Are any two infinite-dim. V.Spaces isomorphic?

[Bacle]

Hi, everyone:

Just curious: Are any two infinite-dimensional vector spaces A,B over the
same field F isomorphic?


It is straightforward to show any two finite-dim. V.Spaces (over the
same field F, of course) are isomorphic. If V,W have dim. n< oo , and respective
bases {v1,..,vn}, {w1,...,wn} ; both V.Spaces
over F, then we can represent any v, any w in V,W respectively, by:

v=f1*v1+f2*v2+...+fn*vn

w=f1'*w1+f2'*w2+...+f'n*wn

where fi,fi' are in F; i in {1,2,..,n}.

And then the maps:

h: v-> (f1,..,fn) :V->F^n

h': w->(f1',..,fn'): W->F^n

are isomorphisms. Then the composition h'^-1o h

gives us an iso. between V,W.

Are A,B infinite-dimensional of the same cardinality (cardinality of dimension, of course)
isomorphic? Does the trick above also work for the infinite-dimensional case?

Thanks.

 

[micromass]

Yes! Two vector spaces are isomorphic if and only if their dimensions are equal. Even if the dimension is an infiite cardinal!

 

[Bacle]

Thanks, Micromass.

Is the proof for the infinite-dim. case similar to that for the finite-dim. case?

I know we can always find a Hamel basis, using AC, and that in Hamel bases,

linear combinations have finite support. But in the oo-dimensional case we don't

have the equivalent isomorphism to F^n that we do in the f.dim. case.

Would you suggest something?

 

[micromass]

Well, let V and W be vector spaces. Let E be a basis of V and W be a basis of W. Let f:E--> F be a bijection. All we need to do is extend this bijection to an isomorphism of V and W.

Alternatively, let V be a vector space with a basis of cardinality [\tex]\kappa[/tex]. Then one can show that V is isomorphic to $$\mathbb{R}^\kappa$$. Just like in the finite case (the proof is also just like in the finite case). We can exchange $$\mathbb{R}$$ with any field we like...

 

[Hurkyl]

The claim requires the axiom of choice. In fact, in ZF, I think it is equivalent to the axiom of choice (but I'm not sure).

IMO the most straightforward way to generalize these sorts of inductive proofs from the finite-dimensional to the infinite-dimensional case is to use transfinite induction.

(but then again, I think I'm unusual in that I prefer to use the well-ordering theorem rather than things like Zorn's lemma)

 

[micromass]

Are you sure it requires AC? If you state the question as follows: "If V and W are vector spaces such that V and W have a basis of the same cardinality, then V and W are isomorphic". I don't think that requires choice at all!

Of course, if you state it as follows "Let V and W be vector space of the same dimension, then V and W are isomorphic", then it requires choice. For the simple reason that the dimension of a vector space is not well defined. But even this is not equivalent to the full AC, but is already implied by the ultrafilter lemma.

 

[Bacle]

Well, let V and W be vector spaces. Let E be a basis of V and W be a basis of W. Let f:E--> F be a bijection. All we need to do is extend this bijection to an isomorphism of V and W.

Alternatively, let V be a vector space with a basis of cardinality [\tex]\kappa[/tex]. Then one can show that V is isomorphic to $$\mathbb{R}^\kappa$$. Just like in the finite case (the proof is also just like in the finite case). We can exchange $$\mathbb{R}$$ with any field we like...

Just curious: how do we turn $$\mathbb{F}^\kappa$$ into a vector space?

Do you use a weak product ( i.e., vectors have finite support)? . Or do you just

define v_i+v_j component-wise?

 

[Hurkyl]

Are you sure it requires AC? If you state the question as follows: "If V and W are vector spaces such that V and W have a basis of the same cardinality, then V and W are isomorphic". I don't think that requires choice at all!

I misread what he was asking.


It still took me some thought to be sure that the existence of a bijection between bases implies an isomorphism (to make sure I didn't accidentally invoke choice) -- but I've convinced myself that the vector space is the union of those subspaces that have a basis which is a subset of the basis for the total space, and once I have that I'm confident I can avoid choice.

 

[mathwonk]

linear maps between vector spaces are essentially the same thing as functions between their bases, and isomorphisms are the same as bijections. I.e. there is a functor from sets to vector spaces, taking a set to a vector space with that set as basis. as with all functors, it takes isomorphisms (of sets) to isomorphisms (of vector spaces). Since saying two sets have the same cardinality essentially means there is a bijection between them, the answer is yes. (to: "Are A,B infinite-dimensional of the same cardinality (cardinality of dimension, of course)
isomorphic? ")

I don't see the need for AC here, but I am rather clueless about logic. To me, AC is used in results like, if there is a surjective linear map from V to W, then W is isomorphic to a subspace of V. Hurkyl is more expert on this.

 

[yaa09d]

Hey there!
I have the following question:
Q: If we consider R and C as Q-vector spaces, then how can we show they are isomorphic?

I know that if a two vector spaces have bases with the same cardinality, then they are isomorphic. Also, Zorn lemma tells us that every vector space has a basis.

In this case, answering my question amounts to showing that any bases of R and C over Q have the same cardinality. In other words, I need to show dim R= dim C over Q.

Remark: it is not hard to show any basis of R (or C ) over Q is infinite, but how can we know the card of such a basis??

Thank you!

 

[Landau]

If $B$ is a Q-basis of R, then $B\cup iB$ is a Q-basis of C (the proof is trivial). But as B is infinite, $|B\cup iB|=|B|+|B|=|B|$.

More informative, we can determine that in fact |B|=|R|. Indeed, $\mathbb{R}\cong \mathbb{Q}^{(B)}$ (functions B\to Q with finite support) is a bijection, or even an isomorphism of Q-vector spaces if the latter is turned into one by pontwise operations. As B is infinite,
[tex]\left|\mathbb{Q}^{(B)}\right|=|\mathbb{Q}|\cdot|B|=|B|[/itex]

[For those who are wondering why B is necessarily infinite: a vector space over Q with finite basis has cardinality |Q|^n, so is countable.]

 

[yaa09d]

Thank you for your reply.

 

[Landau]

You're welcome. I hope it was sufficiently clear?

 

[yaa09d]

Yes, it is clear. I was just confused why
[tex]\left|\mathbb{Q}^{(B)}\right|=|\mathbb{Q}|\cdot|B|[/itex].
However, I found a proof for that on another forum, so I am ok with that now. Do you know if there is any textbook where that relation is proved?
Is it a standard relation in algebra or set theory?

Thank you.

 

[Landau]

I just produced the following proof. Perhaps there is a simpler argument, but I think this works.

For a set A, let

$$A^*=\bigcup_{n\in\mathbb{N}}A^n$$

denote the set of all finite sequences in A.

Lemma: If A is infinite, then $|A|=|A^*|$.
Proof: We know $|A\times A|=|A|$. By induction, $|A^n|=|A|$, say via a bijection $\phi_n: A\to A^n$. Then there is a bijection
$\mathbb{N}\times A\to A^*$

$(n,a)\mapsto \phi_n(a).$
Hence $|A^*|=|\mathbb{N}\times A|=\max(|\mathbb{N}|,|A|)=|A|$.

Proposition: Let A and B be infinite. Fix $a_0\in A$, and let $A^{(B)}$ be the set of functions B\to A with "finite support", i.e. all but finitely many elements in B have a0 as image. Then $\left|A^{(B)}\right|=|A|\cdot|B|$.
Proof: Define a function
$$A^{(B)}\to B^*\times (A-\{a_0\})^*$$

as follows. Endow B with some well-order. Given $f\in A^{(B)}$, consider the support $\{b_1,...,b_n\}$ of f (i.e. $f(b_i)\neq a_0$), and order it such that $b_1<...<b_n$. Now map

$$f\mapsto ((b_1,...,b_n),(f(b_1),...,f(b_n)).$$
It should be clear that this is injective, for f is uniquely determined by the b_i and the images. This is also surjective: given any

$$((b_1,...,b_n),(a_1,...,a_n)\in B^*\times (A-\{a_0\})^*,$$

define $g\in A^{(B)}$ by $g(b_i)=a_i$ and $g(b)=a_0$ for the other b's. We conclude, using the Lemma:

$$\left|A^{(B)}\right|=|B^*|\cdot |(A-\{a_0\})^*|=|B^*|\cdot|A^*|=|B|\cdot|A|=\max(|A|,|B|).$$

In particular, if B is infinite, then:

$$\left|\mathbb{Q}^{(B)}\right|=\max(|\mathbb{Q}|,|B|)=|B|.$$

 

[mathwonk]

This may be stupid but it seems obvious that a Q vector space with a countable basis is countable. Thus R is an uncountable dimensional Q vector space. The same argument should show the dimension of R equals card(R). Since C has the same cardinality as R, it cannot have larger dimension.

I think the missing details are somewhere around the words "obvious that..."

 

[Landau]

Yes, I think your "it is obvious that" is more or less what I proved in my previous answers, altough I'd be interested in an easier argument!

A Q-vector space with countable basis B can be written

$$\{\sum_{i=1}^nq_ib_i|q_i\in \mathbb{Q},b_i\in B, 1\leq i\leq n\}$$

and this is indeed "obviously" countable as Q,B and N are. To prove this, I would say given n, there are Q^n B^n linear combinations of n basis elements. Hence the cardinality of this set is
$$\bigcup_{n\in\mathbb{N}}\mathbb{Q}^n\times B^n$$
which is a countable union of countable sets, hence countable.

I guess this is the same as saying

$$|\mathbb{Q}^{(B)}|=|\mathbb{Q}||B|=|\mathbb{Q}|.$$

 

[disregardthat]

and this is indeed "obviously" countable as Q,B and N are. To prove this, I would say given n, there are Q^n B^n linear combinations of n basis elements.

Q^nB^n is an upper bound at least, counting them in this way does not give an injection. But of course, as you know the space is at least countable, we are done.

 

[Landau]

True. But you probably mean "at most countable" instead of "at least countable".

 

[disregardthat]

True. But you probably mean "at most countable" instead of "at least countable".

I meant at least countable (countable basis). My point was that you gave an argument for why it is at most countable, so we conclude it is countable.

 

[Landau]

Sorry, I didn't read carfully. I agree.

 

[russel]

Hello there, I'd like to ask if the Φ space (the one where each element is a sequence of finite non-zero terms) with norm 1 is isomorphic to Φ space with norm 2. Is it or not? And why? Has this to do with the fact that Φ is never Banach?