Let $V,W$ are vector spaces over $\mathbb Q$ of same cardinality $>\aleph _0$ ; then how to show that $V,W$ are isomorphic i.e. they have bases of same cardinality ?
Due to the cardinality of them being $>\aleph_0$, I can see that both of them are infinite dimensional. If $A$ is a basis of $V$ over $\mathbb Q$, then we can write $V \cong \oplus_A \mathbb Q$ so $V$ is bijective with the set of all functions from $A$ to $\mathbb Q$ with finite support . But $\mathbb Q^*$ is bijective with $\mathbb N^*$ hence we can say that they are the set of all functions from$A$ to $\mathbb N$ with finite support . I am unable to say anything else. Please help . Thanks in advance
What you're missing is the fact that if $A$ is infinite, then the number of functions from $A$ to $\mathbb{N}$ with finite support is $|A|$.
This is a combination of two easy facts:
If $A$ is infinite, then the number of finite subsets of $A$ is $|A|$. Why? Let $\binom{|A|}{n}$ be the number of $n$-element subsets. When $n\geq 1$, $|A|\leq \binom{|A|}{n} \leq |A|^n = |A|$. So the number of finite subsets is $\sum_{n\in\omega} |A| = |A|$.
For any finite set $X$, the number of functions from $X$ to $\mathbb{N}$ is $\aleph_0^{|X|} = \aleph_0$ (or $1$ if $X$ is empty).
So to pick a function from $A$ to $\mathbb{N}$ with finite support, you can pick the support $X$, and then pick the function $X\to \mathbb{N}$. The number of choices is $|A|\cdot \aleph_0 = |A|$.
Now, to conclude, we find that the cardinality of an infinite dimensional vector space over $\mathbb{Q}$ is equal to the cardinality of its basis. So if $|V| = |W|$, then bases for $V$ and $W$ have the same size, and they're isomorphic.