4 ms·
Suppose you choose such a rectangle, let's say it incorporates the 'fringe' of the main cardioid and the fringe of the biggest circle. Within that rectangle, co
by neilkk 3y ago
Suppose you choose such a rectangle, let's say it incorporates the 'fringe' of the main cardioid and the fringe of the biggest circle. Within that rectangle, color green all the pieces which connect to the main cardioid and blue all the pieces which connect to the circle. Local connectedness means that there won't be any points in that rectangle which have both blue and green points arbitrarily close. So there are places where 'locally non-connected parts' of the set can be close together, but there must be a border between them, rather than them being hopelessly entangled.
Strictly speaking, you would do this coloring with all connected components of the intersection of your rectangle and M. (And the rectangle could be any region.)
The example messes this up, although it is a correct example, the square on the diagram showing the local piece containing non-connected parts is wrong. The comb has more and more teeth, infinitely many in a bounded space, on the left side. Only a rectangle which includes the left edge properly shows why the set isn't locally connected. The rectangle pictured includes finitely many teeth which have a separation between them. A rectangle overlapping the left edge of the comb would include separate components which get arbitrarily close to that left edge and so can't be separated by a border.
- hermitcrab 3y agoThanks for taking the time to explain that. So, in terms I find easier to understand, MLC would mean that if I: -take any rectanglular section of the complex plain that includes part or all of the Mandelbrot set -draw the Mandelbrot set in black -pick an arbitary black point and colour it red -recursively colour every black point touching a red point (flood fill) Then every black point would be recoloured red. And this would work with a pixel based image of the mandelbrot if the image had a high enough resolution. Is that right?
- neilkk 3y agoNo, that's not right. Do those first four steps. You wouldn't (necessarily) cover every black point in your rectangle. Choose a remaining black point and flood fill from that, say green. Keep on doing this with different colours until you've covered every black point in your rectangle. You have a bunch of regions of different colours. Now, if the different coloured regions are all nicely separate, then your set is locally connected. Because each point is either cleanly in one component or cleanly in the other. If on the other hand your drawing looks like https://commons.m.wikimedia.org/wiki/File:Julia_set_for_the_rational_function.png https://commons.m.wikimedia.org/wiki/File:Julia_set_for_the_... with mixed up boundaries where some points are infinitesimally close to more than one colour, then it's not locally connected. The difficulty with intuition is that in our intuition, coloured regions always have reasonable boundaries (think countries in a map: the border can be wiggly but there's never infinitely many tiny bits of one country mixed up in the boundary of two others). In fractal geometry, things like the Newton fractal picture above are quite usual.
- hermitcrab 3y agoI think I understand now. Much appreciated! 'Locally connected' seems like quite poor terminology.