TITLE: Minimal bridging sets in infinite connected graphs
QUESTION [5 upvotes]: Let $G=(V, E)$ be a connected, simple, undirected graph. We say that $B\subseteq V$ is a bridging set if $B\neq V$ and removing $B$ makes the graph disconnected, or more formally: $$G \setminus B := (V\setminus B\;, \; \{e\in E: e\cap B = \varnothing\})$$ is not connected any more.
Is there an infinite connected graph $G=(V,E)$ such that for every bridging set $B\subseteq V$ there is a bridging set $B_1$ with $B_1\subseteq B$ and $B_1\neq B$?

REPLY [6 votes]: Yes.
Take for $V = S \coprod T$ with bijections $s : \mathbb N \to S$ and $t : \mathbb N \to T$ ; and have the edge set $K_T \cup \{(s(i),t(j)) \quad\mathtt{ iff }\quad i \le j\}$.

First, see that $\forall n$, $t(\mathbb N + n)$ is a bridging set. Indeed, $s(k)$ for $k \ge n$ is isolated. This gives an infinite strict chain of bridging sets, whose limit is empty (so not a bridging set).
Let $B$ be a bridging set. Let's prove that $B \supset t(\mathbb N + k)$ for a $k$. Suppose it's not the case, that is : $\forall k \in \mathbb N, \exists N(k), t(N(k)) \notin B$. Then take two nodes $u,v$, those nodes are in $T \cup s(\mathbb N + n)$, so both have $t(N(n))$ as a neighbourg.