% analbase.tex % Analytic subspace of the reals without an analytic basis. % Beros and Miller - Feb 2009 minor revisions June 2014 \documentclass[12pt]{article} \usepackage{amsmath, amssymb, amsthm} \def\rr{{\mathbb R}} \def\qq{{\mathbb Q}} \def\al{\alpha} \def\si{\sigma} \def\es{\emptyset} \def\su{\subseteq} \def\ga{\gamma} \def\be{\beta} \def\ep{\epsilon} \def\om{\omega} \def\span{\mathrm{span}} \newtheorem{thm}{Theorem} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \renewcommand{\qedsymbol}{$\boxtimes$} \begin{document} \begin{flushright} K. Beros\\ A. Miller\\ February 2009\\ \end{flushright} \begin{center} Analytic subspace of the reals without an analytic basis. \end{center} A Hamel basis is a basis for the reals $\rr$ considered as a vector space over the field of rationals $\qq$. \begin{thm}(Erdos and Sierpinski \cite{erdos}) There is no analytic Hamel base. \end{thm} \begin{proof} Suppose on the contrary that $B$ is such a basis. $\rr$ is the countable union of sets of the form $q_1 B + \ldots q_n B$, where $q_1,\ldots,q_n \in\qq$. These sets are all analytic, hence have the property of Baire and thus, by the Baire category theorem, for some $q_1, \ldots , q_n \in \qq$, the set $A = q_1 B + \ldots q_n B$ is non meager. There is an interval $I \su \rr$ such that $A$ is comeager in $I$. Pick any distinct $x_1, \ldots , x_n \in B$. Let $q \in \qq $ and $J \su I$ be a subinterval such that $q( x_1 + \ldots + x_n ) + A$ is comeager $J$. Note that $q$ can be chosen so that $q \neq q_i$, for each $i$. This will ensure that there will be none of the $x_i$ will be canceled out by elements of $A$. Then \[ W = (q( x_1 + \ldots + x_n ) + A) \cap A \neq \es \] \noindent because both terms of the intersection above are comeager in $J$. Any element of $W$ will be at the same time a linear combination of at least $n+1$ elements of $B$ and also $n$ elements of $B$. This contradicts the linear independence of $B$ \end{proof} \bigskip \begin{thm} Every proper analytic subspace of $\rr$ is measure zero and meager. \end{thm} \begin{proof} For the first claim, suppose that $A$ is an analytic subspace of $\rr$ has positive measure. Then by Steinhaus' theorem, $A - A$ (the set of differences of elements of $A$) contains a nontrivial interval. Hence $A$ must be all of $\rr$. For the second claim, suppose that $A$ is an analytic subspace which is non-meager. $A$ has the property of Baire and hence there is an open interval $I$ in which $A$ is comeager. Fix any $\al \in \rr$. Let $q \in \qq $ be small enough that $q\al$ is less than the length of $I$. Then $J = (q\al + I) \cap I$ is non empty and $q\al + A$ and $A$ are both comeager in $J$. In other words, there exist $x, y \in A$ such that $q\al+x = y$. Hence $\al = \frac{1}{q}(y - x) \in A$. We see that $A = \rr$. \end{proof} \bigskip The following answers a question raised by Ashutosh Kumar. \begin{thm} There exists a proper (and hence meager) analytic subspace of $\rr$ with no analytic basis. \end{thm} \bigskip We begin by describing the subspace in question. \par Let $\ep_n$ be a decreasing sequence of positive rational numbers such that for every $k$ and each $N \geq k$, $$\sum_{n > N} k\ep_n \leq \frac{1}{4} \ep_N$$ This condition requires the $\ep_n$ to be a very rapidly decreasing sequence. Now let $P$ be the set defined by $$P = \{ \sum_{n \in \om} x_n \ep_n : x_n \in \{-1 , 0 , 1\}\}$$ $P$ is essentially a very sparse Cantor set. We now take the subspace $A$ to be $\span(P)$. Our first objective is to show that $A$ is a proper subspace of $\rr$. To this end, we make the following observations: $A$ is the union of all sets of the form $q_1 P + \ldots + q_nP$, where the $q_i$ are rational numbers. By taking common denominators, we can write such sets as $\frac{1}{m}(p_1 P + \ldots + p_n P)$, for some $p_1, \ldots , p_n \in \om$. If we let $k = p_1 + \ldots + p_n$, then $$(p_1 P + \ldots + p_n P) \subset \underbrace{P + P + \ldots + P}_{k\mbox{ times}}$$ We give this latter set the name $Q_k$. Observe that $Q_k$ can be described by $$Q_k = \{ \sum_{n \in \om} x_n \ep_n : x_n \in \{-k, -k+1, \ldots , k-1 , k\}\}$$ $p_1 P + \ldots + p_n P \subset Q_k$ and hence $\frac{1}{m}(p_1 P + \ldots + p_n P) \subset \frac{1}{m}Q_k$. Note that of course each $\frac{1}{m}Q_k$ is also a subset of $A$. \par Before proceeding, note that throughout we will use the notation $\hat \si$ for the rational number $\sum_{n < |\si|} \si (n) \ep_n$, where $\si$ is a finite sequence of integers. \par We now show that $A$ is a proper subspace via the following two lemmas. \bigskip \begin{lem} Suppose that $\si, \tau \in \{-k, \ldots , k\}^{< \om}$ such that $|\si| = |\tau| > k$. If $\si <_{\mbox{lex}} \tau$, then every point in $Q_k^\si$ is less than every point in $Q_k^\tau$. \end{lem} \begin{proof} It suffices to prove this lemma for the case in which there exists $\ga$ such that $\si = \ga \hat \ i$ and $\tau = \ga \hat \ (i + 1)$, for some $i \in \{-k, \ldots , k-1\}$. Let $M = |\ga|$. The greatest element of $Q_k^\si$ is $$\al = \hat \si + \sum_{n > M} k \ep_n = \hat \ga + i \ep_M + \sum_{n > M} k \ep_n$$ and the least element of $Q_k^\tau$ is $$\be = \hat \tau - \sum_{n > M} k \ep_n = \hat \ga + (i \ep_M - \sum_{n > M} k \ep_n)$$ Therefore, $$\be - \al = \ep_M - \sum_{n>M} 2k \ep_n \geq \frac{1}{2} \ep_M >0$$ \end{proof} \bigskip \begin{lem} Each $Q_k^\si$ is nowhere dense, for $|\si| > k$. \end{lem} \begin{proof} Fix any interval $I$ such that $I \cap Q_k^\si \neq \es$. Choose $\tau \supseteq \si$ such that $\hat{\tau\hat \ i}, \hat{\tau \hat \ (i + 1)} \in I$, for some $i \in \{-k , \ldots , k-1\}$. Then every element of $Q_k^{\tau\hat \ i}$ is less than every element of $Q_k^{\tau\hat \ (i+1)}$ by the previous lemma. Therefore, between all $Q_k^\si$ are closed sets, we may take an interval $J$ between $Q_k^{\tau\hat \ i}$ and $Q_k^{\tau\hat \ (i+1)}$. $J$ is disjoint from $Q_k^\si$, because $Q_k^\si$ is the disjoint union of $Q_k^\ga$ for $|\ga| = |\tau| + 1$ and by the previous lemma, no such $Q_k^\ga$ intersects $J$. \end{proof} \bigskip This shows that each $Q_k^\si$ is nowhere dense. Hence $Q_k$ is as well, being a finite union of such $Q_k^\si$. It follows that each $\frac{1}{m} Q_k$ is nowhere dense and hence $A = \bigcup_{m,k \in \om} \frac{1}{m}Q_k$ is meager. $A$ is therefore proper. \par Now we get to our main claim. \bigskip \begin{lem} $A$ has no analytic basis as a vector space over $\qq $. \end{lem} We begin with some remarks about the set $P$. As in the above lemma, for $\si \in \{-1, 0, 1\}^{< \om}$, we define \[ N_\si = \{ \sum_{n \in \om} x_n \ep_n : x_n \in \{ -1, 0 , 1\} \ \& \ \si (n) = x_n \mbox{ for } n < |\si| \}\] Note that although the $N_\si$ are closed sets in $\rr$ (and hence in $P$), they are also relatively open in $P$. In fact, they form a base for the relative topology on $P$. \begin{proof}[Proof of Lemma 6] Suppose towards a contradiction that $B$ is an analytic basis for $A$. We may assume, without loss of generality, that $1 \in B$. Otherwise, suppose that $x_1, \ldots , x_n \in B$ and $q_1, \ldots q_n \in \qq $ are such that $1 = q_1 x_1 + \ldots + q_n x_n$. Then $[(q_1 - 1)x_1 + q_2 x_2 + \ldots q_n x_n] + x_1 = 1$. Hence $[(q_1 - 1)x_1 + q_2 x_2 + \ldots q_n x_n] + B$ is an analytic basis for $A$ which contains 1. \par Since $B$ is a basis, the generating set $P$ of $A$ must be covered by a union of set of the form $$q_1 (B \cap I_1) + \ldots + q_n (B \cap I_n)$$ \noindent where $q_1, \ldots , q_n \in \qq $ and $I_1, \ldots , I_n$ are pairwise disjoint intervals with rational endpoints. To avoid confusion later on, we assume here that all $q_j$ are nonzero and that each $B \cap I_j$ is nonempty. $P$ is an uncountable closed set and hence a Baire space when regarded as a topological subspace of $\rr$. Because the union described above is countable, the Baire category theorem yields that there are $q_1, \ldots , q_n , I_1 , \ldots , I_n$ as above such that $$W = q_1 (B \cap I_1) + \ldots + q_n (B \cap I_n)$$ \noindent is non-meager in $P$. $W$ is analytic, hence has the Baire property. We therefore obtain $\si \in \{-1, 0 , 1\}^{ \om}$ such that $W$ is comeager in $N_\si$. (Because the $N_\si$ are a base for the relative topology on $P$.) We now define a homeomorphism $\pi$ of $N_\si$ as follows: If $z \in N_\si$, then $z = \hat \si + \sum_{n \geq |\si|} x_n \ep_n$, for some sequence $\langle x_n : n \in \om \rangle \in \{-1,0,1\}^\om$. We define $\pi (z) = \hat \si - \sum_{n \geq |\si|} x_n \ep_n$. It is clear that $\pi$ is an autohomeomorphism of $N_\si$. It follows that $\pi^{-1} (W)$ is also comeager in $N_\si$ and hence $W \cap \pi^{-1} (W) \neq \es$. Let $z \in W \cap \pi^{-1} (W)$. Then $z , \pi(z) \in W$. Note that we may assume that $z$ (and hence $\pi (z)$) are irrational. This follows from the fact that the rationals are meager in $N_\si$. We may now take $x_j, y_j \in B \cap I_j$ such that \[ z = q_1 x_1 + \ldots + q_n x_n \] \[ \pi (z) = q_1 y_1 + \ldots + q_n y_n \] Thus \[ z + \pi (z) = q_1 (x_1 + y_1) + \ldots + q_n (x_n + y_n) \] By the definition of $\pi$, $z + \pi (z) = \hat \si \in \qq $. Note that since the $I_j$ are disjoint, for each $j$ and $i \neq j$ $x_j \neq x_i, y_i$. Further, because $z \notin \qq $, $z \neq \pi(z)$, we have that for at least one $j$, $x_j \neq y_j$. We have therefore expressed a rational number (namely $\hat \si$) as a sum of $n + 1$ distinct elements of the basis $B$. On the other hand, $1 \in B$ and any rational can be expressed as a rational scalar multiple of 1, i.e. a linear combination of length 1. By independence, such linear combinations are unique and so the above leads to a contradiction. \end{proof} \bigskip We conclude with some further notes. \bigskip \begin{thm}\label{seven} For all $\al > 2$ there exists a $\qq$--subspace $A$ of $\rr$ which is $\mathbf \Sigma_\al^0$, but not $\mathbf \Pi_\al^0$. \end{thm} \begin{proof} Let $C \su \rr$ be a perfect, linearly independent set. Choose $B \su C$ which is $\mathbf \Sigma_\al^0$, but not $\mathbf \Pi_\al^0$. Take $A$ to be the linear span of $B$. First of all, $A$ is not $\mathbf \Pi_\al^0$. To see this, observe that, by the independence of $C$, $A\cap C = B$. If $A$ were $\mathbf \Pi_\al^0$, then $B$ would be as well. Secondly, $A$ is $\mathbf \Sigma_\al^0$. Observe that $A$ is the union of sets of the form. \[ q_1(B \cap I_1) + \ldots + q_n(B \cap I_n) \] Where the $q_i$ are nonzero rational numbers and the $I_i$ are disjoint intervals with rational endpoints. We can define a homeomorphism \[ \prod_{i = 1}^n (C \cap I_i) \rightarrow q_i (C \cap I_i) + \ldots + (C \cap I_n) \] by $\langle x_1, \ldots , x_n \rangle \mapsto q_1x_1 + \ldots + q_nx_n$. Under this map, $\prod_{i = 1}^n (B \cap I_i)$ maps onto $q_1(B \cap I_1) + \ldots + q_n(B \cap I_n)$. Hence this latter set of is of the same Borel class as $B$, namely $\mathbf \Sigma_\al^0$. Since the union above is countable, $A$ is also $\mathbf \Sigma_\al^0$. \end{proof} Theorem \ref{seven} is also a consequence of Theorem 2.5 of Farah and Solecki \cite{farah} but has a shorter proof. \begin{thm} For all $\alpha > 3$, there exists a $\mathbb Q$-subspace $W$ of $\mathbb R$ which is $\mathbf \Pi^0_\alpha$ and not $\mathbf \Sigma^0_\alpha$. \end{thm} \begin{proof} Let $C \subset \mathbb R$ be a perfect, independent set over $\mathbb Q$. Let $A_0 \supset A_1 \supset \ldots$ be subsets of $C$ which are $\mathbf \Sigma^0_{<\alpha}$ and such that $A = \bigcap_{n \in \omega} A_n$ is $\mathbf \Pi^0_\alpha \setminus \mathbf \Sigma^0_\alpha$. Let $W_n = \mathrm{span}_{\mathbb Q}(A_n)$ and $W = \bigcap_{n \in \omega} W_n$. Then each $W_n$ is $\mathbf \Sigma^0_{<\alpha}$ as in the proof of Theorem \ref{seven}. Thus $W$ is $\mathbf \Pi^0_\alpha$, but not $\mathbf \Sigma^0_\alpha$. If $W$ were $\mathbf \Sigma^0_\alpha$, then $A = W \cap C$ would be as well. \end{proof} \begin{thebibliography}{99} \bibitem{erdos} Erd\"{o}s, P.; Some remarks on set theory. Proc. Amer. Math. Soc. 1, (1950). 127-141. \bibitem{farah} Farah, Ilijas; Solecki, S{\l}awomir; Borel subgroups of Polish groups. Adv. Math. 199 (2006), no. 2, 499-541. \end{thebibliography} \end{document}