Let $n \in \{1, 2, \dots\}$ and let $C \in \mathbb{R}_{n, n}$ be a real, symmetric and positive definite $n \times n$ matrix. Define $B \in \mathbb{R}_{n, n}$ to be the real, lower triangular matrix with positive diagonal entries, such that $C = BB^T$ is $C$'s Cholesky's decomposition.
Is it true that, for every pair of real column vectors $x, y \in \mathbb{R}_{n, 1}$ of length $n$, we have $$ x \leq y \implies B^{-1}x \leq B^{-1}y $$ where the inequalities are to be interpreted component-wise?
Is it true that $$ \lim_{y \rightarrow \infty} B^{-1}y = \infty $$ where $y \in \mathbb{R}_{n, 1}$ "lives" in the set of real column vectors of length $n$, and convergence to infinity is to be interpreted component-wise?
1) Is not true. This is only true if all entries of $B^{-1}$ are non-negative. Assume the $(i,j)$-th entry of $B^{-1}$ is negative. Then choosing $x = e_j$, $y=2e_j$ yields $B^{-1}x\not\le B^{-1}y$.
2) Is not true. Not all components of $B^{-1}y$ will converge to infinity: Take $B^{-1}=\pmatrix{1 & -1 \\ 0&1}$. Then $$ B^{-1} \pmatrix{ k \\ k } = \pmatrix{ 0 \\ k}. $$
The claim is true if convergence to infinity is meant as $\|y\|\to\infty$.