This question mentions that it's an open question whether the Mandelbrot set is path-connected and the answer conflated it with the more famous open question of whether the Mandelbrot set is locally connected. But locally connected and path-connected are different notions and it's not obvious that they are equivalent in this case. Are they equivalent for the Mandelbrot set?
Here's a possible proof for the Mandelbrot set being path-connected: There is a homeomorphism between the complement of the Mandelbrot set and the complement of a closed disk. This homomorphism can be used to define a continuous function from a circle onto the boundary of the Mandelbrot set, which witnesses that the boundary of the Mandelbrot set is path connected. Any set with path-connected boundary has path-connected closure, and the closure of the Mandelbrot set is the Mandelbrot set itself since it is closed. Where does this argumant go wrong?
Best Answer
Answering the question in the second paragraph, the problem is that you can't necessarily extend a map of an open disk to its boundary. The fact that the complement is an open disk is enough to conclude connectedness but not path connectedness. For a specific example, consider a version of the Warsaw circle with two places that have the infinitely many compressed oscillations. Consider the bounded component of the complement. This is a simply connected open region of the plane, so is homeomorphic to an open disk, but the boundary is not path connected.
As far as the title question is concerned, it is clearly true that the Mandelbrot Set is path connected if it can be shown to be locally path connected. (We know it is connected since its boundary is connected, and locally path connected plus connected implies path connected.) Whether locally connected is sufficient, I am not 100% sure.