Let $S$ be an arbitrary set and $B(S)$ denote the space of all real-valued bounded functions on $S$. Then we know $B(S)$ is a lattice with pointwise maximum or minimum as the lattice operations. We equip $B(S)$ with the topology of uniform convergence too. Let $\mathcal{A}$ be a closed sublattice of $B(S)$.
Now, define $W=$ {$f \in B(S)\colon f= g+\alpha\mbox{ for some }g \in \mathcal{A}\mbox{ and }\alpha \in \mathbb{R}$}. Then how to prove that it is a closed lattice cone, that is, to prove that $W$ is a convex cone in $B(S)$ and closed under the topology of uniform convergence and lattice operations?
It is easy to prove that $W$ is a convex cone and that it is closed under the topology of uniform convergence. I am stuck at showing that $W$ is closed under lattice operations. Given $x_1, x_2 \in W$, consider $x_1 \wedge x_2 = (y_1 + \alpha_1) \wedge (y_2 + \alpha_2)$, for some $y_1,y_2 \in \mathcal{A}$ and $\alpha_1, \alpha_2 \in \mathbb{R}$. Then it is easy to see that $x_1 \wedge x_2 \geq (y_1 \wedge y_2) + (\alpha_1 \wedge \alpha_2)$. Is the reverse inequality possible?
Any help/hint is truly appreciated.