed The Lemmas show that if we define$$D_l=\{q\in \po: \sum_{nl\}$$and for each x\in X and \ep\in\rat^+$$D_x^{\ep}=\{q\in \po: x\in F^q \rmand \exists k \;(x,k,\ep)\in q\}$$then these sets are dense. Applying MA we get a \po-filter G meeting them all. Then letting$$z=\cup\{s^p:p\in G\}$$we have that z is not in l_2 but is orthogonal to every element of X. This proves Theorem \ref{ma} in the case that X contains no elements of l_2. \medskip Next we indicate how to modify our partial order in case X contains finitely many elements of l_2. Let H=X\cap l_2 be finite. First we replace \rat by any countable field \rat^* which contains the rationals and all the coefficients of elements of H. We make the following two additional demands for$$p=(s:N\to\rat^*,F,P)$$to be an element of \po. \begin{enumerate} \item H\su F, and \item for each x\in H we have \sum_{nN_0 and y\in F\se \{x\}$$|\sum_{N_0\leq n N_0$be so that$z(j)\not=0$. Choose$\de>0$so that $$\de\cdot\max\{|y(j)|:y\in F\}\leq \ep_0/2.$$ Now choose$N_1>j>N_0$so that for every$m>N_1$$$|\sum_{N_1\leq n N_1 and |\rh|\leq 1 so that$$\sum_{nN_0$ so that $V=\{z\res{[N_0,N_1)}: z\in H_0\}$ are linearly independent. \medskip Claim. For any $\ep_1>0$ there exists $\de>0$ so that for any $\la \be_v:v\in V\ra$ with $|\be_v|<\de$ there exists $t\in\rr^{[N_0,N_1)}$ with $||t||_2<\ep_1$ and $(v,t)=\be_v$ for every $v\in V$. Here $||\cdot||_2$ is the usual $l_2$ norm and $(v,t)$ the usual inner product in the finite dimensional vector space $\rr^{[N_0,N_1)}$. The proof of the Claim is an elementary exercise in Linear Algebra. Let $V=\{v_1,\ldots,v_m\}$ and let $W$ be the span of $V$. Define the linear map $$T:W\to \rr^m \mbox{ by } T(t)=\la (t,v_i):i=1,\ldots,m\ra.$$ It follows from the Gram-Schmidt orthogonalization process that the kernel of $T$ is trivial. Hence the range of $T$ is $\rr^m$. The existence of $\de$ now follows from the continuity of $T$. This proves the Claim. \medskip We leave the value of $\ep_1$ to be determined latter. We find $N_2>N_1$ so that for any $m>N_2$ and $y\in F\se \{x\}$ $$|\sum_{N_0\leq n N_2 and define t\res{[N_2,N_3)} to be a small scaler multiple of x\res{[N_2,N_3)} in such a way as to make the inner product of x and t zero. We now use the Claim to extend t on the interval [N_0,N_1) to make sure that for every y\in H_0 the inner product of t and y is zero. If we choose \ep_1 small enough so that$$\ep_1\cdot ||x\res{[N_1,N_2)}||_2<\ep_0/2$$for every x\in F, then this t works (when jiggled to have range \rat^*). Since ||t\res{[N_1,N_2)}||_2 \cdot ||x\res{[N_1,N_2)}||_2\leq \ep_0/2 for every x\in F all commitments from P are honored. Also t is orthogonal to all elements of H. The reason is that t\res{N_0}=s is already orthogonal to every z\in H and every element of \{z\res{[N_0,\om)}: z\in H\} is contained in the span of \{z\res{[N_0,\om)}: z\in H_0\} and t is orthogonal to everything in \{z\res{[N_0,\om)}: z\in H_0\}. This concludes the proof of Theorem \ref{ma}. \pfof{support} For any h<\om define the partial order \po_{h} as follows. A condition in \po_{h} has the following form:$$p=((s_i:N\to\rat^{\not=0}:i0$such that$(N^q-N^p)\ep^l>1$and for every$n$with$N^p\leq n < N^q$and$i1$. \qed Remark. We could have avoided the use of Lemma \ref{sup2} by taking a pair of weights$\ep>0$and$\de>0$and the column vectors of the$h \times h$matrix with$-\de$on the diagonal and$\ep$off the diagonal. The columns will be orthogonal provided $$-2\ep\de +(h-2)\ep^2=0 \rmor \de={{h-2}\over 2}\ep$$ In this case we would get that$|s^q_i(n)|\geq\ep$in the conclusion of Lemma \ref{sup3}. \bigskip Finally, to prove Theorem \ref{support} we construct a sequence$p_n \in \po_{h_n}$where$h_n < h_{n+1}$. Start with any$p_0$and$h_0$. At stage$n$given$p_n$apply Lemma \ref{sup1} ($h_n$times) to obtain$p\leq p_n$so that for each$\{i,j\}\in [h_n]^2$for some$k0$such that$(N^q-N^p)\ep^n>1$and for every$m$with$N^p\leq m < N^q$and$i0$with the property that$(s_n,s_n)=\de_n\cdot b_n$and then letting $$T_{n+1}=\{t\concat 0: t\in T_n, t\not=s_n\}\cup \{s_n\concat\de_n,s_n\concat -b_n\}.$$ Taking$P=\{x\in\rr^\om:\forall n\; x\res n\in T_n\}$, the$s_n$are chosen so that$P$has no isolated points and hence is perfect. The only remaining things to be picked are the$\de_n$and$b_n$. Let$\de_1=b_1=1$. Given$T_n\su\rr^n$for$n\geq 2$let $$m=\min\{\max\{|(s,x)|:s\in T_n\}: x\in\rr^n,\; {3\over 4}\leq ||x||_2\leq 1\}.$$ Here$(s,x)$refers to the ordinary inner product in$\rr^n$and$||x||_2$the$l_2$norm of$x$. By compactness it is clear that$m>0$, so we can let $$\de_n={1\over 2}\cdot\min\{{m\over 2},\de_{n-1}\}$$ and then choose$b_n$so that$(s_n,s_n)=\de_n\cdot b_n$. Just as in the proof of Theorem \ref{thm1-5} the family$P$is a maximal orthogonal family in$\rr^\om$. Now define$E\su P$as follows: $$E=\{x\in P:\exists n\forall m>n \;(x(m)\in \{0,\de_m\})\}.$$ Since$\de_{m+1}\leq {1\over 2^m}$, it is clear that$E\su l_2$. We claim$E$is a maximal orthogonal family in$l_2$. Suppose$x\in l_2$is nontrivial and by taking a scalar multiple (if necessary) assume$||x||_2=1$. We will find an element of$E$which has nonzero inner product with$x$. Choose$n$sufficiently large so that $$||x\res n||_2\geq {3\over 4}.$$ Choose$s\in T_n$so that$\de_n\leq {1\over 4}|(s,x\res n)|$. Let$y\in E$be defined by$y\res n=s$and$y(m)\in\{0,\de_m\}$for all$m\geq n$. Note that since$||x||_2=1$,$|x(m)|\leq 1$and thus by our choice of$\de_m$'s ($\de_m\leq{1\over 2^{m-n}}\de_m$) $$|(x\res n, y\res n)| > \sum_{m\geq n} |x(m)y(m)|$$ and consequently$(x,y)\not=0$. This proves Theorem \ref{l2}. \qed %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%% Open questions %%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} Open Questions. \end{center} \begin{enumerate} \item (Abian) Does there exists a model of ZFC with no maximal orthogonal family of cardinality$\om_1$? In particular, what happens under MA or PFA? \item Is it always the case that for any uncountable$\ka$there is a maximal orthogonal family of cardinality$\ka$iff there exists a maximal almost disjoint family of subsets of$\om$of cardinality$\ka$? \item (Kunen) If there is a maximal orthogonal family of cardinality$\ka$, then does there exists one of cardinality$\ka$with almost disjoint supports? \end{enumerate} \begin{thebibliography}{9} \bibitem{H} S.Hechler, Short complete nested sequences in$\be N\se N$and small maximal almost-disjoint families, General Topology and Its Applications, 2(1972), 139-149. \bibitem{K} K.Kunen, {\bf Set Theory}, North-Holland 1980, Thm 2.3 p 256. \bibitem{M} A.Miller, Infinite combinatorics and definability, Annals of Pure and Applied Mathematical Logic, 41(1989), 179-203, \bibitem{R} J.R.Retherford, {\bf Hilbert space: Compact Operators and the Trace Theorem }, London Mathematical Society Student Texts, 21(1993), Cambridge University Press. \end{thebibliography} \bigskip \begin{center} Addresses \end{center} \begin{flushleft} Arnold W. Miller \\ University of Wisconsin-Madison \\ Department of Mathematics Van Vleck Hall \\ 480 Lincoln Drive \\ Madison, Wisconsin 53706-1388, USA \\ e-mail: miller@math.wisc.edu \\ home page: http://www.math.wisc.edu/$\sim\$miller \\ \end{flushleft} \medskip \begin{flushleft} Juris Steprans \\ York University \\ Department of Mathematics \\ North York, Ontario M3J 1P3, Canada \\ e-mail: juris.steprans@mathstat.yorku.ca \\ \end{flushleft} \bigskip \begin{flushright} June 1996. \\ \end{flushright} \end{document}