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Let $E$ be a infinite dimensional vector space. Let $F\subset E$ be a infinite dimensional subspace. Is it possible that $F$ is isomorphic to $E$?

I think this is not possible, but I can't see a straightforward proof of it. Does anyone have any ideas?

Thanks.

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Sure, let $E$ have a countable basis indexed by $\Bbb N$, and let $F$ be spanned by all but the first basis vector of $E.$ Then the isomorphism would be $e_i\mapsto e_{i+1}.$ –  Andrew Oct 5 '12 at 18:41
    
So easy. thanks. –  Tomás Oct 5 '12 at 18:45

3 Answers 3

up vote 4 down vote accepted

It is not only that there is, but in fact a vector space is infinitely dimension if and only if it has a proper subspace which is isomorphic to it.

One direction is easy, if the dimension is finite then this is impossible, since the isomorphism is injective and the sum of ranks of the kernel and image implies that the image of the isomorphism has to be everything.

If the space is infinitely dimensional, let $B=\{v_i\mid i\in I\}$ be a Hamel basis of the space, this is an infinite set therefore there is a bijection between $B$ and $B\setminus\{v\}$ for some $v\in B$. This bijection extends unique to an isomorphism (note that the extension is unique, linear, and injective). Clearly the image of such isomorphism is a proper subspace, since $v$ is not in the image.

(Note the axiom of choice was used extensively here for choosing a basis and for the bijection with a proper subset; without the axiom of choice it is consistent that there are very strange vector spaces which are not finitely generated but do not have this property.)

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Nice one. I was really wrogn about my thoughts. Thanks –  Tomás Oct 5 '12 at 18:58

This is similar to the question whether an infinite set can mapped bijectively to a proper subset. Indeed, this is true, e.g. $\mathbb N_{\geq 0} \mapsto \mathbb N_{\geq 1}, n \mapsto n+1$ is such a bijection. This can be turned into an example for the corresponding statement for vector spaces: Let $E$ be the vector space freely generated by $\mathbb N_{\geq 0}$ and $F$ be the subspace freely generated by $\mathbb N_{\geq 1}$. The map $n \mapsto n+1$ maps the basis of $E$ bijectively to the basis of $F$, hence induces an isomorphism of vector spaces. $E$ may be viewed as the space of finite sequences $(a_i)_{i\geq 0}$ in $k$, then $F$ is the subspace of sequences $(a_i)_{i\geq 0}$ with $a_0 = 0$ and the map induced by $n \mapsto n+1$ is the right shift $$(a_0,a_1,a_2,\ldots) \mapsto (0,a_0,a_1,\ldots).$$

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Consider the vector space of arbitrary sequences of reals, and the right shift map $$ (x_1,x_2,x_3,\ldots) \mapsto (0,x_1,x_2,\ldots) $$

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