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(Gonna answer my own question, as is encouraged.)

To set the stage: an arc (or a Jordan arc) is a non-self-intersecting curve with two distinct endpoints. (For those who are familiar with topology, it's a subset of the plane homeomorphic to $[0,1]$.) So any of these, basically; you're allowed fractals if you want, as long as they don't intersect themselves and have two definite endpoints.

enter image description here

(I'm calling them "arcs" rather than something simple like "curves" is because (a) in mathematics they commonly go under the name Jordan arcs and (b) I want to emphasize that I don't count loops.)

The multiset sum of two sets is like their union, but counting multiplicity. As an example, the multiset sum of a / shape and a \ shape would be a X shape where the center has multiplicity two and all other points have multiplicity one. The multiset sum of > and < could make the same shape.

Show that it is possible for the multiset sum of two arcs to equal the multiset sum of three arcs.

(Note that breaking a line into two pieces won't work because the multiplicity won't be correct where they join.)

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Here's one solution. See this album for larger images. The multiset sum of the two arcs on the left, when placed on top of each other, equals the multiset sum of the three arcs on the right when placed together.

enter image description here

Here's the final configuration (click here for larger). (Unlike in the three images above,) dots mark points of multiplicity two. (Note that the center has multiplicity two as well. The fact that the multiplicity of the center is the same on both sides of the equation is, to me, the most remarkable part of this.)

enter image description here

In words: shoot a line through the heart of a double spiral.

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  • $\begingroup$ The most astounding part of this is there's no reasonable "finite approximation" to this. If you take a double spiral that doesn't go spiral infinitely inwards, such as a Fermat spiral, and draw a line through it, you'll have trouble dividing it into three pieces. $\endgroup$
    – Akiva Weinberger
    Commented Dec 14, 2022 at 5:42
  • $\begingroup$ Note that the total length of your spiral needs to be finite in spite of the infinitely spiraling inwards. This can be arranged if the diameter of the spirals shrinks sufficiently quickly. $\endgroup$
    – quarague
    Commented Dec 14, 2022 at 20:19
  • $\begingroup$ @quarague There isn't anything in the phrasing of the question that requires the length to be finite (in fact, one of the example arcs was a fragment of the Koch snowflake, which has infinite length). That said, I used a logarithmic spiral in the picture, which does indeed have finite length $\endgroup$
    – Akiva Weinberger
    Commented Dec 14, 2022 at 21:56
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    $\begingroup$ @quarague In fact you can. Homeomorphism is all about continuity. Unlike, say, differentiable functions, continuous functions can have infinite speed. (Checking the definition of homeomorphism, we'll have to show that can define a continuous bijection between the unit interval and the Koch snowflake segment (with bijective inverse). One way to do this is to try to formalize the following function: $f(t)$ equals the point that is $t$ along the Koch segment ($0\le t\le1$).) $\endgroup$
    – Akiva Weinberger
    Commented Dec 15, 2022 at 7:22
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    $\begingroup$ (We still require the two endpoints to be there, though. So like an infinite straight line doesn't work because it has no real endpoints. All arcs are bounded.) $\endgroup$
    – Akiva Weinberger
    Commented Dec 15, 2022 at 7:47

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