\input fontmac \input mathmac \input epsf \input color \input soul.sty \def\norm#1{|\!|#1|\!|} \def\threenorm#1{|\!|\!|#1|\!|\!|} \def\bignorm#1{\big|\!\big|#1\big|\!\big|} \def\Norm#1{\Big|\!\Big|#1\Big|\!\Big|} \def\normm#1{\bigg|\!\bigg|#1\bigg|\!\bigg|} \def\B{{\cal B}} \def\supp{\op{supp}} \widemargins \bookheader{TSIRELSON'S SPACE}{MARCEL K. GOH} \maketitle{Tsirelson's space and other exotic constructions}{}{Marcel K. Goh}{16 December 2021} \floattext4.5 \ninebf Abstract. \ninepoint This set of expository notes was written as a final report for the class MATH 567 Introduction to Functional Analysis, taught by Prof.~Gantumur Tsogtgerel at McGill University in the Fall 2021 semester. It builds up the necessary theory to describe Tsirelson's example of an infinite-dimensional Banach space with no subspace isomorphic to $c_0$ or $l_p$ for $1\le p<\infty$, and modifications thereof. \advsect Preliminary notions In this section we state definitions and well-known results that we shall require later on. Proofs are not given for many of the facts given here; they can be found in introductory textbooks on linear analysis (see, e.g.,~\ref{bollobas},~\ref{lindenstrauss},~\ref{schechter}). The reader comfortable with the terminology of Banach space theory may wish to skip this section, returning only upon finding an unfamiliar concept or definition. Let $K$ denote either the real or complex field and let $V$ be a vector space over $K$. A {\it norm} is a function $\norm{\cdot}:V\to\RR$ satisfying \medskip \item{i)} $\norm v \ge 0$ for all $v\in V$ and $\norm v = 0$ if and only if $v = 0$; \smallskip \item{ii)} $\norm{\alpha \cdot v} = |\alpha|\cdot\norm v$ for all $\alpha\in K$ and $v\in V$; and \smallskip \item{iii)} $\norm{v+w}\le \norm v + \norm w$ for all $v,w\in V$. \medskip A vector space equipped with such a function is called a {\it normed vector space}, and one can define a metric $d$ on the space by setting $d(x,y) = \norm{x-y}$. Thus one has a notion of Cauchy sequences in these spaces, and if a normed vector space $X$ is complete (every Cauchy sequence in $X$ has a limit in $X$), we say that $X$ is a {\it Banach space}. Any two norms on a finite-dimensional normed vector space are equivalent, in that they induce the same topology. This fact can be used to show that any finite-dimensional normed vector space over $\RR$ or $\CC$ is a Banach space, since these fields are complete. The field $\RR$ or $\CC$ under the norm given by the absolute value $|\cdot|$ is of course the simplest Banach space. Other basic examples include the following. \medskip \item{i)} Consider an $n$-dimensional vector space over $K$. Writing an element $x$ of this space as $x = (x_1, \ldots, x_n)$, this vector space can be endowed with the {\it $p$-norm} $$\norm{x}_p = \Bigl(\sum_{i=1}^n |x_i|^p\Bigr)^{1/p}.$$ for $1\le p<\infty$. This defines a Banach space which we shall denote by $l_p^n$. As $p\to\infty$, this approaches the {\it maximum norm} $$\norm{x}_\infty = \max_{1\le i\le n} |x_i|,$$ and we denote this Banach space by $l_\infty^n$. \smallskip \item{ii)} For $1\le p< \infty$, the $p$-norm for infinite sequences $x = (x_n)_{n=1}^\infty$ is given by $$\norm{x}_{l_p} = \Bigl(\sum_{i=1}^\infty |x_i|^p \Bigr)^{1/p}.$$ As $p\to\infty$, this approaches the {\it supremum norm} $$\norm{x}_{l_\infty} = \sup_{i\ge 1} |x_i|.$$ For $1\le p\le \infty$, the set $l_p$ of all infinite sequences $x$ with $\norm{x}_{l_p}$ finite is a Banach space; when $p=\infty$, these are the bounded sequences. \smallskip \item{iii)} Let $c_0$ be the normed vector subspace of $l_\infty$ consisting of sequences that tend to zero. It can be shown that this subspace is closed and thus itself a Banach space. \medskip The latter two examples are infinite-dimensional, meaning that in these cases one can exhibit an infinite sequence of vectors that are all linearly independent. \medskip\boldlabel Hamel bases. Assuming the axiom of choice, one can find a basis for any vector space. This is a set of elements $\{e_i\}_{i\in I}$ such that any $x\in X$ can be expressed uniquely as $\sum_{i\in F} a_ie_i$ for a finite set $F\subseteq I$ and scalars $a_i$, both depending on $x$. When dealing with infinite-dimensional spaces, such a basis is called a {\it Hamel basis} and are not awfully useful. To see why, we first need to establish (a variant of) the Baire category theorem, as well as simple lemma. \parenproclaim Theorem B (Baire category theorem). Let $X$ be a metric space that equals the union of countably many closed sets. If $X$ is complete, then at least one of the closed sets has nonempty interior.\slug \proclaim Lemma \advthm. Let $X$ be a normed vector space and let $Y$ be a subspace of $X$. If $Y$ has nonempty interior, then $Y=X$. \proof Suppose that for some $r>0$ and $y\in Y$, the ball $B_r(y) = \{x\in X : \norm{x-y} < r\}$ is contained in $Y$. Now let $x\in X$ and note that the vector $$z = y + {r\over 2}\cdot{x\over\norm x}$$ is at distance $r/2$ from $y$, meaning that $z\in B_r(y)\subseteq Y$. Since $Y$ is a subspace, we find that $$x = {2\norm x\over r}(z-y)$$ is in $Y$, completing the proof.\slug We now show that Hamel bases of Banach spaces cannot be countably infinite. \proclaim Lemma \advthm. Let $X$ be an infinite-dimensional normed vector space. If $X$ is also complete (and thus a Banach space), then any Hamel basis of $X$ is uncountable. \proof Suppose that $\{e_1\}_{i=1}^\infty$ is a countable Hamel basis for $X$. For integers $n\ge 1$, let $X_n$ denote the linear span of $\{e_1,\ldots,e_n\}$. Each normed vector space $X_n$ is finite-dimensional and thus complete, which implies that each $X_n$ is closed in $X$. But each of these subspaces is proper, so by the lemma above, so all of these subspaces have empty interior, which contradicts the Baire category theorem.\slug Since many infinite-dimensional Banach spaces have the cardinality of the continuum to begin with, the above lemma tells us that Hamel bases can essentially be as complicated as the underlying space. (In fact, it has been shown that the cardinality of the Hamel basis of a Banach space always equals the cardinality of the space itself~\ref{halbeisen2000}.) \medskip\boldlabel Completions. Consider now the infinite-dimensional vector space $c_{00}$ of sequences that are eventually zero. For all integers $i\ge 1$, let $e_i$ denote the sequence that is $1$ at index $i$ and $0$ elsewhere. The set $\{e_i\}_{i=1}^\infty$ is countable and clearly a Hamel basis for $c_{00}$, so we conclude that $c_{00}$, under any norm, cannot be a Banach space. However, for every normed vector space $V$, we can always find a Banach space $X$ such that $V$ is dense in $X$. We do this by letting $X$ be the completion of $V$ as a metric space and then defining scalar multiplication and addition as follows. If $x,y\in X$ are such that $x_n\to y$ and $y_n\to y$, then $\lambda x+\mu y$ shall simply be defined as $\lim_{n\to \infty} (\lambda x_n + \mu y_n)$. We leave it as an easy exercise for the reader to check that this makes $X$ a normed vector space under the norm $\norm x = \lim_{n\to\infty}\norm{x_n}$, and that $X$ is complete with respect to this norm. \medskip\boldlabel Dual spaces and the weak topology. Let $X$ be a normed vector space. A linear operator from $X$ to its base field is called a {\it linear functional}, and is continuous if and only if it is bounded. The set of bounded linear functionals on a normed vector space $X$ is a Banach space under the operator norm $$\norm{T} = \sup_{x\in X} {\norm {Tx}\over \norm{x}}.$$ This space is called the {\it dual space of $X$} and is denoted $X^*$. If $(x_n)$ is a sequence of vectors in $X$ such that for some $x\in X$, $x^*(x_n)\to x^*(x)$ for all $x^*\in X^*$, then we say that $(x_n)$ {\it converges weakly to $x$}. If $(x_n)$ converges weakly to $0$, then we call $(x_n)$ {\it weakly null}. The topology that this definition of convergence induces on $X$ is called the {\it weak topology}; it is the coarsest topology such that every element of $X^*$ is still a continuous function. % When distinguishing the topology given by the norm from the % weak one, we shall call the norm topology the {\it strong topology}. % Since $x_n\to x$ implies that $x_n$ converges weakly % to $x$, the weak topology is, in general, strictly coarser than the strong topology. However, there are % many weak-to-strong'' principles in the theory of Banach spaces, the most celebrated of which is the following. % % \parenproclaim Theorem U (Uniform boundedness principle). Let $X$ be a Banach space, $Y$ a normed vector space, % and ${\cal T}$ a collection of linear operators from $X$ to $Y$. If % $$\sup_{T\in{\cal T}} \bignorm{T(x)}_Y < \infty,$$ % then % $$\sup_{T\in {\cal T}} \sup_{x\in X} {\bignorm{T(x)}_Y\over \norm{x}_X} <\infty.\noskipslug$$ % % The uniform boundedness principle can be used to show that any set of vectors % that is bounded in the weak topology is also bounded in the strong topology. % This fact will come in handy in a proof later on. \advsect Schauder bases We saw in the previous section that Hamel bases are not very useful for performing analysis on infinite-dimensional normed vector spaces. A better notion in the infinite-dimensional setting is that of a {\it Schauder basis}. This is a countable sequence of vectors $(e_i)_{i=1}^\infty$ such that every $x\in X$ has a unique representation $$x = \sum_{i=1}^\infty a_ie_i$$ for some sequence $(a_i)_{i=1}^\infty$ of scalars, where convergence of the infinite sum is defined in terms of the metric induced by the norm. From here on out, we shall sometimes write basis'' to mean Schauder basis'', for brevity's sake. A basis $(e_i)$ for which there exists $C$ such that $1/C \le \norm{e_i} \le C$ for all $i$ is called {\it seminormalised} and a seminormalised basis whose corresponding constant $C$ equals $1$ is said to be {\it normalised}. A sequence of vectors $(x_i)$ that is a basis for the closure of its linear span is called a {\it basic sequence}. \medskip\boldlabel Equivalent bases. A basis $(x_i)$ for a Banach space $X$ and a basis $(y_i)$ for a Banach space $Y$ are said to be {\it equivalent} if the map $T$ sending $x_i$ to $y_i$ extends to a linear homeomorphism between $X$ and $Y$. Of course, this relation is reflexive, symmetric, and transitive. One can also show this to be the same as saying that there are constants $00$ such that $f(x_i) \ge \delta$ for all $i$. (If not, then $(x_i)$ would have a subsequence that is weakly null, and thus by subsymmetry the sequence itself is weakly null.) By scaling $f$ we can assume that it has operator norm equal to $1$. For any $n$, and scalars $(b_1,\ldots,b_n)$ we have $$\Norm{\sum_{i=1}^n |b_i| x_i} =\norm f\cdot \Norm{\sum_{i=1}^n |b_i| x_i} \ge f\Bigl(\sum_{i=1}^n |b_i|x_i\Bigr) = \sum_{i=1}^n |b_i|f(x_i)\ge \delta \sum_{i=1}^n |b_i|.$$ Now let $C$ be the constant given by the unconditionality of $(x_i)$ and setting $a_i = |b_i|$, we have $$\Norm{\sum_{i=1}^n |b_i|x_i} \le C \cdot\Norm{\sum_{i=1}^n b_ix_i}.$$ Putting everything together establishes $$\Norm{\sum_{i=1}^n b_ix_i} \ge {1\over C} \Norm{\sum_{i=1}^n |b_i|x_i}\ge {\delta\over C} \sum_{i=1}^n |b_i|$$ for all $n$, and taking $n$ to infinity gives a corresponding lower bound on all infinite sequences $(b_i)$. The upper bound is a simple consequence of the triangle inequality. Let $M$ be the constant bounding $(x_i)$ from above. Then since $\norm{x_i}\le M$ for all $i$, for an arbitrary infinite sequence $(b_i)$ of scalars we have $$\Norm{\sum_{i=1}^\infty b_ix_i}\le \sum_{i=1}^\infty |b_i|\cdot\norm{x_i}\le M \sum_{i=1}^\infty |b_i|,$$ which combines with the lower bound to give $${\delta\over C} \sum_{i=1}^\infty |b_i| \le \Norm{\sum_{i=1}^\infty b_ix_i} \le M \sum_{i=1}^\infty |b_i|.$$ Hence $(x_i)$ is equivalent to the unit vector basis of $l_1$. \slug \advsect Tsirelson's space and its dual For many decades, the infinite-dimensional Banach spaces known to functional analysts all contained a subspace linearly homeomorphic to $c_0$ or $l_p$ for $1\le p<\infty$. The first infinite-dimensional space without such a subspace was constructed by B.~S.~Tsirelson in 1974~\ref{tsirelson}, and that same year T.~Figiel and W.~B.~Johnson gave a more explicit characterisation of its dual~\ref{figieljohnson}. In time, the dual of the original space has come to be known as $T$, with Tsirelson's original space denoted $T^*$. Properties of Tsirelson's space and variations thereof are collected in a book by P.~G.~Casazza and T.~J.~Shura~\ref{casazzashura}. We shall describe the space $T$ of Figiel and Johnson in this section, following Casazza and Shura's exposition but supplying some of the proofs that are omitted there. Let $E$ and $E'$ be subsets of the positive integers. We write $E0$ there is $n\ge 1$ such that $\norm{x}< \norm{x}_n + \eps$ and $\norm{x}_n > \norm{x}_{n-1}$. This means there is some $k'$ and $E_1,\ldots,E_{k'}$ such that $$\norm {x}_n = {1\over 2} \sum_{i=1}^{k'} \norm{E_i x}_{n-1}.$$ It follows that \eqalign{ \norm x - \eps &< \norm{x}_n = {1\over 2} \sum_{i=1}^{k'} \norm{E_i x}_{n-1} \le {1\over 2} \sum_{i=1}^{k'} \norm{E_i x} \le \sup{1\over 2} \sum_{i=1}^{k} \norm{E_i x}. } For an upper bound on the supremum, we pick $k''$ and an admissible sequence $E_1,\ldots, E_{k''}$ such that $$\sup{1\over 2} \sum_{i=1}^{k} \norm{E_i x} < {1\over 2} \sum_{i=1}^{k''} \norm{E_i x} + \eps.$$ We then pick $n'$ such that $$\sup{n\ge 1}{1\over 2} \sum_{i=1}^{k''} \norm{E_i x}_n < {1\over 2} \sum_{i=1}^{k''} \norm{E_i x}_{n'} + \eps.$$ This yields the upper bound \eqalign{ {1\over 2} \sum_{i=1}^{k} \norm{E_i x} &< {1\over 2} \sum_{i=1}^{k''} \norm{E_i x} + \eps \cr &= {1\over 2} \sum_{i=1}^{k''} \sup_{n\ge 1} \norm{E_i x}_n + \eps \cr &= \sup_{n\ge 1} {1\over 2} \sum_{i=1}^{k''} \norm{E_i x}_n + \eps \cr &< {1\over 2} \sum_{i=1}^{k''} \norm{E_i x}_{n'} + 2\eps \cr &\le \norm{x}_{n'+1} + 2\eps\cr &\le \norm{x} + 2\eps.\cr } We have shown that $$\norm{x}-\eps < \sup{1\over 2} \sum_{i=1}^{k} \norm{E_i x} < \norm{x} + 2\eps,$$ which, since $\eps$ was arbitrary, gives us equality.\slug This recursive identity is the only definition one really uses when performing any serious computations with Tsirelson's norm. In particular, it asserts that we have, for any $x\in T$ and admissible sequence $E_1,\ldots,E_k$, $$\sum_{i=1}^k \norm{E_i x} \le 2 \norm x.$$ As a demonstration of the utility of the identity, consider the proof of the following proposition, which gives us an idea of how Tsirelson's norm behaves on subsets of a seminormalised block basis. \newcount\tsirelsonblock \tsirelsonblock=\thmcount \proclaim Proposition \advthm. Let $(t_i)$ denote the usual unit vector basis of Tsirelson's space and let $k$ be a positive integer and let $M\ge 1$. For any $k$ blocks $(y_j)_{j=1}^k$, where each $y_j$ satisfies $1/M \le \norm{y_j}\le M$ and is of the form $$y_j = \sum_{i=n_j+1}^{n_{j+1}} a_i t_i$$ for some scalars $a_i$ and $k-1\le n_10$, there is a normalised block basic sequence $(u_j)_{j=0}^\infty$ such that $$(1-\delta)\sum_{j=0} |a_j| \le \Norm{\sum_{j=0}^\infty a_ju_j} \le \sum_{j=0}^\infty |a_j|$$ for every sequence $(a_j)_{j=0}^\infty$ of scalars such that at least one $a_j$ is nonzero. \proof The statement holds trivially when $\delta\ge 1$, so let $0<\delta<1$. The subspace $B$ isomorphic to $l_1$ contains a sequence $(x_i)_{i=0}^\infty$ for which there exist constants $m$ and $M$ such that $$m\sum_{i=0}^\infty |a_i| \le \Norm{\sum_{i=0}^\infty a_i x_i} \le M\sum_{i=0}^\infty |a_i|$$ for all sequences $(a_i)_{i=0}^\infty$ of scalars. For $n\ge 1$, let $$K_n = \inf \biggl\{ \Norm{ \sum_{i=n}^\infty a_ix_i } : \sum_{i=0}^\infty |a_i| = 1\biggr\}$$ and set $K$ to $\lim_{n\to \infty} K_n$. We of course have $m\le K\le M$. Let $\phi >1$ and $0<\theta<1$ be such that $1-\delta \le \theta/\phi$. Choose $n_0 \{n_0\}$, so by the definition of $K_{n_0+1}$, $$\Norm{\sum_{j=0}^\infty a_jy_j} = \Norm{\sum_{j=n_0+1}^\infty a_jy_j} \ge K_{n_0+1} \sum_{j=0}^\infty |a_j| \ge \theta K \sum_{j=0}^\infty |a_j|.$$ Then since $\norm{y_j} \le \phi K$ for all $K$ and $\theta/\phi \ge 1-\delta$, we have $$\Norm{\sum_{j=0}^\infty a_j u_j} = \normm{\sum_{j=0}^\infty a_j {y_j\over \norm{y_j}}} \ge {1\over \phi K} \Norm{\sum_{j=0}^\infty a_j y_j} \ge (1-\delta)\sum_{j=0}^\infty |a_j|,$$ which proves the lower bound.\slug Note that in James' original statement of the lemma, the $u_j$ are just vectors contained in the closed unit ball. However, the construction in his proof actually produces a block basic sequence, and we normalised it in our own version of the proof above. We are now set to show that $l_1$ does not embed into $T$. \newcount\nolone \nolone=\thmcount \proclaim Theorem \advthm. Tsirelson's space $T$ does not contain a subspace isomorphic to $l_1$. \proof Suppose, towards a contradiction, that $T$ did contain a subspace isomorphic to $l_1$. Then, applying Lemma~J with $\delta = 1/9$, there exists a normalised block basic sequence $(y_j)_{j=0}^\infty$ such that for all sequences $(a_j)_{j=0}^\infty$ of scalars, $${8\over 9} \sum_{j=0}^\infty |a_j| \le \Norm{\sum_{j=0}^\infty a_jy_j} \le\sum_{j=0}^\infty |a_j|.$$ In particular, if we let $r\ge 1$ be an integer and let $(a_j)_{j=0}^\infty$ be the sequence with $$a_j = \cases{ 1,& if j=0;\cr 1/r,& if 1\le j\le r;\cr 0,& if j>r,}$$ then we have $\sum_{j=0}^\infty |a_j| = 2$. This yields $$\Norm{ y_0 + {1\over r}\sum_{j=1}^r y_j} = \Norm{\sum_{j=0}^\infty a_jy_j} \ge {8\over 9}\sum_{j=0}^\infty |a_j| = {16\over 9}$$ for all integers $r\ge 1$. Now let $\{k\}\le E_1 < E_2 < \cdots < E_k$ be an arbitrary admissible sequence of sets, and let $r_0 =\max\supp y_0$. If $k > r_0$, then $$\sum_{i=1}^k \normm{ E_i \Bigl( y_0 + {1\over r}\sum_{j=1}^r y_j\Bigr)} =\sum_{i=1}^k \normm{ E_i \Bigl({1\over r}\sum_{j=1}^r y_j\Bigr)} \le 2\cdot\Norm{{1\over r}\sum_{j=1}^r y_j} \le 2.$$ If $k\le r_0$, then we set $$S = \bigl\{1\le j\le r : \norm{E_i y_j}\ne 0\ \hbox{for at least two values of}\ i\bigr\}$$ and $$T = \bigl\{1\le j\le r: \norm{E_i y_j}\ne 0\ \hbox{for at most one value of}\ i\bigr\}.$$ Note that if $j\in S$, then $y_j$ straddles the border between $E_i$ and $E_{i+1}$ for some $i$, and no other $y_{j'}$ can do so, since the supports of the $y_j$ are disjoint. So the cardinality of $S$ is at most $k-1$. It is also clear that $S$ and $T$ are disjoint and $|S|+|T| = r$, so $$2|S| + |T| \le 2(k-1) + r- k+1 = r+k-1.$$ Since the $y_j$ are unit vectors, we have \eqalign{ \sum_{i=1}^k \normm{E_i \Bigl({1\over r}\sum_{j=1}^\infty y_i\Bigr)} &= \sum_{j\in S} \sum_{i=1}^k \norm{E_iy_j} + \sum_{j\in T} \sum_{i=1}^k \norm{E_iy_j}\cr &\le \sum_{j\in S} 2\cdot \norm{y_j} + \sum_{j\in T} \norm{y_j},\cr &\le r+k-1\cr } by rearranging sums. Dividing by $r$ and adding the unit vector $y_0$ into the mix, we have \eqalign{ \sum_{i=1}^k \normm{ E_i \Bigl( y_0 + {1\over r} \sum_{j=1}^r y_j\Bigr)} &\le\sum_{i=1}^k \norm{E_i y_0} + {1\over r}\sum_{i=1}^k\normm{E_i\Bigl({1\over r}\sum_{j=1}^\infty y_i\Bigr)}\cr &\le 2 + {r+k-1\over r} \cr &\le 3 + {n_0\over r}.\cr } Selecting some $r\ge 2n_0$, the right-hand side is at most $7/2$, meaning that $$\Norm{y_0 + {1\over r} \sum_{j=1}^r y_j} \le 7/4,$$ and we have $7/4 \ge 16/9$, the contradiction we sought.\slug In the proof that $c_0$ and the other $l_p$ spaces do not embed into $T$ either, we shall make use of the following theorem, a version of what is called the Bessaga--Pe\l czy\'nski Selection Principle (see~\ref{bessagapelczynski}, Theorem 3). \parenproclaim Theorem S (Bessaga--Pe\l czy\'nski, {\rm 1958}). Let $X$ be a Banach space with basis $(e_i)$ and let $(e_i^*)$ be the coefficient functionals given by $$e_j^*\Big(\sum_{i=1}^\infty a_i e_i\Bigr) = a_j.$$ If $(y_i)_{i=1}^\infty$ is a sequence of vectors in $X$ such that \medskip \item{i)} $\lim_{i\to\infty} \norm{y_i} > 0$; and \smallskip \item{ii)} $\lim_{i\to\infty} e_j^*(y_i) = 0$ for all indices $j\ge 1$, \medskip\noindent then there exists a subsequence of $(y_i)$ which is equivalent to a block basis with respect to the original basis $(e_i)$.\slug To prove that $c_0$ and $l_p$ do not embed into $T$, we shall actually prove the following stronger result. \proclaim Theorem \advthm. Tsirelson's space $T$ does not contain a seminormalised subsymmetric basic sequence. \proof Towards a contradiction, suppose that $(y_i)$ is a seminormalised subsymmetric basic sequence in $T$. Since the sequence is seminormalised, it is bounded from above, so by Lemma~{\the\dichotomy}, either $(y_i)$ is equivalent to the canonical basis for $l_1$ or $(y_i)$ is weakly null. We already proved Theorem~{\the\nolone} above, which deems the first scenario impossible, so $(y_i)$ must be weakly null and thus in particular satisfies condition (ii) of Theorem S. In addition, since $(y_i)$ is seminormalised, it is uniformly bounded away from zero, so we are in the position to conclude from Theorem S that there is a subsequence of $(y_i)$ which is equivalent to a block basic sequence $(x_i)$ against the unit vector basis $(t_i)$ of $T$. We shall now pass to subsequences both in $(x_i)$ and in $(y_i)$. First, pass to a subsequence and reindex $(x_i)$ to ensure that $\supp x_i > \{i\}$ for all $i$. Now we pass to a subsequence of $(y_i)$ that is equivalent to this new $x_i$ and reindex; this is equivalent to the original sequence by subsymmetry. Since $y_i$ is seminormalised and there is a linear homeomorphism mapping each $y_i\mapsto x_i$, $(x_i)$ is seminormalised as well; that is, there is $M\ge 1$ such that $1/M \le \norm{x_i}\le M$. Now since $\supp x_i > \{i\}$ for all $i$, we can apply Proposition~{\the\tsirelsonblock} with the constant $M$ on any $n$ blocks $x_{i_1}, x_{i_2}, \ldots, x_{i_n}$ of $(x_i)$ to find that for all $n$-tuples $(b_1,\ldots, b_n)$ of scalars, $${1\over 2M}\sum_{k=1}^n |b_k| \le \Norm{\sum_{k=1}^n b_kx_{i_k}} \le M\sum_{k=1}^n |b_k|.$$ We thus find that for any $n$, $(x_i)$ (and hence $(y_i)$) contains a subsequence of length $n$ which is equivalent to the standard basis of $l_1^n$. Complete each of these finite sequences to infinite subsequences. Recall that $(y_i)$ is equivalent to all of them, so for all $n$, the first $n$ elements of $(y_i)$ are equivalent to the standard basis of $l_1^n$. Thus $(y_i)$ is equivalent to the standard basis of $l_1$ and we know this cannot happen.\slug \proclaim Corollary \advthm. Tsirelson's space $T$ does not contain $c_0$ or $l_p$ for $1 1$, we say that $X$ is {\it $\lambda$-distortable} if there exists an equivalent norm $\threenorm\cdot$ on $X$ such that for each infinite-dimensional subspace $Y$ of $X$, we have $$\sup\biggl\{{\threenorm{y_1}\over\threenorm{y_2}}:y_1,y_2\in Y,\,\norm{y_1} = \norm{y_2} = 1\biggr\}\ge\lambda.$$ The norm $\threenorm\cdot$ is called a {\it $\lambda$-distortion} if this is true. The space $X$ is called {\it distortable} if it is $\lambda$-distortable for some $\lambda>1$, and {\it arbitrarily distortable} if it is distortable for every $\lambda>1$. In 1991, T.~Schlumprecht proved the existence of an arbitrarily distortable Banach space~\ref{schlumprecht1991}. Schlumprecht's construction relies on a function $f : [1,\infty)\to [1,\infty)$ satisfying \medskip \item{i)} $f(1) = 1$ and $f(x)1$; \smallskip \item{ii)} $f$ is strictly increasing to infinity; \smallskip \item{iii)} $lim_{x\to\infty} f(x)/x^q = 0$ for all $q>0$; \smallskip \item{iv)} the function $g(x) = x/f(x)$ is concave for $x\ge 1$; and \smallskip \item{v)} $f(x)f(y) \ge f(xy)$ for all $x,y\ge 1$. \medskip Of course, we need such a function to actually exist, but it is easily seen that $f(x) = \log_2(x+1)$ does. We then define a sequence $\threenorm{\cdot}_n$ of norms on $c_{00}$ by setting $\threenorm{x}_0 = \norm{x}_{l_\infty}$ and for $n\ge 1$ we put $$\threenorm{x}_{n} = \max_{k\ge 2} \max {1\over f(k)} \sum_{i=1}^k \threenorm{E_i x}_{n-1},$$ where the inner maximum is over all subsets $E_10$, and every infinite-dimensional subspace $Y$ of $X$, there are unit vectors $y_1, y_2\in Y$ such that $$\norm{y_1}_k \ge 1-\eps\qquad\hbox{\rm and}\qquad \norm{y_2}_k \le {1+\eps\over f(k)}.$$ In particular, this means that $\norm\cdot _k$ is an $f(k)$-distortion on $X$.\slug \medbreak \boldlabel The unconditional basic sequence problem. In Section~2 we defined what it means for a basis or a basic sequence to be unconditional. Both Tsirelson's space and Schlumprecht's space have unconditional bases, and so do all classical Banach spaces. It was unknown for many years whether every infinite-dimensional Banach space contains an unconditional basic sequence. Soon after Schlumprecht presented his example in the summer of 1991, W.~T.~Gowers and B.~Maurey independently constructed Banach spaces without an unconditional basic sequences. They subsequently discovered that both examples were fundamentally the same, and ended up publishing jointly~\ref{gowersmaurey1993}. Recall that we say that $X$ is the {\it direct sum} of $Y$ and $Z$ and write $X = Y\oplus Z$ if the map $Y\times Z\to X$ sending $(y,z)\mapsto y+z$ is a linear homeomorphism. An infinite-dimensional Banach space $X$ is said to be {\it decomposable} if $X$ can be written as $X = Y\oplus Z$ where both $Y$ and $Z$ are also infinite-diimensional. An infinite-dimensional Banach space $X$ without any decomposable subspaces is called {\it hereditarily indecomposable}. If a space $X$ has an unconditional basic sequence $(x_i)_{i=1}^\infty$, then letting $Y$ be the closed linear span of $(x_{2i-1})_{i=1}^\infty$ and $Z$ be the closed linear span of $(x_{2i})_{i=1}^\infty$, we see that $X$ is not hereditarily indecomposable. So to produce an infinite-dimensional space without an unconditional basic sequence, it suffices to produce a hereditarily indecomposable one, and that is what Gowers and Maurey did. Before describing their construction, we need to establish a fair bit of terminology and notation. Let $J$ be a set of positive integers such that if $m,n\in J$ with $m0$ such that $\norm{Tx} \ge c\norm x$ for all $x\in X$. Gowers and Maurey show that every operator from $X$ to $X$ can be written as a scalar multiple of the identity, plus a strictly singular operator. This implies that there are, in some sense, very few operators on $X$. A subsequent paper of Gowers and Maurey, published in 1997, describes a more general method of producing Banach spaces whose spaces of operators are small~\ref{gowersmaurey1997}. Recall that an operator is said to be {\it compact} if it sends bounded sets in the domain to sets with compact closure in the image. Every compact operator is strictly singular. In their original 1993 paper, the authors ask if there exists a Banach space $X$ such that every operator from $X$ to $X$ is a scalar multiple of the identity, plus a compact operator. This question was answered positively S.~A.~Argyros and R.~G.~Haydon in 2009; the paper was published in 2011~\ref{argyroshaydon}. \medskip\boldlabel The Banach hyperplane problem. In a book published in 1932, S.~Banach~\ref{banach} asked whether an infinite-dimensional Banach space $X$ over the real numbers is always isomorphic to $X\oplus \RR$. This amounts to asking whether every infinite-dimensional Banach space $X$ is such that that every subspace of codimension $1$ is isomorphic to $X$ itself. Since a subspace of codimension $1$ is sometimes called a {\it hyperplane}, this question came to be known as the hyperplane problem. It turns out that the answer to the question is no, and there is an infinite-dimensional Banach space that is not isomorphic to any of its hyperplanes. This problem was solved by W.~T.~Gowers soon after he solved the unconditional basic sequence problem. His construction, which appeared in print in 1994, is extremely similar~\ref{gowers1994}. We shall thus retain the definitions from the previous subsection and immediately define this space. As the reader probably suspects at this point, the construction is inductive. Set $X_0 = (c_{00}, \norm\cdot_0)$. For $n\ge 1$, we let $X_n = (c_{00}, \norm\cdot _n)$, where $\norm{x}_n$ be the maximum of $$\sup_{m\in \NN} \sup {1\over f(m)} \sum_{i=1}^m \norm{E_i x}_{n-1},$$ where the inner supremum is over all sequences of subsets $E_1 < \cdots < E_m$, and $$\sup_{k\in K} \sup_{g\in B^*_k(X_{N-1})} \bigl|g(x)\bigr|.$$ The notation we use here is not exactly the same as in the original paper, but using the same letters as before makes it clear that the only change from the space without an unconditional basic sequence is the replacement of $\sup_{E\subseteq \NN} \bigl|g(Ex)\bigr|$ has changed to the simpler $\bigl|g(x)\bigr|$. The space $X$ is the completion of $c_{00}$ with respect to the limit of the norms $\norm{x}_n$ above. Perhaps surprisingly, this new space $X$ has an unconditional basis. It is noted in Gowers' paper that the Gowers--Maurey space of the previous section is also a counterexample to the hyperplane problem, though is is more difficult to prove. The proof that $X$ is not isomorphic to any of its hyperplanes is done by showing that $X$ satisfies the hypothesis of the following lemma, which is attributed to P.~G.~Casazza. \proclaim Lemma C. If $X$ is a Banach space in which no two equivalent sequences $(y_i)$ and $(z_i)$ satisfy $y_i < z_i < y_{i+1}$ for all $i$, then $X$ is not isomorphic to any proper subspace.\slug \section References \bye