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Let $V,W$ be two countably infinite dimensional vector space over the same field , then are $V,W$ isomorphic as vector spaces ? And please give example of two non-isomorphic uncountable dimensional vector space over a same field . Thanks in advance

EDIT : If $V,W$ are infinite dimensional vector spaces over the same field such that $V,W$ has same cardinality as sets , then are $V,W$ isomorphic as vector spaces ?

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If $V$ and $W$ have bases $B_V$ and $B_W$, the axiom of choice proves that the cardinality of such a basis is unique. If $|B_V|=|B_W|$, then any bijection between the two sets can be extended (uniquely) to an isomorphism between $V$ and $W$. Just like in the finite case.

If both spaces have a countable basis, then this is immediate.

"Uncountable" just means "not countable" just like "having at least two elements" does not mean anything except that the set has more than one element.

However, it should be noted, that if $V$ and $W$ have the same cardinality it does not mean that they have the same dimension. $\Bbb R$ and $\Bbb R[x]$ have the same cardinality, but $\Bbb R$ is only one dimensional, whereas $\Bbb R[x]$ has a countably infinite dimension. Moreover, the dual space of $\Bbb R[x]$, which is $\Bbb{R^N}$ (all the infinite sequences) has the same cardinality as well, but its dimension is already uncountable.

If, however, the spaces have the same cardinality and it is larger than the cardinality of the field, then they have the same dimension and are therefore isomorphic. For example, $\Bbb R$ and $\Bbb C$ have the same cardinality and are both vector spaces over $\Bbb Q$, so they are isomorphic as such vector spaces. But as vector spaces over $\Bbb R$ they no longer satisfy the second requirement and indeed they are not isomorphic as $\Bbb R$-vector spaces.

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    $\begingroup$ in 2nd paragraph do oyu mean to say $\mathbb R[x]$ and $\mathbb R^{\mathbb N}$ has same cardinality ? $\endgroup$ – user217921 Mar 5 '15 at 14:39
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    $\begingroup$ : could yo please elaborate (proof) on why " if the spaces have the same cardinality and it is larger than the cardinality of the field , then they are isomorphic as vector spaces " ? $\endgroup$ – user217921 Mar 5 '15 at 15:00
  • $\begingroup$ I am unable to understand those facts also @Asaf Karagila $\endgroup$ – Learnmore Mar 5 '15 at 15:39
  • $\begingroup$ @Saun Dev: Yes to the first comment. To the second comment, the proof is practically in martini's answer. $\endgroup$ – Asaf Karagila Mar 5 '15 at 17:03
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Yes, because $\mathbb N \times \mathbb N$ is countable. Just pick any bases of the two spaces and any bijection $\mathbb N \times \mathbb N$ with $\mathbb N$ and it extends to an isomorphism.

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    $\begingroup$ What do I have to do with $\mathbb N \times \mathbb N$ ? $\endgroup$ – user217921 Mar 5 '15 at 14:18
  • $\begingroup$ Suppose $X$ and $Y$ are two vector spaces each with countable dimension. Find $f:\mathbb N\hookrightarrow X$ such that the image $f(\mathbb N)$ is a basis of $X$ (which you can do simply because of the definition of $X$ being a countably infinite dimensional space). Do the same for $Y$, $g:\mathbb N\hookrightarrow Y$. Then $f\times g:\mathbb N \times \mathbb N\hookrightarrow X\times Y$ is given by $(m,n) \mapsto (f(m),g(n))$. We know the image $(f\times g)(\mathbb N\times\mathbb N)$ is a basis of $X\times Y$ and we know $\mathbb N\times\mathbb N$ is countable. QED. $\endgroup$ – Gregory Grant Mar 5 '15 at 15:24
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    $\begingroup$ But the question didn't ask about $W\times V$, just about $W$ and $V$. $\endgroup$ – Asaf Karagila Mar 5 '15 at 17:04
  • $\begingroup$ Oh weird I think the question has been edited, originally I think it was about the product. $\endgroup$ – Gregory Grant Mar 5 '15 at 19:35
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Let $k$ be a field, and $V$, $W$ vector spaces over $k$ and $B_V$, $B_W$ $k$-bases of $V$ resp. $W$. Note that any isomorphism $V \to W$ restricts to a bijection $B_V \to B_W$ and vice versa any bijection $B_V \to B_W$ extends by linear extension to an isomorphism $V \to W$.

Therefore, if $B_V$, $B_W$ are countably infinite, $V$ and $W$ are isomorphic. But, for example, as $X$ and $P(X)$ are never of the same size, $k^{(\mathbb R)}$ and $k^{(P(\mathbb R))}$ are non-isomorphic uncountable dimensional vector spaces, where $k^{(X)} := \{f \colon X \to k\mid f^{-1}[k - \{0\}] \text{ is finite} \}$.

To the edit: Note that always $V \cong k^{(B_V)}$ (any vector can uniquely be respresented as a linear combination. Now the cardinality of $k^{(B_V)}$ is $$ \sum_{A \in [B_V]^{<\omega}} k^{|A|} $$ If $k$ is infinite, $k^{|A|} = k$ for any finite $|A|$, hence this sum equals $$ |k| \cdot |[B_V]^{<\omega}| = |k| \cdot |B_V| $$ If $k$ is finite, this sum equals $|[B_V]^{<\omega}| = |B_V|$, so in any case $|V| = \max(|B_V|, |k|)$. So the answer is: Only if the dimension is larger than the cardinality of the field.

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  • $\begingroup$ Could you please elaborate on the calculation of the cardinaltiy of $k^{B_V}$ ? $\endgroup$ – user217921 Mar 5 '15 at 14:49

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