Non-translation invariant Gibbs state of the Ising model In the book "Statistical mechanics of lattice systems" by Sacha Friedli and Yvan Velenik, on page 157 we have the following theorem concerning non translation invariant Gibbs states of the Ising model in dimension $d$:

Theorem 3.58: Assume $d\geq 3$, then for all $\beta>\beta_c(d-1)$, there exists a sequence of integers $n_k\uparrow\infty$ along which
$$\langle\cdot\rangle^{Dob}_{\beta,0}=\lim_{k\rightarrow\infty}\langle\cdot\rangle^{Dob}_{\Lambda^d(n_k);\beta,0}$$
is a well-defined Gibbs state that satisfies $$\langle\sigma_0\rangle^{Dob}_{\beta,0}>0>\langle\sigma_{\overline{0}}\rangle^{Dob}_{\beta,0}$$

where the $\beta$ is the inverse tenperature, $\beta_c$ is the critical temperature, and the Dobrushin boundary condition and other relevant notation can be found on page 157.
On page 159, we have the proof:

The construction of $\langle\cdot\rangle^{Dob}_{\beta,0}$ along some subsequence can be done as in exercise 3.8 (I think by Prokhorov's theorem). Observe that, by symmetry, $$\tag{3.88} \langle\sigma_0\rangle^{Dob}_{\Lambda^d(n_k);\beta,0}=-\langle\sigma_{\overline{0}}\rangle^{Dob}_{\Lambda^d(n_k);\beta,0}$$ which gives after $k\rightarrow\infty$, $$\tag{3.89}\langle\sigma_0\rangle^{Dob}_{\beta,0}=-\langle\sigma_{\overline{0}}\rangle^{Dob}_{\beta,0}.$$ Observe that, by the FKG inequality, applying a amgentic field $h\uparrow\infty$ on the spins living in $B^d(n_k)-\Lambda$ yields $$\langle\sigma_0\rangle^{Dob}_{\Lambda^d(n_k);\beta,0}\geq\langle\sigma_0\rangle^{Dob}_{B^d(n_k);\beta,0}.$$

My question is: what does it mean by "applying a magnetic field $h\uparrow\infty$"? How is the FKG inequality applied here? Maybe the authors mean conditioning on "all spins in $B^d(n_k)-\Lambda$ being 0"?
 A: If I understand correctly, the question is really about the FKG inequality. If you'd like me to also address some other aspects, just tell me.
Let us consider an Ising model in a finite set $\Lambda\Subset\mathbb{Z}^d$ and some arbitrary boundary condition $\eta$. I assume that there is a site-dependent magnetic field $h_i$, that is, the Hamiltonian in $\Lambda$ is defined on configurations in $\Omega_\Lambda^\eta$ by
$$
\mathcal{H}_\Lambda = - \beta \sum_{\{i,j\}\in\mathcal{E}_\Lambda^{\rm b}} \sigma_i \sigma _j - \sum_{i\in\Lambda} h_i \sigma_i,
$$
where $\beta\geq 0$ and, for each $i\in\Lambda$, $h_i$ is a real number (possibly negative). Let $f$ be some nondecreasing local function. A simple computation shows that
$$
\frac{\partial}{\partial h_j} \langle f \rangle^\eta_{\Lambda;\beta,(h_i)_{i\in\Lambda}} = \langle f\sigma_j \rangle^\eta_{\Lambda;\beta,(h_i)_{i\in\Lambda}} - \langle f \rangle^\eta_{\Lambda;\beta,(h_i)_{i\in\Lambda}} \langle \sigma_j \rangle^\eta_{\Lambda;\beta,(h_i)_{i\in\Lambda}} \geq 0 ,
$$
where the inequality is a consequence of the FKG inequality. In particular, the expectation $\langle f \rangle^\eta_{\Lambda;\beta,(h_i)_{i\in\Lambda}}$ is a nondecreasing function of the magnetic fields $(h_i)_{i\in\Lambda}$.
Coming back to your question: in the application you mention, we apply to all the vertices of the set $B^d(n_k)\setminus \Lambda^d(n_k)$ a magnetic field $h\geq 0$. By the monotonicity described above, this increases the expectation value (notice that $\sigma_0$ is a nondecreasing local function):
$$
\langle \sigma_0 \rangle^{\rm Dob}_{B^d(n_k);\beta,0}
\leq
\langle \sigma_0 \rangle^{\rm Dob}_{B^d(n_k);\beta,(h_i)_{i\in B^d(n_k)}}
$$
where $h_i=h\geq 0$ if $i\in B^d(n_k)\setminus \Lambda^d(n_k)$ and $h_i=0$ otherwise. Now, we can take the limit $h\to+\infty$. In this limit, the Gibbs measure concentrates on configurations $\omega\in\Omega_{B^d(n_k)}^{\rm Dob}$ such that $\sigma_i(\omega)=1$ for all $i\in B^d(n_k)\setminus \Lambda^d(n_k)$. Therefore, the previous inequality yields
$$
\langle \sigma_0 \rangle^{\rm Dob}_{B^d(n_k);\beta,0}
\leq
\langle \sigma_0 \rangle^{\rm Dob}_{\Lambda^d(n_k);\beta,0},
$$
since, under the Dobrushin boundary condition applied to the box $\Lambda^d(n_k)$, all spins in $B^d(n_k)\setminus \Lambda^d(n_k)$ indeed take the value $+1$.

Notice that all these shenanigans with the boxes $B^d(n)$ and $\Lambda^d(n)$ are necessary, because van Beijeren's argument requires one $n+1$ layers of spins with $+$ boundary condition (all the layers at height from $0$ to $n$), but only $n$ layers with $-$ boundary condition (those at height from $-n$ to $-1$). This makes the situation asymetrical w.r.t. $+$ and $-$ spins in the boundary condition. The above manupulations allow one to recover a symmetrical situation (in the box $\Lambda^d(n)$).
