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Given $X$, $Y$ two real Banach spaces, let's say that $(X,\ Y)$ is a Borsuk pair if for any continuous mapping $T$ : {$x$ $\in$ $X$ ; $||x||\leq1$} $\rightarrow$ $Y$ s.t. $T$ is odd on {$x$ $\in$ $X$ ; $||x||=1$}, it follows that the set $T^{-1}$( { 0 } ) is nonempty.

My conjecture is : $(X,\ Y)$ Borsuk pair $\Longleftrightarrow$ dim $Y$ < $\infty$ and dim $Y$ $\leq$ dim $X$. [Here "dim" stands for the algebraic dimension; it may be a natural number or $\infty$.]

In other words, there is no Borsuk pair with $X$, $Y$ both infinite-dimensional.

Any thoughts ?

[There exist several infinite-dimensional versions of the Classical Borsuk Theorem, e.g., when $X$ is reflexive, $Y$ is its dual, and $T$ is a demicontinuous mapping of monotone type.]

[I think the "big" problem (if it truly exists, of course ;-)) would be when both $X$, $Y$ are infinite dimensional, and the density character dens($X$) $\gg$ dens($Y$), i.e., when $X$ is "huge".]

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I have thought about a related problem, Ady. Suppose that $X$ is a separable subspace of $Y$ and the density character of $Y$ is large. Must there exist a unit vector $y$ in $Y$ whose distance to $X$ is one? – Bill Johnson Feb 2 '10 at 1:44
This is a big one. I'll think on it. Thank you, Professor Johnson. – Ady Feb 2 '10 at 23:45
up vote 13 down vote accepted

Without loss of generality you can think that $Y=\ell^1$ (just pick an infinite sequence of linearly independent unit vectors in $Y$ and multiply them by the coefficients decaying so fast that no non-trivial $\ell^1$ combination will have any chance to be $0$). Now, construct a locally finite cover of $X$ with opposite points identified that has small diameters of the covering sets (every metric space is paracompact). Construct the corresponding continuous partition of unity and lift it back to $X$ to get symmetric continuous locally finite partition of unity. Decouple symmetric open sets consisting of 2 symmetric components. Now, for each open set in the new cover, choose a number so that the opposite open sets intersecting the sphere correspond to opposite numbers and all numbers are non-zero, but otherwise arbitrarily. Multiply the numbers by the partition of unity. Fix also any linear ordering of the symmetric pairs of neighborhoods. All we need now is a continuous mapping $F$ from the metric space of ordered finite sets of numbers with the metric that allows both finitely many small insertions and small perturbations (to find the distance between two sets, you can place their elements in the sequence in the given order adding any number of zeroes anywhere and take the minimum of the $\ell^1$ norm of the difference that can be attained in this way) to $\ell^1$ that is symmetric ($F(-x)=-F(x)$) and never $0$ on for $x\ne 0$. This is a great relief because the space $X$ we want to map is not huge anymore, it is just a fancy metric space of finite sets of numbers. The downside is that we need much more symmetry than before but still not too much ($x$ and $-x$ are still clearly distinguishable though can come arbitrarily close). Now we can raise the level of abstraction again and consider any separable metric space $X$ with an isometric involution $U$ without fixed points and try to map it to $\ell^1$ so that $F(Ux)=-F(x)$. This is trivial. Just create a countable symmetric (with respect to $U$) dense set, for each pair of symmetric points $x,Ux$ take the balls around $x$ and $Ux$ of radius $1/3$ times the distance between them, and construct an antisymmetric function that vanishes outside these 2 balls but not inside them (the plus-minus distance to the boundary, say). Multiply these functions by small numbers and put them in a sequence. You are done.

Of course, the main part of this proof is the paracompactness result. So, if you strongly dislike AC, then the question remains open (but how can you do abstract Banach spaces and dislike AC?)

OK, if you so strongly prefer formal language, here is a sequence of claims. For each claim, tell me if you agree with it, have a counterexample, would like to see the proof/reference, or just do not understand what I mean. I'll respond to your reaction.

Claim 1. We can take $Y=\ell^1$

Claim 2. If $Z$ is a separable metric space with an isometric involution $I$ without fixed points, then we can find a continuous maping $F$ from $Z$ to $\ell^1$ such that $F(Iz)=-F(z)$ for all $z\in Z$ and $F(z)\ne 0$ for all $z\in Z$.

Claim 3. Define the equivalence relation on the set of finite or infinite sequences of reals with finitely many non-zero terms as follows. Two sequences are equivalent if two finite sequences obtained by deleting all zero elements in the original sequences (so 0,1,0,0,2,0,3,0,0,0,...) becomes just (1,2,3)) coincide. The space $Z$ of equivalence classes of non-zero sequences with the distance defined as the infimum (actually, minimum) of the $\ell^1$ distances between class representatives is a separable metric space.

Claim 4. The mapping $I$ that maps the class of a sequence $z$ to the class of the sequence $-z$ is well defined and is an isometric involution on $Z$ without fixed points.

Claim 5. Every set can be linearly ordered.

Claim 6. There is a symmetric locally finite covering $\mathcal U$ of the unit ball in $X$ by open sets $U$ of diameter less than $1/10$ each (symmetric means that if $U$ is in the cover, then so is $-U$).

Claim 7. There exists a mapping $G:\mathcal U\to\{\pm 1\}$ such that $G(-U)=-G(U)$ when $U$ intersects the unit sphere.

Define the functions $f_U(x)=dist(x,X\setminus U)$.

Claim 8. For each point x at least one $f_U(x)\ne 0$ and one can find a neighborhood $V$ of $x$ and a finite subset $\mathcal V_x\subset\mathcal U$ such that $f_U$ is identically zero on $V$ for all $U\notin\mathcal V_x$.

Claim 9. There exists a linear order $L$ on $\mathcal U$ such that $U_1<U_2$ implies $(-U_1)<(-U_2)$ when both $U_1,U_2\in \mathcal U$ intersect the unit sphere in $X$ and have non-empty intersection.

Claim 10: The mapping $H$ from the unit ball in $X$ to $Z$ that maps every point $x$ to the equivalence class of the sequence $\{G(U)f_U(x)\}_{U\in V_x}$ arranged according to $L$ is well-defined, continuous, and satisfies $H(-x)=I(H(x))$ on the unit sphere.

Claim 11: The composition mapping $F\circ H$ is a continuous mapping from $X$ to $\ell_1$ that is antisymmetric on the unit sphere and the pre-image of zero under this mapping is empty.

Response round 1.

Proof of claim 2. Consider a countable dense set $S$ in $Z$. For each $y\in S$, define $r=r_y$ to be one third of the distance between $y$ and $Iy$.

Claim 2.1: The open ball $B(Iy,r)$ equals $IB(y,r)$ and these 2 balls are disjoint.

Claim 2.2: The function $f_y(z)$ defined as $r-d(z,y)$ on $B(y,r)$, as $d(z,Iy)-r)$ on $B(Iy,r)$ and 0 everywhere else is continuous and satisfies $f_y(Iz)=-f_y(z)$.

Claim 2.3: For each $z\in Z$ there exists $y\in S$ such that $f_y(s)\ne 0$.

Claim 2.4. There exists a bijection $\psi:\mathbb N\to S$ and a mapping $\eta:\mathbb N\to(0,+\infty)$ such that the sequence $\eta(n)r_{\psi(n)}$ belongs to $\ell^1$.

Claim 2.4. The mapping $F$ that maps $z$ to the sequence $\eta(n)f_{\psi(n)}(z)$ has the properties proclaimed in Claim 2.

Metric space of equivalence classes of sequences:

Definition 3.1. A sequence is a mapping from any linearly ordered set S to reals.

Definition 3.2. A sequence $\psi$ is non-zero and has finitely many non-zero elements if the set of $s\in S$ such that $\psi(s)\ne 0$ is finite and non-empty.

Definition 3.3. Non-zero sequences $\psi_1:S_1\to\mathbb R$ and $\psi_2:S_2\to\mathbb R$ with finitely many non-zero elements are equivalent if there is an order preserving bijection $\varphi$ between the sets $S'_j=\{s\in S_j:\psi_j(s)\ne 0\}$ such that $\psi_2\circ \varphi=\psi_1$.

Claim 3.4. Thus introduced equivalence relation is reflexive, symmetric, and transitive.

Definition 3.4. The distance between equivalence classes $\Psi_1$ and $\Psi_2$ is defined as the infimum over all pairs $\psi_1\in \Psi_1$ and $\psi_2\in \Psi_2$ such that $\psi_1$ and $\psi_2$ are defined on the same finite set $S$ of the (finite) sums $\sum_S|\psi_1(s)-\psi_2(s)|$

Claim 3.5. This distance is well-defined and satisfies three standard distance axioms.

Claim 3.6. The metric space thus constructed is separable.

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Thank you for your answer. Of course you can choose as Y any separable infinite dimensional Banach space, since they are all homeomorphic, due to Kadec. I am sorry, but with "choose a number", and "All we need now", and with the "fancy metric space", I am not sure at all that this is a real [mathematical] proof. Simply, I can't see "The Great Relief". It is not the axiom of choice, but the prose poems i'm disliking. [But this may be just a silly opinion.] – Ady Feb 2 '10 at 23:43
Well, if you think the proof is incorrect or unclear at some places, pinpoint the error or tell what exactly you don't understand. As to the style, everybody has his own and I'm not obliged to present my argument as a formal sequence of lemmas (though I will do it if it really helps you to understand it). My claim is that it is a full and correct argument, though somewhat sketchy. You are welcome to ask as many questions about it as you need. I'll answer to the best of my abilities. – fedja Feb 3 '10 at 6:48
If you are not willing to (minimally) formalize this rather vague argument, then I have nothing to ask. It may be a great idea, it may be an error. – Ady Feb 3 '10 at 17:28
Well, if you prefer "the set of cats that give the positive value to the function that is 1 if a cat is in the room and 0 if it is not has cardinality zero" to "all cats left the room", I translated my "poetry" into your "machine code". For me it doesn't matter which language to speak. They are equally expressive and exact if you know the words. But somehow I'm usually in the "poetic" mood. See if you can understand it better now. – fedja Feb 3 '10 at 19:52
Well, you may apply this theory to the Schroedinger' cat (especially due to your mapping $G$). Claim 1 was already accepted, while Claim 5 is certainly not yours (hence also accepted). Claim 2 is interesting, but it needs a proof, IMHO. However, who is "the set of finite or infinite sequences of reals with finitely many non-zero terms'' ? Is it the vector subspace of $\ell^{1}$ AKA $c{}_{00}$? If so, how that the mapping $H$ is well-defined ? – Ady Feb 4 '10 at 2:10

This is a negative result, heavily relying on P. Dodos' answer to Boundedness of nonlinear continuous functionals.

Let $\kappa$ be a measurable cardinal, let $K$ be the closed unit ball of $\ell_{\infty}\left(\kappa\right)$, and suppose, ad absurdum, that there exists a continuous mapping $F: K \rightarrow$ $\ell^{1} $\ {${ 0} $} that is odd on the unit sphere of $\ell_{\infty}\left(\kappa\right)$.

Let $j:\{\ell}^{1}$ $\rightarrow$ ${\ell}^{2}$ be the natural injection, and let $\rho$ be a metric generating the weak topology on the closed unit ball of ${\ell}^{2}$. Define next the continuous operator $T:K$ $\rightarrow$ ${\ell}^{2}$\ {${ 0} $} by $Tx:=$ $\dfrac{jFx}{\left\Vert jFx\right\Vert _{2}}$.

Define also the continuous function $f:\ K\rightarrow\mathbb{R}$,
expressed by $f(x):=$ $\rho\left(0,Tx\right)^{-1}$. Then there is a subspace $E$ of $\ell_{\infty}\left(\kappa\right)$ that is isomorphic to $c_{0}\left(\kappa\right)$ s.t. $f(K$ $\cap$ $E$ $)$ is bounded. Let $\left(P_{n}\right)$ be a sequence of ortho-projectors on ${\ell}^{2}$, having $n$-dimensional ranges $Y_{n}$, and s.t. $P_{n}$(x)$\ \rightarrow\ x $ for each $x$ in ${\ell}^{2}$. Take also a sequence $\left(X_{n}\right)$ of $n$-dimensional subspaces of $E$, and apply the classical Borsuk Theorem for the mappings $T_{n}$:=$P_{n}T:K\cap X_{n}$ $\rightarrow Y_{n}$, in order to get a sequence $\left(x_{n}\right)$ s.t. $T_{n}$ $x_{n}=$0$.$ Then it is easy to show that $Tx_{n}$ $\rightharpoonup0$ [i.e., weakly] in ${\ell}^{2}$, hence $f$ is not bounded on $K\cap E$, a contradiction.

Therefore, the Conjecture is not true.

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I think you can accept your own answer, if no others are forthcoming. – Yemon Choi Feb 7 '10 at 9:35
@Yemon Thanks, it should be easy to do it, but I'm thinking this would be unfair, since Pandelis is now renegating his own answer... So that, I simply cannot do it [even I'm still disliking the existing proof]. However, this would be some good lesson :-) – Ady Feb 7 '10 at 15:29

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