Two Deletion Theorems

Introduction

For all the normed spaces we consider here, the base field will be \(\rmat.\)\[ \]

How many of us would bet that in a normed space the complementary of a one-point set could be homeomorphic to the whole space? And yet… In an infinite-dimensional space it is always true! This is just a special case of Theorem 2 below. Many similar and equally surprising results flourished in the 1950s and 1960s, dealing with geometry of normed spaces and specially deletion properties (a subset \(A\) of a topological space \(E\) is said to be deletable if \(E\setminus A\) is homeomorphic to \(E).\)

We present here the proofs of the following two results, due to Klee and Bessaga (see ).

It follows that in a non-complete (resp. infinite-dimensional) normed space, the complementary of any complete (resp. compact) subset is arcwise connected, which is not so obvious.

Preliminary 1

Let \((E,d)\) be a metric space. We recall that a map \(f:E\fl E\) is \(k\)-contracting if \(f\) is \(k\)-Lipschitz and \(k\in[0,1).\)

In this case, if \(E\) is complete, then \(f\) has a unique fixed point Banach-Picard theorem.

Additionally, assume that \(E\) is a normed space (hence a Banach space). Then the map \(g:x\mapsto x+f(x)\) is an homeomorphism from \(E\) onto itself. Indeed, on one hand, the condition \(g(x)=y\) is equivalent to \(x\) is a fixed point of \(g_y\), where the map \(g_y:x\mapsto y-f(x)\) is, like \(f,\) \(k\)-contracting. This proves that \(g\) is one-to-one and onto. Moreover, \(g\) is continuous, and denoting \(x=g^{-1}(y)\) and \(x'=g^{-1}(y')\) we have \(x=y-f(x)\) and \(x'=y'-f(x'),\) hence \(\displaystyle||x-x'||≤||y-y'||+||f(x)-f(x')||\)\(\displaystyle≤||y-y'||+k||x-x'||.\) We obtain \(||x-x'||≤{1\over 1-k}||y-y'||,\) which provesWe could as well use Banach-Picard theorem with parameter. See for instance the following reference in French : Ramis E., Deschamps C., Odoux J., Cours de Mathématiques Spéciales, 3 Topologie et éléments d'analyse, Paris : Masson 1976.\(g^{-1}\)'s continuity.

Note finally that in this case, replacing \(x\) by \(g^{-1}(x)\) in equality \(g(x)=x+f(x),\) we obtain \(x=g^{-1}(x)+f(g^{-1}(x)),\) that is \[g^{-1}(x)=x-f(g^{-1}(x))\,.\] This equality will be used at the very end of this post.

Proof of theorem 1

\(E\) denotes a non-complete normed space and \(A\) a complete subset of \(E.\) Let \(E'\) be the completion of \(E,\) and \(a\in E'\setminus E\) be such that \(||a||<1,\) fixed once and for all. We will construct a map \(f:E'\fl E'\) satisfying the following conditions:

1. \(f\) is a contracting map;
2. \(f(A)=\{a\}\);
3. \(f(E'\setminus A)\subset E\);
4. \(\forall x\in E'\,,d(x,A)≥1\impl f(x)=0_E.\)

Then we will just have to take as a deleting homeomorphism the map from \(E\setminus A\) into \(E,\) induced by the map \(h\) defined on \(E'\) by \(h(x)=x+f(x).\) Indeed, condition (i) ensures that \(h\) is an homeomorphism of \(E'\) onto itself (see preliminaries), and (ii) and (iii) imply that we have, for every \(x\in E':\) \(h(x)\in E\ssi x\in E\setminus A.\)

To obtain such a contraction \(f,\) we will define a contracting path \(\gamma:\rmat^+\fl E'\) such that \(\displaystyle\gamma(0)=a\,;\quad\)\(\quad\displaystyle \gamma(\rmat^{+*})\subset E\,;\quad\)\(\quad\displaystyle \forall t≥1\,,\>\gamma(t)=0_E\,.\) Then, by setting \(f(x)=\gamma(d(x,A))\) conditions (i) à (iv) are then satisfied: condition (i) results from the contracting character of \(\gamma\) and from the fact that \(x\mapsto d(x,A)\) is 1-Lipschitz; conditions (ii) and (iv) are obvious; condition (iii) comes from the completeness of \(A:\) the set \(A\) is then closed in \(E',\) since it is complete; it follows that if \(x\in E'\setminus A\) then \(d(x,A)>0,\) and therefore \(f(x)=\gamma(d(x,A))\in E.\)

Let us proceed to the construction of \(\gamma:\) Let \(r\) such that \(0\lt r\lt 1.\) Since \(E\) is dense in its completion \(E',\) there exists a sequence \((x_n)\) of points of \(E\) such that \[\forall n\in\nmat\,,\quad||x_n-a||≤r^n||a||\,;\qquad x_0=0_E\,,\] and it is clear that one can easily construct such a sequence \((x_n)\) in such a way that it is one-to-one.

We have, by triangle inequality: \(\forall n\in\nmat\,,\>||x_{n+1}-x_n||≤(r+1)r^n||a||.\) Hence \(\sum||x_{n+1}-x_n||\) is a convergent series, and its sum is not greater than \({r+1\over 1-r}||a||.\) This quantity tends to \(||a||<1\) when \(r\tv0,\) hence we can choose \(r\) such that \[S=\sum_{n=0}^\pinfty||x_{n+1}-x_n||<1\,.\]

Let the sequence \((t_n)\) be defined by induction by \[t_0=1\,;\qquad \forall n\in\nmat\,,\quad t_{n+1}=t_n-{1\over S}||x_{n+1}-x_n||\,.\tag{1}\] Then \((t_n)\) is a decreasing sequence, with limit 0 since \(\displaystyle t_n=1+\sum_{k=0}^{n-1}(t_{k+1}-t_k)\)\(\displaystyle =1-{1\over S}\sum_{k=0}^{n-1}||x_{k+1}-x_k||\)\(\displaystyle ={1\over S}\sum_{k=n}^{\infty}||x_{k+1}-x_k||\tv0\,.\) The path \(\gamma\) is defined by \(\quad\displaystyle\gamma(0)=a\,;\quad\)\(\quad\displaystyle\gamma(t_n)=x_n\,;\quad\)\(\quad \displaystyle\gamma\hbox{ est affine sur }[t_{n+1},t_n]\,;\quad\)\(\quad\displaystyle\forall t≥1\,,\>\gamma(t)=0_E\,.\quad\) The restriction of \(\gamma\) to \([t_{n+1},t_n]\) is affine, with derivative \((x_{n}-x_{n+1})/(t_{n}-t_{n+1}),\) whose norm equals \(S\) according to \((1).\) But \(S\lt 1,\) which implies the contractance of \(\gamma\) on every interval \([t_{n+1},t_n],\) and therefore on \((0,1]\) by triangle inequality, then on \([0,1]\) by continuity, and finally on all of \(\rmat^+,\) which completes the proof.

Preliminary 2

We recall the following three classical results.

Minkowski gauge Let \(E\) be a nonzero vector space. We will say that a set \(C\subset E\) is admissible if it is a convex, non empty set, symmetric with respect to \(0_E,\) such that for every one-dimensional subspace \(D,\) we have: \(C\inter D≠\{0_E\}\) and \(D\not\subset C.\) In this case, the map \[j_C:x\mapsto \inf\{\lambda\in\rmat^{+*}\tq x\in\lambda C\}\] is well-defined, and \(j_C\) is a norm on \(E,\) called the gauge of \(C.\)

Banach theorem Any continuous linear bijection between two Banach spaces is an homeomorphism.

Banach-Mazur theorem Every separable normed space (i.e. having a dense countable subset) is isomorphic (as a normed space) to a subspace of \(C^0([0,1])\) (endowed with the infinite norm).

Proof of theorem 2

We will need some preliminary results.

Let \(C\) be the convex hull of \(B_1\union B_2,\) where \[B_1=\{x\in E\tq ||x||≤1/k\}\,;\qquad B_2=\{x\in F\tq N(x)≤1\}.\]

1. a Consider the map \(x\mapsto d(x,F)\) (in the sense of the original norm \(||\cdot||).\) It easily satisfies the triangle inequality and it is positively homogeneous. It is zero on \(B_2,\) and on \(B_1\) it is bounded by \(x\mapsto d(x,0_E)=||x||≤1/k.\) It is therefore bounded by \(1/k\) on \(C.\)
2. b We deduce that the set \(C\) is admissible (see preliminary). The only condition which is not totally obvious to verify is the fact that \(C\) does not contain a whole one-dimensional subspace (The set \(B_2,\) and therefore also \(C,\) is in general not bounded for \(||\cdot||).\) Suppose by contradiction that a line \(D\) satisfies \(D\subset C.\) According to a) the map \(x\mapsto d(x,F)\) is therefore bounded on \(D\) and positively homogeneous. We deduce that it is zero on \(D,\) and therefore \(D\subset F\) since \(F\) is closed. But the trace of \(C\) on \(F\) clearly equals \(B_2,\) which is a ball for a norm and and therefore cannot contain a straight line.
3. c The set \(C\) being therefore admissible, we deduce that \(N'=j_C\) is a norm on \(E\) (see preliminary). We clearly have \(N'\restr{F}=N,\) since \(B_1\subset B_2,\) and since \(B_1\subset C\) we have \(N'(x)=j_C(x)≤k||x||\) for every \(x\in E.\) Hence the norm \(N'\) satisfies conditions (i) and (ii).
4. d It remains to show that \(F\) is closed in \((E,N').\) According to a) we have by homogeneity: for every point \(x\in E,\) \(d(x,F)≤N'(x)/k.\) Let \((x_n)\) be a sequence of points in \(F\) which converges for \(N'\) to a point \(a\in E.\) We have \[d(a,F)=d(x_n-a,F)≤N'(x_n-a)/k\tv0\,,\] that is \(d(a,F)=0,\) and therefore \(a\in F\) since \(F\) is closed for \(||\cdot||.\)

If \((E,||\cdot||)\) is not complete there is nothing to do: one just has to take \(N=||\cdot||.\) We will therefore assume that \(E\) is a Banach space.

1^st case: \(E\) is separable. According to Banach-Mazur theorem we can assume that \(E\) is a subspace of \(C^0([0,1]),\) and that \(||\cdot||=||\cdot||_\infty.\) Let's take \(N=||\cdot||_2.\) We have \(N≤||\cdot||.\) It remains to prove that \((E,N)\) is not complete.

Suppose by contradiction that \((E,N)\) is complete. The identity map is continuous from \((E,||\cdot||)\) onto \((E,N),\) and both being Banach spaces by assumption, according to Banach theorem it is bicontinuous. Hence there exists a constant \(C>0\) such that \(\forall f\in E\,,\>||f||_\infty≤C||f||_2.\)

But such an inequality can only occur if \(E\) is finite-dimensional, which contradicts the assumptions. The argument is classical: if \(E\) is of dimension at least \(k\) there exists an orthonormal family \((f_1,\ldots,f_k)\) of \(E\) equipped with the usual scalar product of \(C^0([0,1]).\) Let \(t\in[0,1]\) and \(f=\sum_{i=1}^kf_i(t)f_i.\) We have: \[\sum_{i=1}^kf_i^2(t)=f(t)≤||f||_\infty≤C||f||_2=C\sqrt{\sum_{i=1}^kf_i^2(t)}\,,\] hence \(\sum_{i=1}^kf_i^2(t)≤C^2.\) By integrating this inequality for \(t\) varying from \(0\) to \(1\) we obtain: \(k≤C^2.\)

2^nd case: general case. Since \(E\) is finite-dimensional there exists an algebraically independant sequence \((e_n)_{n\in\nmat}.\) Take \(F=\overline{\vect((e_n)_{n\in\nmat})}.\) According to the first case there exists a norm \(N\) on \(F\) such that \(\forall x\in F\,,\>N(x)≤||x||,\) and \((F,N)\) is not complete. By lemma 1, \(N\) can be extended to a norm \(N'≤||\cdot||\) on \(E\) for which \(F\) is closed. Since the latter is not complete, it follows that \((E,N')\) is not complete.

Let \(K\) be a compact set in an infinite-dimensional normed space \((E,||\cdot||).\) According to lemma 2, there exists a norm \(N≤||\cdot||\) on \(E\) such that \((E,N)\) is not complete. The map \(\id:(E,||\cdot||)\fl(E,N)\) is then continuous hence \(K\) remains compact for \(N,\) and therefore complete. According to theorem 2, there exists an homeomorphism \(h\) of \(E\setminus K\) onto \(E.\)

However, we must be careful not to conclude too quickly, since this homeomorphism is relative to the topology induced by \(N\) (in both domain and codomain). For the initial norm \(||\cdot||,\) neither the continuity of \(h\) nor that of \(h^{-1}\) is self-evident.

Let us resume the construction of \(h\) given by the proof of theorem 1. The map \(h\) is defined by \(h(x)=x+\gamma(d(x,K)),\) where the distance \(d\) and the path \(\gamma\) are built using the norm \(N.\) The path \(\gamma\) is contracting for that norm, which ensures the bijectivity of \(h.\)

The key argument is that on \(E\setminus K,\) \(x\mapsto d(x,K)\) has values in \(]0,1].\) On this interval the path \(\gamma,\) affine on every interval \([t_{n+1},t_n],\) is continuous for every norm (and even for every topological vector space structure) on \(E.\)

The map \(x\mapsto\gamma(d(x,K))\) is then continuous on \(E\setminus K\) as the composition of the diagram \(\displaystyle(E\setminus K,\>||\cdot||){\buildrel\id_E\over{\quad\longrightarrow\quad}}(E\setminus K,\>N)\)\(\displaystyle{\buildrel \quad x\,\mapsto\, d(x,K)\quad\over\longrightarrow}\rmat {\buildrel\gamma\over{\quad\longrightarrow\quad}}(E,\>||\cdot||)\,,\) from which we deduce the continuity of \(h:(E\setminus K,||\cdot||)\fl(E,||\cdot||).\)

Finally, the map \(h^{-1}\) satisfies \(h^{-1}(x)=x-\gamma(d(h^{-1}(x)))\) (see ). Its continuity is obtained by composing the following diagram, where each arrow is a continuous map \(\displaystyle(E,||\cdot||){\buildrel\id_E\over{\quad\longrightarrow\quad}}(E,N){\buildrel h^{-1}\over{\quad\longrightarrow\quad}}(E\setminus K,N)\)\(\displaystyle{\buildrel \quad x\,\mapsto\, d(x,K)\quad\over\longrightarrow}\rmat {\buildrel\gamma\over{\quad\longrightarrow\quad}}(E,||\cdot||)\,,\) which completes the proof.

References and add-ons

Theorem 2 is due to KleeKlee V. L., A Note on Topological Properties of Normed Linear Spaces, Proceedings of the American Mathematical Society Vol. 7, No. 4 (1956), pp. 673-674. but the proofs we present here, specially the non-complete norm technique, are essentially due to Bessaga. This article is inspired by the book by Bessaga and PelczyńskiBessaga C. & Pelczyński A., Selected topics in infinite-dimensional topology, Warszawa : Monografie Matematyczne 1975.. The style of the latter is rather elliptical and it took me a while to convince myself of the correctness of the arguments they used. I also took the opportunity to slightly simplify the proof of lemma 2 by using Banach-Mazur's theorem and an elementary and classical argument used in a theorem of Grothendieck, well known to students in topology: every subspace of \(C^0([0,1])\) on which \(||\cdot||_2\) is finer than \(||\cdot||_\infty\) is finite-dimensional.

Other equally surprising results originate in the same period. For example: the closed unit ball of an infinite-dimensional normed space is deletable; the unit sphere of an infinite-dimensional Hilbert space is homeomorphic (Klee)Klee V. L., Convex bodies and periodic homemorphisms in Hilbert space, Trans. Amer. Math. Soc. vol. 74 (1953) pp. 10-43. to the whole space, and even \(C^\infty\)-diffeomorphic (Bessaga)Bessaga C., Every infinite-dimensional Hilbert Space is diffeomorphic with its unit sphere, Bull. Acad. Polon. Sci., Sér. sci. math. astr. et phys. 14 (1966), pp. 27-31. and even \(\rmat\)-analytically isomorphic (Dobrowolski).Dobrowolski T., Every Infinite-Dimensional Hilbert Space is Real-Analytically Isomorphic with Its Unit Sphere, Journal of Functional Analysis, 134-2 (1995), pp. 350-362. Partial extensions to topological vector spaces have also been obtained.See e.g. : Anderson R.D., On a theorem of Klee, Proceedings of the American Mathematical Society Vol. 17, No. 7 (1966), p. 1401-1404.