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The textbook I use introduced the Taxicab "circle" using the set-notation, $$\{P\mid d_{T}(P,A)=1\}.$$ as the set of points, such that the distance between all the points and the center is equal to $1$ taxicab distance, effectively defining the unit-taxicab circle. A later exercise problem extrapolated this notation, and I am confused as to its meaning. I will show the problem, then give my own interpretation. I am looking for error correction and further explanations, either written or visual, if possible.

Graph $\{P\mid d_{T}(P,A)+d_{T}(P,B)=d_{T}(A,B)\}$.

This is the set of points such that the sum of the distances between $P$ and $A$, and $P$ and $B$ respectively, is equal to the distance between $A$ and $B$. Firstly, this is not a specific computation; rather, it is looking for something in the general case. Secondly, the set of points $P$ that is the answer is dependent on the distance between $A$ and $B$, thus I am to observe some property of the relationship between the sum of circles, which hardly makes sense to me — how does one "add" a geometrical figure? (a question perhaps for another time). The answer to this will be a taxicab circle in the end, arithmetically related to the two centers $A$ and $B$.

The furthest I've gotten is this. I suspect that the taxicab circles of centers $A$ and $B$ will overlap, and their overlapping will constitute some answer, but this is mere postulating, and I haven't any proof for my statement, and no bearing as to whether or not I am even in the right direction.

To reiterate, I am looking for

  1. error correction, and/or
  2. a written/visual explanation.
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…the set of points $P$ that is the answer is dependent on the distance between $A$ and $B$, thus I am to observe some property of the relationship between the sum of circles, which hardly makes sense to me…

Two points with this. Firstly, yes, the solution set depends on the distance $d_T{\left(A,B\right)}$, but it will scale uniformly, so you can set this to any distance you like without loss of generality.

Secondly, $d_T{\left(P,A\right)} + d_T{\left(P,B\right)}$ isn’t a sum of circles; it’s a sum of distances. The circle is the complete set expression.

This isn’t unique to taxicab geometry. A unit circle in standard Euclidean geometry could be stated as $\left\{P \mid d_E{\left(P,A\right)} = 1\right\}$, if we used $d_E$ for the Euclidean metric (Pythagoras’ theorem).

In fact, think about what $d_E{\left(P,A\right)} + d_E{\left(P,B\right)} = d_E{\left(A,B\right)}$ (or in more ordinary notation, $\left\|P-A\right\| + \left\|P-B\right\| = \left\|A-B\right\|$) would mean: “The total distance from $A$ to $P$ and then to $B$, is equal to the distance directly from $A$ and $B$.”

Since the points $A$ and $B$ are constant, the distance between them is constant too, and we can say instead, “The total distance from $A$ to $P$ and then to $B$ is a constant.” This is one definition of an ellipse. So we’re looking for the taxicab equivalent of an ellipse. A “circle” in taxicab geometry is a (Euclidean) square, so it makes a certain amount of sense that an ellipse would be a (Euclidean) rectangle—which is indeed the solution.

(In fact, in the general case it’s a rectangle with bevelled corners, an irregular octagon.)

The fact that this specific choice of distance results in a degenerate ellipse—the line segment between $A$ and $B$—is a slight stumbling block, it’s true. But this happens because the shortest Euclidean path between two points is unique.

Two points surrounded by concentric circles, marked (using colours) as belonging to matched pairs, one around each point. As the circles around one point get larger, their matching partners around the other point get smaller. As a result, a matched pair of circles can only ever touch at a single point.

(In this image, pairs of circles in the same colour have the same total distance from the two centres; bigger circles around one centre match smaller circles around the other. Same-colour circles only ever intersect in one point.)

In contrast, in taxicab geometry, there’s an infinitude of shortest paths, so long as the two points aren’t on the same horizontal or vertical line.

Another way to think of it is that distinct Euclidean circles can only intersect at one or two points, and the distance constraint makes it one point in this case. But taxicab circles are Euclidean squares with axis-aligned diagonals, and these can intersect at one or two points, or along a partial or complete edge. If $A$ and $B$ lie on the same horizontal or vertical line, then once again we get the “one point” case. But if not…

Two points surrounded by concentric squares, marked (using colours) as belonging to matched pairs, one around each point. As the squares around one point get larger, their matching partners around the other point get smaller. Each matched pair of squares intersects along part of one edge, and these intersections fill a rectangular area.

…then the squares intersect along part of an edge, and these intersections form the perimeter and interior of a rectangle.

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This is not about "adding sets" (though such a notion does exist—you might search the term "Minkowski sum"). Rather, the set is defined by an equation, and you are being asked to graph the set of solutions to that equation.

You might start by trying to solve this problem in a more familiar setting: let $d_{\mathrm{E}}$ denote the usual Euclidean metric, i.e. $$ d_{\mathrm{E}}\bigl( (x_1, y_1), (x_2,y_2) \bigr) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}. $$ Then, given two points $A$ and $B$, the set $$\{ P \mid d_{\mathrm{E}}(P,A) + d_{\mathrm{E}}(P,B) = d_{\mathrm{E}}(A,B)\} $$ is the collection of all points $P$ such that the sum of the distances from $P$ to $A$ and $B$ is equal to the distance from $A$ to $B$. Geometrically, this is the collection of all points $P$ which form a "triangle" where the length of the longest side is equal to the sum of the lengths of the other two sides. Such a triangle is degenerate, and the figure formed is simply the segment $\overline{AB}$.

A reasonable exercise would be to verify this algebraically, using the definition of the metric.

The taxicab metric is defined differently. The intuition is that it is the distance that a cab would have to travel from one point to another if it is restricted to traveling only in directions parallel to the axes (it is restricted to driving on streets that form a grid). This can be written as

$$ d_{\mathrm{T}}\bigl( (x_1, y_1), (x_2, y_2) \bigr) = \lvert x_1 - x_2\rvert + \lvert y_1 - y_2\rvert. $$

Observe that \begin{align} &d_{\mathrm{T}}(P,A) + d_{\mathrm{T}}(P,B) = d_{\mathrm{T}}(A,B) \\ &\qquad\iff \left( \lvert x_P - x_A\rvert + \lvert y_P - y_A\rvert \right) + \left( \lvert x_P - x_B \rvert + \lvert y_P - y_B\rvert \right) = \lvert x_A - x_B \rvert + \lvert y_A - y_B \rvert, \end{align} where $A = (x_A, y_A)$, $B = (x_B, y_B)$, and $P = (x_P, y_P)$. It can be shown that this holds if and only if $x_P$ is between $x_A$ and $x_B$ and $y_P$ is between $y_A$ and $y_B$. That is, either $$ x_A \le x_P \le x_B \qquad\text{or}\qquad x_A \ge x_P \ge x_B, $$ with a similar condition on the $y$-coordinates. In other words, if $P$ is located both horizontally and vertically between $A$ and $B$, then the taxicab distance from $A$ to $B$ will be equal to the sum of the taxicab distances from $P$ to each of the other two points.

To get some intuition for this, suppose that $A$ is to the northwest of $B$. To get from $A$ to $B$, a taxi can drive east until it is due north of $B$, and then drive south directly to $B$. If $P$ is any point southeast of $A$ and northwest of $B$, then the another cab could drive due east to a point directly north of $P$, drive due south to $P$, then go due east to a point due north of $B$, and finally drive due south to $B$. Both taxis will have driven the same distance: the distance east from $A$ to $B$, and the distance east from $A$ to $B$.

Thus any point $P$ in a rectangle with opposite vertices at $A$ and $B$, and sides parallel to the axes, will satisfy the given identity. So the desired set is a rectangle.

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As @XanderHenderson notes in his excellent answer, this is not about adding sets, it's about adding distances - in this case taxicab distances. It's not about circles either.

One way to think of the set in your question is that given fixed points $A$ and $B$, it contains all the points $P$ you can visit on a trip from $A$ to $B$ without making the total trip any longer.

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    $\begingroup$ “all the points $P$ you can visit on a trip from $A$ to $B$ without making the total trip any longer”—A clever intuitive explanation! $\endgroup$ Commented 2 days ago

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