If $\dim(V) $ is infinite, show that $V\oplus V$ is isomorphic to $V$ For a vector space $V$ of infinite dimension, to show that $V\oplus V$ is isomorphic to $V$ is to show that there exists an invertible linear transformation between $V \oplus V $ and $V$.
Every vector space have a basis.
If $B$ is the infinite set of basis for $V$, then the set $B\oplus0 \cup 0\oplus B$ is the basis for $V \oplus V$.  Using axiom of choice the cardinality $|B\oplus0 \cup 0\oplus B|= |B\oplus0|+|0 \oplus B|=|B|+ |B| = \max\{|B|,|B|\}= |B|$ and then I was trying to argue that this implies that there is an isomorphism between the bases of $V\oplus V$ and $V$, so they are isomorphic. But I'm not sure how $|B|+|B|=\max\{|B|,|B|\}= |B|$ is true by the axiom of choice. How could we show that if $V$ is infinite dimensional vector space, then $V\oplus V \cong V$?
 A: Your argument is good, you just need to finish it up.
Assuming choice, you can consider a basis $\mathcal{B}$ of $V$. Then the set $\mathcal{C}=\{(v,0):v\in\mathcal{B}\}\cup\{(0,v):v\in\mathcal{B}\}$ is a basis for $V\oplus V$.
Since $\mathcal{B}$ is infinite, there is a bijection $f\colon\mathcal{C}\to\mathcal{B}$, because $\mathcal{C}$ is just the disjoint union of two copies of $\mathcal{B}$. Then the linear map $T\colon V\oplus V\to V$, defined on $\mathcal{C}$ by $T(w)=f(v)$ and extended by linearity, is an isomorphism, because it sends a basis onto a basis.
Note that choice is needed in two places: to obtain a basis for $V$ and to use that $|X|+|X|=|X|$ when $X$ is an infinite set.
A: Since your vector space $V$ is infinite-dimensional, let $b_1, b_2, \dots, b_n, \dots$ be a basis of $V$. Then the set
$$(B \oplus 0) \cup (0 \oplus B) = \{b_1 \oplus 0, b_2 \oplus 0, \dots, 0 \oplus b_1, 0 \oplus b_2, \dots\}$$
forms a basis of of $V \oplus V$.
Assuming the dimension of $V$ is countable, the dimension of $V \oplus V$ is countable too. Can you come up with an invertible linear transformation yourself, by thinking of how you show that $|\mathbb{N}| = |\mathbb{Z}|$?
A: I think there's a clearest way of stating this :
The Axiom of choice gives you a basis $B=(b_i)_{i\in I}$ where $I$ (and incidentally $B$) is infinite and may or may not be countable.
Now given $J_1,J_2\subset I$ such that $J_1\cup J_2=I$ and $J_1\cap J_2=\emptyset$, it is clear that $X_1=(b_j)_{j\in J_1}$ and $X_2=(b_j)_{j\in J_2}$ are such that $span(X_1)\oplus span(X_2)=V$.
Now is it possible to find $J_1,J_2$ such that $span(X_1)$ and $span(X_2)$ are each isomorphic to $V$? (In which case $V\oplus V\cong span(X_1)\oplus span(X_2)=V$)
This boils down to the following question :


*

*Given an infinite set $I$, is it possible to find $J_1,J_2\subset I$ such that $J_1\cup J_2=I$ , $J_1\cap J_2=\emptyset$ and $J_1\sim I$ and $J_2\sim I$ ($\sim$ being here used as a (non-standard) way of saying "there exists a bijection between the two")


Now if $I$ is infinite, it is in bijection with $I\times\{0,1\}$ (using AC).
Now if $f$ is your bijection just set $J_1=f^{-1}(I\times\{0\}),J_2=f^{-1}(I\times\{1\})$.
Qed
