$V = X \oplus Y$. Will dim$V$ = dim$X$ + dim$Y$ Hold If $V$ Is Infinitely-Dim? I don't understand much about the infinite-dimensional vector space.
Will the statement $V = X \oplus Y \rightarrow$ dim$V$ = dim$X$ + dim$Y$ hold in a infinitely-dimensional case?
Let $V$ be infinitely-dimensional, so either $X$ or $Y$ is infinitely-dimensional.
$B_X = \{x_1, x_2 ....\}$ is a basis of $X$, $B_Y = \{y_1, y_2 ...\}$ a basis of $Y$.  Show that $B_X \cup B_Y$ be a basis of $V$.  
Let $v \in V$ be given.  $v = x + y$ with $x \in X$, $y \in Y$.  So, $v = \sum_{k=1}^m a_kx_k + \sum_{j=1}^n a_ky_k \in$ span$\{x_1, x_2 ... y_1, y_2 ...\}$.  Conversely, since $B_X, B_Y \subset V$, a vector space, span$(B_X \cup B_Y) \subset V$.
Now, let $\{x_1, x_2 ....\}, \{y_1, y_2 ....\}$ be any subsets of $B_X, B_Y$ respectively.
Since $V = X \oplus Y$, $0 = 0 + 0$, with $0 \in X\cap Y$.  So $0 = 0 + 0 = \sum_{k=1}^m a_kx_k + \sum_{j=1}^n a_ky_k \leftrightarrow a_k = 0, b_j = 0$ for all $k = 1$ to $m, j = 1$ to $n$. Since $\{x_1, x_2 ....\}, \{y_1, y_2 ....\}$ are arbitrary, $B_X \cup B_Y$ is linearly independent.
$B_X \cup B_Y$ is a basis of $V$, entailing that dim$V = |B_x \cup B_Y|$.  Because $X \cap Y = \{0\}$, we have dim$V = |B_x \cup B_Y| = |B_X| + |B_Y|$.
I cannot see why the proof will fail if $V$ is infinitely-dimensional.  Since we are not giving dim$V$ a specific number, I do not see a problem in the last line either, dim$V = |B_x \cup B_Y| = |B_X| + |B_Y|$.

Yes, this works. The dimension of an infinite dimensional vector space is an infinite cardinal, and the sum operation is the usual cardinal addition ($|A| + |B| = |A\sqcup B|$, where $\sqcup$ is disjoint union).
