\section{Maximal Connected \color{black}Level Sets of the IP Value Function}
A level set of the IP value function may consist of multiple subsets of the hyperrectangle $\cB$ that are not connected (see \thref{Tstep} and Figures \ref{fig:MC-level-A} and \ref{fig:MC-level-B}). Hence, the structure of optimal solutions may vary greatly within the same level set. Having access to connected \color{black}subsets of level sets for the recourse value function in these problems could allow for optimization with respect to a connected set \color{black} of right-hand sides in each subset as a subproblem with fixed second-stage value and known allowable variability with respect to those bounds. We explore connected \color{black}subsets of level sets, and in particular, we define and discuss properties of maximal connected \color{black}subsets of level sets.
Maximal connected \color{black}level lattices (MC-level lattices) are MC-level sets intersected with the lattice $\mathbb{Z}^{m}_+$\color{black}. Note that because we are interested in the connections between sets of right-hand sides, we assume throughout this section that $\cB = \mathbb{R}^m_+$ (or, equivalently, that each component of $\mathbf{b}$ is positive infinity). The removal of this assumption affects only \thref{MC_Adjacent}, which does not necessarily hold for MC-level sets that intersect an upper boundary plane of $\cB$.\color{black}
\begin{defn}\thlabel{ivcurve}
Given $\bbeta_1$, $\bbeta_2 \in \cB$, a continuous function $d:[0,1]\to\cB$ is a \textbf{continuous isovalue curve} from $\bbeta_1$ to $\bbeta_2$ if $d(0) = \bbeta_1$, $d(1) = \bbeta_2$, and $z(d(t)) = z(\bbeta_1)$ for all $t\in [0,1]$.
\end{defn}
Given disjoint connected subsets $M_1, M_2 \subset [0,1]$, we say that $M_1$ precedes $M_2$ in the ordering of $[0,1]$ if for any $t_1 \in M_1$, $t_2 \in M_2$, $t_1 < t_2$.\color{black}
\begin{defn}\thlabel{Tstep} For all ${\bm{\beta}} \in \cB$, define the \textbf{MC-level set} $T({{\bm{\beta}}}) \coloneqq \left\{\bar{\bm{\beta}}\in\cB\,\vert\right.$ there exists a continuous isovalue curve $d$ from ${\bbeta}$ to $\left.\bar{\bbeta}\right\}$. For all ${\bm{\beta} \in }$ \wenxin{$\bar{\mathcal{B}}$}, \color{black} define the \textbf{MC-level lattice} $C({\bm{\beta}}) \coloneqq T({\bm{\beta}})\cap \mathbb{Z}^m_+ $.
\end{defn}
\begin{continueexample}{exExampleIP}
In (EXIP), with $\cB$ unbounded above, $T((1,1)^{\top})$ is the unbounded set $\left\{\bbeta\,\vert\, \bbeta_1 \geq 1, \ 1\leq \bbeta_2 < 2\right\}$, and $C((1,1)^{\top}) =\left\{(v,1)\,\vert\,v \in \mathbb{N}\right\}$\color{black}.\hfill\qedsymbol\end{continueexample}
\begin{myl}\thlabel{easyiso}
Given $\bbeta_1\in \cB$, $\bbeta_2 \in T(\bbeta_1)$, there exists an isovalue curve $d': [0,1]\to\cB$ from $\bbeta_1$ to $\bbeta_2$ such that $\lfloor d'(\zeta)\rfloor = \lfloor d'(\theta)\rfloor$ implies $\lfloor d'(\zeta)\rfloor = \lfloor d'(\eta)\rfloor$ for all $0\leq \zeta<\eta<\theta\leq 1$  - that is, $\lfloor d'\rfloor$ takes on any given value for at most a single connected subset of $[0,1]$.
\end{myl}\color{black}

\begin{defn}
Given $\bbeta_1,\bbeta_2 \in \mathbb{Z}^m_+$, $\bbeta_1$ and $\bbeta_2$ are \textbf{adjacent} if there exists $j \in \{1, ..., m\}$ such that $\bbeta_1-\bbeta_2 = \pm\mathbf{e}_j$.
\end{defn}

\thref{Csteppro} shows that there exists a sequence of adjacent points in an MC-level lattice which connects any two of its members.
\begin{mypro}\thlabel{Csteppro}
Given ${\bm{\beta}}\in \bar{\mathcal{B}}$ and $\bm{\beta}_1$, $\bm{\beta}_2 \in C({\bm{\beta}})$, there exists a finite sequence of points, $V = \left\{v_1 = \bm{\beta}_1, ..., v_r = \bm{\beta}_2\right\}$, such that for all $i\in \{1, ..., r\}$, $v_i\in C({\bm{\beta}})$ and for all $i \in \{1, ..., r-1\}$, $v_i$ and $v_{i+1}$ are adjacent.
\end{mypro}
\begin{proof}
By \thref{Tstep} and \thref{easyiso}, there must exist a continuous isovalue curve $d':[0,1]\to\cB$ from $\bbeta_1$ to $\bbeta_2$ for which $\lfloor d'\rfloor$ takes on any given value for at most a single connected subset of $[0,1]$. Let $\Gamma = \left\{\bm{\gamma}^1 = \bbeta_1, ..., \bm{\gamma}^s = \bbeta_2\right\}$ be the ordered sequence of values taken on by $\lfloor d'\rfloor$, so that $\lfloor d(t_i)\rfloor = \bm{\gamma}^i$, $\lfloor d(t_j)\rfloor = \bm{\gamma}^j$, $t_i, t_j \in [0,1]$, and $i < j$ collectively imply $t_i < t_j$. Let $\mathbf{M} = \{M_1, ..., M_s\}$ be the ordered sequence of connected subsets of $[0,1]$ corresponding to the members of $\Gamma$ - that is, for any $i \in \{1,\dots,s\}$, $t \in M_i$ is equivalent to $\lfloor d'(t)\rfloor = \bm{\gamma}^i$. Denote the closure of $M_i$ as $\overline{M}_i$, and define {\tajayi{ the singleton $\tau_i$ such that $\{\tau_{i}\} = \overline{M}_i\cap\overline{M}_{i+1}$; because the members of $\mathbf{M}$ are disjoint and ordered, and their union is $[0,1]$, distinct $\tau_i$ will exist for each $i\in\{1, ..., s-1\}$.}} Note that either $\tau_i \in M_i$ or $\tau_i \in M_{i+1}$, and that since $d'$ is a continuous function over a bounded domain, each component of $d'$ is also bounded, so that $\Gamma$ is finite. Further, since $d'$ is continuous, $\Vert \bm{\gamma}^{i+1}-\bm{\gamma}^i\Vert_{\infty} = 1$ for all $i \in \{1, ..., s-1\}$.\par
Suppose there exists $i \in \{1, ..., s-1\}$ for which for some $j \in \{1, ..., m\}$, $\bm{\gamma}^{i+1}_j > \bm{\gamma}^i_j$ and for some $k \in \{1, ..., m\}$, $\bm{\gamma}^i_k > \bm{\gamma}^{i+1}_k$. There are two possibilities:

\vspace{12pt}
\setlength{\leftskip}{1cm}
Case 1: Suppose that both $M_i$ and $M_{i+1}$ include more than a single point. Since $d'$ is continuous, there must exist $\epsilon > 0$ such that for all $t \in (\tau_i-\epsilon, \tau_i)$, $\lfloor d'(t)\rfloor = \bm{\gamma}^i$ and for all $t \in (\tau_i, \tau_i+\epsilon)$, $\lfloor d'(t)\rfloor = \bm{\gamma}^{i+1}$. For any $t$ such that $\lfloor d'(t)\rfloor = \bm{\gamma}^i$, we must have $(\bm{\gamma}^{i+1}-d'(t))_j > 0$, so by the continuity of $d'$, we must have $d'(\tau_i)_j = \bm{\gamma}^{i+1}_j$. Similarly, for any $t$ such that $\lfloor d'(t)\rfloor = \bm{\gamma}^{i+1}$, we must have $(\bm{\gamma}^i-d'(t))_k > 0$, so that by the continuity of $d'$, we must have $d'(\tau_i)_k = \bm{\gamma}^i_k$. However, this implies that $\tau_i \notin M_i\cup M_{i+1}$, a contradiction. 

Case 2: Suppose on the other hand that $M_i$ is a singleton; then $M_{i} = \{\tau_i\}$. {\tajayi{Observe that, because $M_{i} \cap M_{i+1} = \emptyset$, if $M_{i+1}$ is a single point, $\overline{M_{i+1}} \cap \overline{M_{i}} = \emptyset$. This contradicts the continuity of $d'$; thus, $M_{i+1}$ contains more than a single point. Fix $\epsilon > 0$ such that $(\tau_{i}, \tau_{i} + \epsilon) \subseteq M_{i+1}$. For any $\delta \in (0, \epsilon)$, $d'(\tau_{i} + \delta)_{j} \geq \lfloor d'(\tau_{i} + \delta)_{j}\rfloor = \bm{\gamma}^{i+1}_{j} = \bm{\gamma}^{i}_{j} + 1$. By continuity, $d'(\tau_{i})_{j} = \lim\limits_{t \to \tau_{i}^{+}} d'(t)_{j} \geq \gamma^{i}_{j} + 1 = \lfloor d'(\tau_{i})_{j}\rfloor + 1 > d'(\tau_{i})_{j}$, a contradiction. An analogous argument leads to a similar contradiction when $M_{i+1}$ is a singleton. Thus, we conclude that for all $i \in \{1,\dots,s\},$ either $\gamma^{i} \leq \gamma^{i+1}$ or $\gamma^{i} \geq \gamma^{i+1}$.}}\par
\vspace{12pt}
\setlength{\leftskip}{0cm}
Define $\hat{\bm{\gamma}}^j$ such that $\hat{\bm{\gamma}}^j_k = \mathbbm{1}_{k\leq j}\bm{\gamma}^i_k + \mathbbm{1}_{k> j}\bm{\gamma}^{i+1}_k$ for all $k \in \{1, ..., m\}$, and for all $j \in \{0, ..., m\}$. Then for all $j \in \{0, ..., m\}$, $\bm{\gamma}^i\leq \hat{\bm{\gamma}}^j\leq \bm{\gamma}^{i+1}$ or $\bm{\gamma}^i\geq\hat{\bm{\gamma}}^j\geq\bm{\gamma}^{i+1}$, and for all $j \in \{0, ..., m-1\}$, either $ \hat{\bm{\gamma}}^j-\hat{\bm{\gamma}}^{j+1} =\pm \mathbf{e}_j$ or $\hat{\bm{\gamma}}^j-\hat{\bm{\gamma}}^{j+1} = \mathbf{0}$. By the monotonicity of the value function, $\hat{\bm{\gamma}}^j \in C(\bbeta)$ for all $j \in \{0, ..., m\}$. Inserting $\left\{\hat{\bm{\gamma}}^1, ..., \hat{\bm{\gamma}}^{m-1}\right\}$ in $\Gamma$ between $\bm{\gamma}^i$ and $\bm{\gamma}^{i+1}$ for each $i \in \{1, ..., s-1\}$, then removing sequentially adjacent duplicate values, will yield a sequence $V$ satisfying the conditions of the proposition statement.
\end{proof}\color{black}
\thref{finseq} shows an analogous result for any pair of points in an MC-level set.
\begin{cor}\thlabel{finseq}
Given ${\bm{\beta}}\in\cB$, $\bm{\beta}_a$ and $\bm{\beta}_b\in T({\bm{\beta}})$, there exists a finite sequence of points,\\$\left\{\bm{\gamma}^0 = \bm{\beta}_a, ..., \bm{\gamma}^r = \bm{\beta}_b\right\}$, such that $\bm{\gamma}^i\in T({\bm{\beta}})$ and $\vert \bm{\gamma}^{i+1}-\bm{\gamma}^i\vert = \lambda_i\mathbf{e}_j$ for all $i$, where $\lambda_i \in (0,1]$ for all $i$ and $\mathbf{e}_j$ is a unit vector for $j \in \{1, ..., m\}$.
\end{cor}
Note that other than the steps necessary to move from $\bbeta_a$ and $\bbeta_b$ to the nearest isovalue lattice points, all the intermediate points in the above sequence may be made to be lattice points, as can be seen in Figure \ref{fig:MC-level-C}.
\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{isopathbw.png} 
    \vspace{-12pt}
    \caption{An isovalue path from $\bbeta_a$ to $\bbeta_b$}
    \label{fig:MC-level-C}
\end{figure}\par
\begin{defn}\thlabel{adjhc}
An $m$-dimensional unit hypercube $H$ is \textbf{anchored} if all of its vertices are in $\mathbb{Z}^m$. In particular, $H(\bbeta)$, $\bbeta \in \mathbb{Z}^m$, is the anchored unit hypercube with $\bbeta$ as its vertex with all components minimal. 
\end{defn}
\begin{defn}
The \textbf{open-ceiling} unit hypercube anchored at $\bbeta \in \bar{\cB}$ is defined as $G(\bbeta) \coloneqq$ $\{\bar{\bbeta} \in$ $\cB\, \vert\,\lfloor \bar{\bbeta}\rfloor = \bbeta\}$.
\end{defn}
\begin{remark}
Given an open-ceiling unit hypercube $G(\bbeta)$, the closure of $G(\bbeta)$ is $H(\bbeta)$.
\end{remark}
\begin{defn}
Two anchored $m-dimensional$ unit hypercubes $H_1$ and $H_2$ are \textbf{adjacent} if $H_1\cap H_2$ is an anchored unit hypercube of dimension $m-1$.
\end{defn}
\thref{MC_Adjacent} shows that the closure of any MC-level set can be written as the union of a set of $m$-dimensional unit hypercubes anchored at integer points.
\begin{mypro}\thlabel{MC_Adjacent}
Given $\bbeta \in \cB\subset \mathbb{R}^m$, there exists a unique \color{black}set $P$ of anchored \color{black}unit hypercubes of dimension $m$ with integer-valued vertex coordinates such that $\bigcup\limits_{H \in P}H = \overline{T(\bbeta)}$, the closure of $T(\bbeta)$. Further, for all pairs $H_1$, $H_2\,\in P$, there exists a finite sequence $U = \{U_1, ..., U_t\}$ of hypercubes in $P$ such that $U_1 = H_1$, $U_t = H_2$, and $U_i$ and $U_{i+1}$ are adjacent for all $i \in \{1, ..., t-1\}$.\color{black}
\end{mypro}
\begin{proof}
Fix $\bbeta' \in T(\bbeta)$. Then the open-ceiling unit hypercube $G(\lfloor \bbeta'\rfloor)$ is a subset of $T(\bbeta)$, and $\bbeta' \in G(\lfloor \bbeta'\rfloor)$. As such, there exists a unique countable set $Q$ of $m$-dimensional open-ceiling anchored unit hypercubes, $Q\coloneqq \{Q_{r}\}_{r \in R}$, where $R$ is the index set for $Q$, such that $T(\bbeta) = \bigcup\limits_{r\in R}Q_r$ - that is, for all $\bar{\bbeta}\in T(\bbeta)$, there exists $r \in R$ such that $Q_r = G(\lfloor \bar{\bbeta}\rfloor)$. Let $P = \{\overline{Q_{r}}\}_{r \in R}$, and observe that $\overline{T(\bbeta)} = \bigcup\limits_{r \in R} \overline{Q_{r}}$ - that is, $P$ is a unique set of $m$-dimensional anchored unit hypercubes whose union is the closure of $T(\bbeta)$.\par
Next, fix $H_1$, $H_2$ in $P$. Let $\bbeta_1$ and $\bbeta_2$ anchor $H_1$ and $H_2$, respectively. Let $V = \{v_1 = \bbeta_1, ..., v_s = \bbeta_2\}$ be the sequence of adjacent points whose existence is guaranteed by \thref{Csteppro} given $\bbeta_1$ and $\bbeta_2$ as input points to connect, and define a finite sequence $U = \{H(v_1), ..., H(v_s)\}$, where $H(v_1) = H_1$ and $H(v_s) = H_2$. Since each of these points is in $T(\bbeta)$, the $m$-dimensional hypercube anchored at each point is a subset of $\overline{T(\bbeta)}$ and thus is an element of $P$. Further, since $v_i-v_{i+1}$ is a unit vector for all $i$, $H(v_i)\cap H(v_{i+1})$ is a unit anchored hypercube of dimension $m-1$, so $H(v_i)$ and $H(v_{i+1})$ are adjacent for all $i \in \{1, ..., s-1\}$.
\end{proof}
\color{black}\thref{MC_Adjacent} shows that the closure of an MC-level set can be constructed using anchored \color{black}unit hypercubes of dimension $m$\color{black}. In addition, given any two anchored unit hypercubes in that closure, we can find a finite sequence of adjacent anchored unit hypercubes connecting the two. In Figure \ref{fig:MC-level-A}, there is a sequence of adjacent dark squares between any two dark squares, since $T(\bbeta_1) = T(\bbeta_2)$. On the other hand, in Figure \ref{fig:MC-level-B}, although the closures of $T(\bbeta_4)$ and $T(\bbeta_5)$ intersect, they do not actually share a point - and therefore there is no sequence of isovalue adjacent squares from $T(\bbeta_4)$ to $T(\bbeta_5)$.
\begin{figure}[H]{
    \begin{minipage}[b]{0.43\textwidth}
    \centering
    \includegraphics[width=1\textwidth]{leftcompbw.png} 
    \caption{$T({\bm{\beta}}_1)= T({\bm{\beta}}_2)$}
    \label{fig:MC-level-A}
\end{minipage}
\hspace{0.5cm}
    \begin{minipage}[b]{0.43\textwidth}
    \centering
    \includegraphics[width=1\textwidth]{rightcompbw.png}
    \caption{$T({\bm{\beta}}_4)\cap T({\bm{\beta}}_5)= \emptyset$}
    \label{fig:MC-level-B}
    \end{minipage}
}\end{figure}
\par
We next demonstrate some membership properties of MC-level sets based on level-set minimal vectors.
\begin{cor}\thlabel{segment}
Given $\bar{\bm{\beta}}\in \cB$, for all $\hat{\bbeta}\in \cB$ such that $z(\bar{\bm{\beta}}) = z(\hat{\bbeta})$ and $\hat{\bbeta}\geq\bar{\bm{\beta}}$, we have $\hat{\bbeta}\in T(\bar{\bm{\beta}})$.
\end{cor}
\begin{proof}
By the monotonicity of $z$, $z(\hat{\bbeta}) = z(\bar{\bbeta}) \leq z(\bbeta') \leq z(\hat{\bbeta})$, for all $\bbeta' \in \left\{\bbeta \ \vert \ \bbeta = \lambda\bar{\bbeta} + (1 - \lambda)\hat{\bbeta}, \lambda \in (0,1)\right\}$ because $\bbeta' \in (\bar{\bbeta}, \hat{\bbeta})$. As such, $d(t) \coloneqq t\bar{\bbeta}+(1-t)\hat{\bbeta}$ is a continuous isovalue curve from $\bar{\bbeta}$ to $\hat{\bbeta}$, so $\hat{\bbeta}\in T(\bar{\bbeta})$.
\end{proof}
\begin{mypro}\thlabel{LSMcor}
For any $\bar{\bm{\beta}}\in\cB$, $\bm{\beta}\in T(\bar{\bm{\beta}})$ if and only if $z(\bm{\beta}) = z(\bar{\bm{\beta}})$ and there exists $\hat{\bm{\beta}} \in $ \wenxin{$\bar{\mathbf{B}}$} \color{black} $\cap T(\bar{\bm{\beta}})$ such that $\hat{\bm{\beta}}\leq \bm{\beta}$.
\end{mypro}
\begin{cor}\thlabel{LSMseq}
Let \wenxin{$\bar{\bbeta} \in \bar{\mathbf{B}}$} \color{black} such that there exists $\hat{\bbeta} \in \bar{\mathbf{B}}\cap T(\bar{\bm{\beta}})$, $\hat{\bbeta}\neq \bar{\bbeta}$. Then there exists a sequence $V = \{v_1, ..., v_s\}$ of distinct vectors in $\bar{\mathbf{B}}\cap T(\bar{\bm{\beta}})$ such that $\bar{\bbeta} = v_1$, $\hat{\bbeta} = v_s$, and for all $i \in \{1, ..., s-1\}$, there exists $\bbeta_i \in T(\bar{\bbeta})$ such that $v_i\lneq \bbeta_i$ and $v_{i+1}\lneq \bbeta_i$.
\end{cor}
\color{black}
\thref{LTstepdownt} discusses the relationship between a sequence of right-hand sides when ``stepping down" towards MC-level sets with smaller function value.
\begin{remark}\thlabel{optstat}
For any ${\bbeta} \in \cB$, let $\mathbf{x}^*\in \opt({\bbeta})$. Then for all $\bar{\bbeta} \in T({\bbeta})$ such that $\bar{\bbeta}\geq{\bbeta}$, $\mathbf{x}^*\in\opt(\bar{\bbeta})$.
\end{remark}
\begin{mypro}\thlabel{LTstepdownt}
For any $\bar{\bm{\beta}}\in$ \wenxin{$\bar{\mathbf{B}}$} \color{black} such that $z(\bar{\bm{\beta}}) > 0$, given ${\mathbf{x}}^* \in \opt(\bar{\bm{\beta}})$ and $\bbeta \in T(\bar{\bbeta})$ such that $\bar{\bbeta}\leq \bbeta$, then for all $j$ such that ${x}^*_j > 0$ and all $t \leq {x}^*_j$, $\bm{\beta}-t\mathbf{a}_j\in T(\bar{\bm{\beta}}-t\mathbf{a}_j)$.
\end{mypro}
\begin{proof}
Note that if $\bbeta = \bar{\bbeta}$ the result holds trivially. Suppose there exists $\hat{\bm{\beta}}\in T(\bar{{\bm{\beta}}})$, $\bar{\bm{\beta}}\lneq\hat{\bm{\beta}}$, such that there exists $t:\,1\leq t\leq {x}^*_j$ for which $\hat{\bm{\beta}}-t\mathbf{a}_j\notin T(\bar{\bm{\beta}}-t\mathbf{a}_j)$. Then there exists $\bm{\pi}\in\bar{\mathbf{B}}$ such that $\bar{\bm{\beta}}-t\mathbf{a}_j\lneq \bm{\pi}\leq \hat{\bm{\beta}}-t\mathbf{a}_j$ and \wenxin{$z(\bar{\bm{\beta}}-t\mathbf{a}_j)<z(\bm{\pi})=z(\hat{\bm{\beta}}-t\mathbf{a}_j)$}. \color{black}Let $\hat{\mathbf{x}}\in\opt(\bm{\pi})$. Then \wenxin{$z(\hat{\bm{\beta}}) \geq z(\bm{\pi}+t\mathbf{a}_j) \geq z(\bm{\pi})+tc_j$}. \color{black} However, by \thref{equal} and \thref{down},  $z(\bar{\bm{\beta}}-t\mathbf{a}_j) = z(\bar{\bm{\beta}})-tc_j = z(\hat{\bm{\beta}})-tc_j$, a contradiction since we have now claimed that $z(\hat{\bm{\beta}})-tc_j < z(\bm{\pi}) \leq z(\hat{\bm{\beta}})-tc_j$.
\end{proof}
\begin{continueexample}{exExampleIP}
    In (EXIP), with $\cB$ unbounded above, $(1,1)^{\top} \in \bar{\mathbf{B}}_6$ and $z((1,1)^{\top}) = 3$, with the unique optimal solution $\mathbf{e}_4$. Further, $(2,1)^{\top}, (3,1)^{\top}\in T((1,1)^{\top})$. So by \thref{LTstepdownt}, $(1,0)^{\top}, (2,0)^{\top} \in T((0,0)^{\top})$.\hfill\qedsymbol\end{continueexample}


\thref{bk=b} indicates that if, for any $k \in \{1, ..., n\}$, $\bar{\mathbf{B}}_k=\bar{\mathcal{B}}$, then for all $k' \geq k$, all MC-level sets will be unit hypercubes.
\begin{mypro}\thlabel{bk=b}
If there exists $k\in\{1,\dots,n\}$ such that $\bar{\mathbf{B}}_k=\bar{\mathcal{B}}$, then for all $\hat{k}\geq k,\hat{k}\in \{1,\dots,n\}$, $\bar{\mathbf{B}}_{\hat{k}}=\bar{\mathcal{B}}$.
\end{mypro}
{"config": "arxiv", "file": "2006.10223/betaStep.tex"}