Could two complement spaces of two isomorphic subspace be non-isomorphic? I met this on p.93 of GTM135, S. Roman: "Advanced Linear Algebra", where it says for vector spaces:


*

*It is possible that $A\oplus B=C\oplus D$, with $A\approx C$ but $B\not\approx D$.


*It is possible that $(S\oplus B)\approx(S\oplus D)$ but $B\not\approx D$.

I cannot figure out any examples.
For (1), Since $A\oplus B=C\oplus D$, so $\dim(A)+\dim(B)=\dim(C)+\dim(B)$, but since $A\approx C$, so $\dim(A)=\dim(C)$, and as a result we get $\dim(B)=\dim(D)$ and since we are talking about vector spaces, we must have $B\approx D$.
Did I make any mistakes? Could anyone give some examples for the above two statements?
 A: The key observation here is that 

two vector spaces are isomorphic if and only if their bases have the same cardinality. 

Take any infinite dimensional space $V$, over a field $\mathbb F$, with basis $\{e_j\}_{j\in J}$. Let $h,k\in J$. Now let 
$$
A=\text{span}\,\{e_j:\ j\ne k\},\ \ \ 
B=\text{span}\,\{e_k\},$$
$$ 
C=\text{span}\,\{e_j:\ j\ne h,\ j\ne k\},\ \ \ \ 
D=\text{span}\,\{e_h,e_k\}. 
$$
Then $A\oplus B=C\oplus D$, $A\simeq C$, $B\not\simeq D$. 

Similarly, take $S$ any infinite-dimensional space. Take $T,U$ be vector spaces, over the same field, of dimensions 1 and 2 respectively. By the observation, 
$$
S\oplus T\simeq S\oplus U,
$$
while $T\not\simeq U$. 
A: As has been noted in the comments, the issue with your argument is the implicit assumption that all the vector spaces in question are finite-dimensional. So it makes sense to look for a counterexample involving infinite-dimensional vector spaces.
For $(1)$, let $V=F^{\infty}$ be the vector space of all sequences $\{x_n\}_{n=1}^{\infty}$ taking values in a given field $F$, and consider the subspaces $A=\{x\in V:x_1=0\}$, $B=\{x\in V:x_j=0$ for $j\geq 2\}$, $C=\{x\in V:x_1=x_2=0\}$, and $D=\{x\in V:x_j=0$ for $j\geq 3\}$.
Then $A\oplus B=V=C\oplus D$, and the map $T:V\to V$ given by
$$ T(x_1,x_2,\dots)=(0,x_1,x_2,\dots)$$
restricts to an isomorphism $A\simeq C$. But $B$ and $D$ are not isomorphic because they have different dimensions.
