Where is the "thread" of a river?

Question

Lawyers speak of the "thread" of a river. When the boundary between two counties or states is a river, it is usually the "thread" of the river, a path running along the center of the river. (In England, I suspect this has applied primarily to boundaries between counties.) If a line is drawn from a point on one bank to a corresponding point on the opposite bank, one can take the midpoint to be a point on the "thread", but which point on the opposite bank is the corresponding point when the river meanders erratically, as they all do (but some more than others)? I am inclined to doubt that lawyers have clearly answered this, and to suspect that some substantial components of any reasonable answer can be contributed only by mathematicians.

Exceptions include:

• The northern boundary of Kentucky. The whole of the Ohio River is in Kentucky; the states north of it, Ohio, Indiana, and Illinois, begin at the north bank. I think this goes back to an act of Parliament after the Seven Years' War in 1763, declaring the Ohio to be the southern boundary of Canada, which became an English colony at that time. (This is alluded to in the language about a "fit instrument" in the Declaration of Independence.)
• The boundary between Delaware and New Jersey, which puts the Delaware River entirely in Delaware, so that New Jersey begins at the west bank.
• The boundary between Vermont and New Hampshire (surely the most colorful story of weird politics in the history of the U.S.A., if not of the Universe). The Connecticut River is in New Hampshire; Vermont begins at the west bank. On July 20, 1764, King George III decreed this after a scandalously ex-parte hearing, with the later result that the half of the Constitution of Vermont before its amendment-in-its-entirety in 1793 was a list of grievances against the government of New York, and the further later result that in a lawsuit that lasted from 1915 through 1933, a federal court ended up upholding the King's decision.

But dozens of boundaries between states in the U.S.A. follow the "thread" of a river, if I'm not mistaken.

Answer 1

Notation

Let $d(x,y)$ be the distance between $x$ and $y$ for any $x,y \in \mathbb{R}^3$.

Let $d(x,S) = \inf_{y \in S} d(x,y)$ for any $x \in \mathbb{R}^3$ and $S \subseteq \mathbb{R}^3$.

Let $d(S,T) = \inf_{(x,y) \in S \times T} d(x,y)$ for any $S,T \subseteq \mathbb{R}^3$.

A possible definition of a river's thread

Take any non-empty closed disjoint sets $A,B \in \mathbb{R}^3$ [corresponding to the two banks].

Let $T = \{ x : d(x,A) = d(x,B) \}$ for any $r \in [0,1]$.

Then $T$ can be considered the middle boundary between $A$ and $B$.

Properties of this thread

Regions

Let $A' = \{ x : d(x,A) < d(x,B) \}$.

Let $B' = \{ x : d(x,B) < d(x,A) \}$.

Then $A',B'$ are clearly open since $d$ is continuous and hence $T$ is closed.

Closest-Point

For any $x \in \mathbb{R}^3$:

  $d(x,A) = d(x,y)$ for some $y \in A$ because:

    Let $(y_n)$ be a sequence in $A$ such that $d(x,y_n) \to d(x,A)$ as $n \to \infty$.

    Then $(y_n)$ is bounded and has some cluster point $y$ [by Bolzano-Weierstrass].

    Thus $y \in A$ [because $A$ is closed].

  [This means that there is a point in $A$ closest to $x$. Similarly for $B$.]

Disjoint-Separation

$T,A',B'$ are clearly disjoint.

$A \subseteq A'$ because:

  For any $x \in A$:

    If $d(x,B) = 0$:

      Let $y \in B$ such that $d(x,y) = d(x,B)$ [by Closest-Point].

      Then $x = y \in B$.

      Contradiction.

    Therefore $d(x,B) > 0 = d(x,A)$ and hence $x \in A'$.

Similarly $B \subseteq B'$ and hence $T$ is disjoint from both $A$ and $B$.

No-holes

For any topological sphere $C \subseteq A'$ having no point of $B$ in its interior:

  Let $D$ be the topological ball corresponding to $C$.

  For any $x \in D$:

    Let $y \in B$ such that $d(x,y) = d(x,B)$ [by Closest-Point].

    Let $L$ be the line segment from $x$ to $y$.

    Let $z = L \cap C$ [which exists because $B$ is outside $C$].

    Then $d(x,A) \le d(x,z) + d(z,A) < d(x,z) + d(z,B) \le d(x,z) + d(z,y) = d(x,y)$

    $= d(x,B)$.

  Therefore $D \subseteq A'$.

Connected-Separation

If $A$ is connected, then $A'$ is path-connected because:

  For any $x \in A'$:

    Let $y \in A$ such that $d(x,y) = d(x,A)$ [by Closest-Point].

    Let $L$ be the line segment from $x$ to $y$.

    If $z \in B'$ for some $z \in L$:

      $d(x,B) \le d(x,z) \le d(x,y) = d(x,A)$.

      Contradiction.

    Therefore $L \subseteq A'$.

    Thus $L \cup A$ is connected.

  Therefore $A'$ is connected and hence path-connected because $A'$ is a open subset of $\mathbb{R}^3$.

Therefore if $A,B$ are both connected, then $A',B'$ will be both path-connected [and the whole space $\mathbb{R}^3$ will be divided into exactly two 'pieces' separated by $T$]. I believe but cannot prove that $T$ will in this case be connected. It is not necessarily path-connected; $T$ can be the Koch snowflake if $A,B$ are the inside and outside of it, in which case no two points in $T$ are path-connected.

Comments

$x \mapsto \dfrac{d(x,A)}{d(x,A)+d(x,B)}$ is a continuous function that is $0$ on exactly $A$ and $1$ on exactly $B$ and $\frac{1}{2}$ on exactly $T$. See Urysohn function for a generalization.

If either $A$ or $B$ is bounded, then $d(A,B) > 0$. Otherwise it is possible that $d(A,B) = 0$ despite $A,B$ being closed and disjoint. But this has nothing to do with the river thread so it belongs in this section.

Whether this definition is 'used' in real life I do not not know.