From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Q: regarding space Date: 12 Jul 1999 16:53:36 GMT Newsgroups: sci.math Keywords: model theory, independence of vector space axioms In article <7mc1n1$7kg$1@vixen.cso.uiuc.edu>, research wrote: >I wonder if there is any definition of the following 'space'. >The space I'm considering is same as the vector space V >except the axiom that: > >For a vector v (which is \in V), >there is a vector notated as -v \in V, for which >v + (-v) = 0 Attention desert-island mathematicians! This one's for you! Recall that a (real) Vector Space is defined to be a set V on which two binary operations "+" : V x V -> V and "*" : R x V -> V are defined, meeting these axioms: 1. forall u,v,w in V: (u+v)+w = u+(v+w) 2. forall u,v in V: u+v = v+u 3. exists z in V forall v in V: v+z = v 4. forall v in V exists w in V: v+w = z 5. forall u,v in V forall r in R: r*(u+v) = r*u+r*v 6. forall r,s in R forall v in V: (r+s)*v = r*v+s*v 7. forall r,s in R forall v in V: (rs)*v = r*(s*v) 8. forall v in V: 1*v = v It is proposed to drop axiom #4. It's a little hard to drop just that axiom, since w = (-1)*v comes pretty close to satisfying what is needed: (-1)*v + v = (-1)*v+1*v=(-1+1)*v= 0*v So in the presence of the other axioms, #4 is equivalent to the condition 0*v = zero-vector z. This is turn can be deduced from any of a number of other conditions, e.g. the axiom forall v in V, v + v = v => v = z. But it appears to be true that _some_ other condition is needed to show that 0*v = z, otherwise all we know is that multiplication by 0 is a homomorphism from the semigroup (V,+) to the sub-semigroup of idempotent elements of V (a sub-semigroup on which R acts trivially: r*v=v). Challenge: find a model (V,+,*) in which the other 7 axioms are satisfied but in which there is a vector v in V for which 0*v is not the zero vector. SPOILER Let V = R union {z}. Define "+" so that its restriction to real numbers is ordinary addition, and then r+z=z+r = r for all real numbers r; only z+z=z gives a sum not in R. Define "*" on R x R to be ordinary multiplication, and define r*z = z for all reals r. Just for clarity let me remark that 0*v is _never_ equal to z unless v=z, and v+w is never equal to z unless v=w=z, so axiom 4 is rather aggressively violated! Bonus challenge: for each of the other seven axioms find a model satisfy all axioms except that one. This is a little unsatisfactory in the case of deleting axiom #3, without which #4 has no meaning. Probably the right replacement for #4 is the cancellation law: 4a. forall u,v,w in V : u+v = w+v => u=w For in this case if you add axiom #3 you can recover axiom #4: simply let w=(-1)*v and prove (v+w)+v = z+v, so that #4a => v+w=z. dave (recent Linear Algebra teacher) ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Q: regarding space Date: 15 Jul 1999 14:05:09 GMT Newsgroups: sci.math This follows up my article <7md6ig$j9i$1@gannett.math.niu.edu>, since I see no one else took the bait. A recent post prompted me to look up a family of examples I have used to annoy my linear algebra students: examples showing the necessity of each of the axioms for Vector Spaces. I summarized those axioms and asked for examples which showed that each was independent of the others. In detail, I first gave the list of axioms: >Recall that a (real) Vector Space is defined to be a set V on which two >binary operations "+" : V x V -> V and "*" : R x V -> V are defined, >meeting these axioms: >1. forall u,v,w in V: (u+v)+w = u+(v+w) >2. forall u,v in V: u+v = v+u >3. exists z in V forall v in V: v+z = v >4. forall v in V exists w in V: v+w = z >5. forall u,v in V forall r in R: r*(u+v) = r*u+r*v >6. forall r,s in R forall v in V: (r+s)*v = r*v+s*v >7. forall r,s in R forall v in V: (rs)*v = r*(s*v) >8. forall v in V: 1*v = v I then issued this challenge: for each of the eight axioms, find a model satisfying all axioms except that one. In order to make sense of axiom 4 when axiom 3 is removed, I replaced it with the cancellation law: >4a. forall u,v,w in V : ( u+v = w+v ) => ( u=w ) This challenge was a little sneaky since in fact axiom #3 is _not_ independent of the others; one can define a vector space using only axioms 1,2,4a, and 5-8! It's also a little arbitrary since in some cases the axioms have to be defined just-so in order to pull off the stunt of creating models satisfying the other n-1 axioms, whereas the set of axioms for a vector space are really very "natural" in their usual form. But here goes my set of non-vector-spaces. Nicer examples welcome. no-1: Let V=R^2, "*" = ordinary scaling, but define u"+"v to be |cos(theta)| (u+v), where theta is the angle between u and v; if either of u or v is zero, take theta to be zero. no-2: Let V=R, u "+" v = u for any two vectors u and v, and (likewise) r*u = u for any vector u and real number r. Note that this only succeeds because I have set cancellation (4a) and the zero law (3) in the appropriate order; of course _if_ commutativity (2) is assumed, the order becomes irrelevant! no-3: Can't happen! For any v and w we have 0*w + (0*w + 0*v) = (0*w + 0*w) + 0*v = (0+0)*w + 0*v = 0*w + 0*v which by symmetry (and commutativity) is then equal also to 0*v + (0*w + 0*v). By cancellation we conclude 0*w=0*v. So if we pick any v and define z = 0*v, then for any other w in V we get w + z = w + 0*v = 1*w + 0*w = (1+0)*w = w. Thus this z is a zero vector. no-4: As in previous post, this model satisfies 1,2,3,5-8 but not 4 or 4a: Let V = R union {z}. Define "+" so that its restriction to real numbers is ordinary addition, and then r+z=z+r = r for all real numbers r; only z+z=z gives a sum not in R. Define "*" on R x R to be ordinary multiplication, and define r*z = z for all reals r. Just for clarity let me remark that 0*v is _never_ equal to z unless v=z, and v+w is never equal to z unless v=w=z, so axiom 4 is rather aggressively violated! no-5: This one is a little tricky. It's easier to violate in the complex case: Let V = C x C with usual vector addition but to define scalar multiplication, let sigma be any nontrivial automorphism of the complex field (e.g. complex conjugation) and then let r*(x,y) = (r.x, r.y) if x <> 0 (0, sigma(r).y) if x = 0. We can use the same trick over any other field, except that R has NO nontrivial automorphisms! And if we just restrict the previous example to the reals inside C if sigma is complex conjugation, then axiom #5 _will_ hold. But we can use the fact that C has automorphisms which do not preserve R (not even setwise), and use the above formulas to define a scalar multiplication on C x C (or if you prefer, on R x C = R^3.) Remark: r*(u+v) = r*u+r*v automatically holds (given the other axioms) for all _rational_ r. So e.g. continuity would force #5, too. no-6: Let V=R, "+" = usual addition, r*v = r^2 . v no-7: Let V=R, "+" = usual addition, r*v = phi(r).v where phi : R -> Q is any homomorphism of (additive) groups no-8: Let V=R, "+" = usual addition, r*v = 0 for all r dave ============================================================================== From: Pertti Lounesto Subject: Re: Q: regarding space Date: Thu, 15 Jul 1999 18:01:24 +0200 Newsgroups: sci.math Dave Rusin wrote: > A recent post prompted me to look up a family of examples I have used > to annoy my linear algebra students: examples showing the necessity of > each of the axioms for Vector Spaces. I summarized those axioms and asked > for examples which showed that each was independent of the others. > In detail, I first gave the list of axioms: > > >Recall that a (real) Vector Space is defined to be a set V on which two > >binary operations "+" : V x V -> V and "*" : R x V -> V are defined, > >meeting these axioms: > >1. forall u,v,w in V: (u+v)+w = u+(v+w) > >2. forall u,v in V: u+v = v+u > >3. exists z in V forall v in V: v+z = v > >4. forall v in V exists w in V: v+w = z > >5. forall u,v in V forall r in R: r*(u+v) = r*u+r*v > >6. forall r,s in R forall v in V: (r+s)*v = r*v+s*v > >7. forall r,s in R forall v in V: (rs)*v = r*(s*v) > >8. forall v in V: 1*v = v > > I then issued this challenge: for each of the eight axioms, find a model > satisfying all axioms except that one. In order to make sense of axiom 4 when > axiom 3 is removed, I replaced 4 with the cancellation law: > >4a. forall u,v,w in V : ( u+v = w+v ) => ( u=w ). Your 4a resembles 3 and should replace 3 rather than 4. Rename your 4a as 3a and replace your 4 by a new 4a: 4a. for all u in V exists v in V: (u+v)+u = u.