Value to use as center of Mandelbrot Set zoom? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A way to determine the ideal number of maximum iterations for an arbitrary zoom level in a Mandelbrot fractalMandelbrot set approximationWhy is the bailout value of the Mandelbrot set 2?Mandelbrot set incorrect picturePerturbation of Mandelbrot set fractalHow to compute a negative “Multibrot” set?Determine coordinates for Mandelbrot set zoom.Why does the Mandelbrot set appear when I use Newton's method to find the inverse of $tan(z)$?finding the periods of miniships in the Burning ShipGolden spirals in the Mandelbrot set?
Example of compact Riemannian manifold with only one geodesic.
My body leaves; my core can stay
Button changing its text & action. Good or terrible?
Would an alien lifeform be able to achieve space travel if lacking in vision?
What happens to a Warlock's expended Spell Slots when they gain a Level?
Can withdrawing asylum be illegal?
Sort list of array linked objects by keys and values
Sub-subscripts in strings cause different spacings than subscripts
Why are PDP-7-style microprogrammed instructions out of vogue?
How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?
How to support a colleague who finds meetings extremely tiring?
should truth entail possible truth
Could an empire control the whole planet with today's comunication methods?
Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?
Is an up-to-date browser secure on an out-of-date OS?
Do I have Disadvantage attacking with an off-hand weapon?
Single author papers against my advisor's will?
1960s short story making fun of James Bond-style spy fiction
Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?
Was credit for the black hole image misappropriated?
Word for: a synonym with a positive connotation?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Are spiders unable to hurt humans, especially very small spiders?
"is" operation returns false even though two objects have same id
Value to use as center of Mandelbrot Set zoom?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A way to determine the ideal number of maximum iterations for an arbitrary zoom level in a Mandelbrot fractalMandelbrot set approximationWhy is the bailout value of the Mandelbrot set 2?Mandelbrot set incorrect picturePerturbation of Mandelbrot set fractalHow to compute a negative “Multibrot” set?Determine coordinates for Mandelbrot set zoom.Why does the Mandelbrot set appear when I use Newton's method to find the inverse of $tan(z)$?finding the periods of miniships in the Burning ShipGolden spirals in the Mandelbrot set?
$begingroup$
I'm wondering what complex number or numbers is preferable to have at the center of a view of a Mandelbrot Set as the viewing range becomes increasingly smaller in order that the complexity of the set becomes visible. On any point in the set, the view will eventually become a solid region of points that are all within the set(infinite iterations), while for many points that aren't in the set, a similar thing will occur for a region of points not in the set. I'm wondering if there is a way to determine what the center of the view of the Mandelbrot Set will give a meaningful view for arbitrary viewing ranges.
complex-numbers recursion fractals coloring
asked Mar 10 '17 at 22:55
user2649681user2649681
373
$endgroup$
add a comment |
$begingroup$
I'm wondering what complex number or numbers is preferable to have at the center of a view of a Mandelbrot Set as the viewing range becomes increasingly smaller in order that the complexity of the set becomes visible. On any point in the set, the view will eventually become a solid region of points that are all within the set(infinite iterations), while for many points that aren't in the set, a similar thing will occur for a region of points not in the set. I'm wondering if there is a way to determine what the center of the view of the Mandelbrot Set will give a meaningful view for arbitrary viewing ranges.
complex-numbers recursion fractals coloring
asked Mar 10 '17 at 22:55
user2649681user2649681
373
$endgroup$
$begingroup$
You are asking if there are points where any neighborhood of that point contains points which are in the Mandelbrot set and points that are not? My guess is that there are infinitely many. I believe one would be at -2, though I'm not sure it would make such an interesting zoom.
$endgroup$
– Josh B.
Mar 10 '17 at 23:33
$begingroup$
Different points of the Mandelbrot set have different small scale behaviors. No single close-up of one point will give a meaningful view of the whole Mandelbrot set. In fact, the more you zoom in to a single point, the more special the view will be, losing more and more information about the global structure of the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:25
$begingroup$
Also, since the Mandelbrot set is closed, its complement is open. Each point of the complement is therefore the center of an open ball disjoint from the Mandelbrot set, and so zooming sufficiently closely into each such point you see only a vacuum entirely disjoint from the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:27
1
$begingroup$
One last thing, for each point $p$ in the topological frontier of the Mandelbrot set, every open ball around $p$ contains points both of the Mandelbrot set and of its complement, and therefore no matter how much you zoom into $p$ you will never see a solid region of points that are all within the set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:30
add a comment |
$begingroup$
I'm wondering what complex number or numbers is preferable to have at the center of a view of a Mandelbrot Set as the viewing range becomes increasingly smaller in order that the complexity of the set becomes visible. On any point in the set, the view will eventually become a solid region of points that are all within the set(infinite iterations), while for many points that aren't in the set, a similar thing will occur for a region of points not in the set. I'm wondering if there is a way to determine what the center of the view of the Mandelbrot Set will give a meaningful view for arbitrary viewing ranges.
complex-numbers recursion fractals coloring
asked Mar 10 '17 at 22:55
user2649681user2649681
373
$endgroup$
I'm wondering what complex number or numbers is preferable to have at the center of a view of a Mandelbrot Set as the viewing range becomes increasingly smaller in order that the complexity of the set becomes visible. On any point in the set, the view will eventually become a solid region of points that are all within the set(infinite iterations), while for many points that aren't in the set, a similar thing will occur for a region of points not in the set. I'm wondering if there is a way to determine what the center of the view of the Mandelbrot Set will give a meaningful view for arbitrary viewing ranges.
complex-numbers recursion fractals coloring
complex-numbers recursion fractals coloring
asked Mar 10 '17 at 22:55
user2649681user2649681
373
asked Mar 10 '17 at 22:55
user2649681user2649681
373
asked Mar 10 '17 at 22:55
user2649681user2649681
373
asked Mar 10 '17 at 22:55
user2649681user2649681
373
asked Mar 10 '17 at 22:55
user2649681user2649681
373
373
$begingroup$
You are asking if there are points where any neighborhood of that point contains points which are in the Mandelbrot set and points that are not? My guess is that there are infinitely many. I believe one would be at -2, though I'm not sure it would make such an interesting zoom.
$endgroup$
– Josh B.
Mar 10 '17 at 23:33
$begingroup$
Different points of the Mandelbrot set have different small scale behaviors. No single close-up of one point will give a meaningful view of the whole Mandelbrot set. In fact, the more you zoom in to a single point, the more special the view will be, losing more and more information about the global structure of the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:25
$begingroup$
Also, since the Mandelbrot set is closed, its complement is open. Each point of the complement is therefore the center of an open ball disjoint from the Mandelbrot set, and so zooming sufficiently closely into each such point you see only a vacuum entirely disjoint from the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:27
1
$begingroup$
One last thing, for each point $p$ in the topological frontier of the Mandelbrot set, every open ball around $p$ contains points both of the Mandelbrot set and of its complement, and therefore no matter how much you zoom into $p$ you will never see a solid region of points that are all within the set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:30
add a comment |
$begingroup$
You are asking if there are points where any neighborhood of that point contains points which are in the Mandelbrot set and points that are not? My guess is that there are infinitely many. I believe one would be at -2, though I'm not sure it would make such an interesting zoom.
$endgroup$
– Josh B.
Mar 10 '17 at 23:33
$begingroup$
Different points of the Mandelbrot set have different small scale behaviors. No single close-up of one point will give a meaningful view of the whole Mandelbrot set. In fact, the more you zoom in to a single point, the more special the view will be, losing more and more information about the global structure of the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:25
$begingroup$
Also, since the Mandelbrot set is closed, its complement is open. Each point of the complement is therefore the center of an open ball disjoint from the Mandelbrot set, and so zooming sufficiently closely into each such point you see only a vacuum entirely disjoint from the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:27
1
$begingroup$
One last thing, for each point $p$ in the topological frontier of the Mandelbrot set, every open ball around $p$ contains points both of the Mandelbrot set and of its complement, and therefore no matter how much you zoom into $p$ you will never see a solid region of points that are all within the set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:30
$begingroup$
You are asking if there are points where any neighborhood of that point contains points which are in the Mandelbrot set and points that are not? My guess is that there are infinitely many. I believe one would be at -2, though I'm not sure it would make such an interesting zoom.
$endgroup$
– Josh B.
Mar 10 '17 at 23:33
$begingroup$
You are asking if there are points where any neighborhood of that point contains points which are in the Mandelbrot set and points that are not? My guess is that there are infinitely many. I believe one would be at -2, though I'm not sure it would make such an interesting zoom.
$endgroup$
– Josh B.
Mar 10 '17 at 23:33
$begingroup$
Different points of the Mandelbrot set have different small scale behaviors. No single close-up of one point will give a meaningful view of the whole Mandelbrot set. In fact, the more you zoom in to a single point, the more special the view will be, losing more and more information about the global structure of the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:25
$begingroup$
Different points of the Mandelbrot set have different small scale behaviors. No single close-up of one point will give a meaningful view of the whole Mandelbrot set. In fact, the more you zoom in to a single point, the more special the view will be, losing more and more information about the global structure of the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:25
$begingroup$
Also, since the Mandelbrot set is closed, its complement is open. Each point of the complement is therefore the center of an open ball disjoint from the Mandelbrot set, and so zooming sufficiently closely into each such point you see only a vacuum entirely disjoint from the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:27
$begingroup$
Also, since the Mandelbrot set is closed, its complement is open. Each point of the complement is therefore the center of an open ball disjoint from the Mandelbrot set, and so zooming sufficiently closely into each such point you see only a vacuum entirely disjoint from the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:27
1
1
$begingroup$
One last thing, for each point $p$ in the topological frontier of the Mandelbrot set, every open ball around $p$ contains points both of the Mandelbrot set and of its complement, and therefore no matter how much you zoom into $p$ you will never see a solid region of points that are all within the set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:30
$begingroup$
One last thing, for each point $p$ in the topological frontier of the Mandelbrot set, every open ball around $p$ contains points both of the Mandelbrot set and of its complement, and therefore no matter how much you zoom into $p$ you will never see a solid region of points that are all within the set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Misiurewicz points, which lie on the boundary of the Mandelbrot set, provide an interesting location to zoom in. There are some illustrations near the end of the Wikipedia article. By definition, Misiurewicz points $M_k,n$ are the roots of equation $f_c^(k)(0) = f_c^(k+n)(0)$ where $f(z) = z^2+c$ and superscipts mean iteration. Two simple examples are $-2$ (which is $M_2,1$) and $i$ (which is $M_2,2$), but the more interesting points (with larger $k$ or $n$) lead to algebraic equations for $c$ that cannot be solved exactly.
$endgroup$
add a comment |
$begingroup$
You want a point on the boundary of the Mandelbrot set. Even so, "simple" ways to choose boundary points are not so interesting, because they eventually loop.
Points at rational internal angles (measured in turns) of hyperbolic components are at cusps, so zooming in will give one or two straight lines from the center, and only then if interior colouring is used.
Points at some irrational internal angles (measured in turns) of hyperbolic components exhibit (approximate) self-similarity, with the periods of the disc-like components increasing according to the Fibonacci sequence:
There is also approximate self-similarity about Feigenbaum points at the tip of period-doubling cascades:
The Mandelbrot set is asymptotically self-similar around Misiurewicz points (pre-periodic points in the boundary of the Mandelbrot set filaments, typically spiral centers and tips of filaments):
Iterative renormalization: a "minibrot" in the filaments of a parent minibrot, repeatedly zooming to the same location relative to the next child minibrot, pruning off the additional decorations by choice of colouring algorithm. This is a generalization of Feigenbaum points.
Ultimately, "interesting" zoom paths require aesthetic choices of where to zoom.
answered Mar 26 at 3:05
ClaudeClaude
2,585523
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2181175%2fvalue-to-use-as-center-of-mandelbrot-set-zoom%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Misiurewicz points, which lie on the boundary of the Mandelbrot set, provide an interesting location to zoom in. There are some illustrations near the end of the Wikipedia article. By definition, Misiurewicz points $M_k,n$ are the roots of equation $f_c^(k)(0) = f_c^(k+n)(0)$ where $f(z) = z^2+c$ and superscipts mean iteration. Two simple examples are $-2$ (which is $M_2,1$) and $i$ (which is $M_2,2$), but the more interesting points (with larger $k$ or $n$) lead to algebraic equations for $c$ that cannot be solved exactly.
$endgroup$
add a comment |
$begingroup$
Misiurewicz points, which lie on the boundary of the Mandelbrot set, provide an interesting location to zoom in. There are some illustrations near the end of the Wikipedia article. By definition, Misiurewicz points $M_k,n$ are the roots of equation $f_c^(k)(0) = f_c^(k+n)(0)$ where $f(z) = z^2+c$ and superscipts mean iteration. Two simple examples are $-2$ (which is $M_2,1$) and $i$ (which is $M_2,2$), but the more interesting points (with larger $k$ or $n$) lead to algebraic equations for $c$ that cannot be solved exactly.
$endgroup$
add a comment |
$begingroup$
Misiurewicz points, which lie on the boundary of the Mandelbrot set, provide an interesting location to zoom in. There are some illustrations near the end of the Wikipedia article. By definition, Misiurewicz points $M_k,n$ are the roots of equation $f_c^(k)(0) = f_c^(k+n)(0)$ where $f(z) = z^2+c$ and superscipts mean iteration. Two simple examples are $-2$ (which is $M_2,1$) and $i$ (which is $M_2,2$), but the more interesting points (with larger $k$ or $n$) lead to algebraic equations for $c$ that cannot be solved exactly.
$endgroup$
Misiurewicz points, which lie on the boundary of the Mandelbrot set, provide an interesting location to zoom in. There are some illustrations near the end of the Wikipedia article. By definition, Misiurewicz points $M_k,n$ are the roots of equation $f_c^(k)(0) = f_c^(k+n)(0)$ where $f(z) = z^2+c$ and superscipts mean iteration. Two simple examples are $-2$ (which is $M_2,1$) and $i$ (which is $M_2,2$), but the more interesting points (with larger $k$ or $n$) lead to algebraic equations for $c$ that cannot be solved exactly.
answered Mar 14 '17 at 18:21
user357151
add a comment |
add a comment |
$begingroup$
You want a point on the boundary of the Mandelbrot set. Even so, "simple" ways to choose boundary points are not so interesting, because they eventually loop.
Points at rational internal angles (measured in turns) of hyperbolic components are at cusps, so zooming in will give one or two straight lines from the center, and only then if interior colouring is used.
Points at some irrational internal angles (measured in turns) of hyperbolic components exhibit (approximate) self-similarity, with the periods of the disc-like components increasing according to the Fibonacci sequence:
There is also approximate self-similarity about Feigenbaum points at the tip of period-doubling cascades:
The Mandelbrot set is asymptotically self-similar around Misiurewicz points (pre-periodic points in the boundary of the Mandelbrot set filaments, typically spiral centers and tips of filaments):
Iterative renormalization: a "minibrot" in the filaments of a parent minibrot, repeatedly zooming to the same location relative to the next child minibrot, pruning off the additional decorations by choice of colouring algorithm. This is a generalization of Feigenbaum points.
Ultimately, "interesting" zoom paths require aesthetic choices of where to zoom.
answered Mar 26 at 3:05
ClaudeClaude
2,585523
$endgroup$
add a comment |
$begingroup$
You want a point on the boundary of the Mandelbrot set. Even so, "simple" ways to choose boundary points are not so interesting, because they eventually loop.
Points at rational internal angles (measured in turns) of hyperbolic components are at cusps, so zooming in will give one or two straight lines from the center, and only then if interior colouring is used.
Points at some irrational internal angles (measured in turns) of hyperbolic components exhibit (approximate) self-similarity, with the periods of the disc-like components increasing according to the Fibonacci sequence:
There is also approximate self-similarity about Feigenbaum points at the tip of period-doubling cascades:
The Mandelbrot set is asymptotically self-similar around Misiurewicz points (pre-periodic points in the boundary of the Mandelbrot set filaments, typically spiral centers and tips of filaments):
Iterative renormalization: a "minibrot" in the filaments of a parent minibrot, repeatedly zooming to the same location relative to the next child minibrot, pruning off the additional decorations by choice of colouring algorithm. This is a generalization of Feigenbaum points.
Ultimately, "interesting" zoom paths require aesthetic choices of where to zoom.
answered Mar 26 at 3:05
ClaudeClaude
2,585523
$endgroup$
add a comment |
$begingroup$
You want a point on the boundary of the Mandelbrot set. Even so, "simple" ways to choose boundary points are not so interesting, because they eventually loop.
Points at rational internal angles (measured in turns) of hyperbolic components are at cusps, so zooming in will give one or two straight lines from the center, and only then if interior colouring is used.
Points at some irrational internal angles (measured in turns) of hyperbolic components exhibit (approximate) self-similarity, with the periods of the disc-like components increasing according to the Fibonacci sequence:
There is also approximate self-similarity about Feigenbaum points at the tip of period-doubling cascades:
The Mandelbrot set is asymptotically self-similar around Misiurewicz points (pre-periodic points in the boundary of the Mandelbrot set filaments, typically spiral centers and tips of filaments):
Iterative renormalization: a "minibrot" in the filaments of a parent minibrot, repeatedly zooming to the same location relative to the next child minibrot, pruning off the additional decorations by choice of colouring algorithm. This is a generalization of Feigenbaum points.
Ultimately, "interesting" zoom paths require aesthetic choices of where to zoom.
answered Mar 26 at 3:05
ClaudeClaude
2,585523
$endgroup$
You want a point on the boundary of the Mandelbrot set. Even so, "simple" ways to choose boundary points are not so interesting, because they eventually loop.
Points at rational internal angles (measured in turns) of hyperbolic components are at cusps, so zooming in will give one or two straight lines from the center, and only then if interior colouring is used.
Points at some irrational internal angles (measured in turns) of hyperbolic components exhibit (approximate) self-similarity, with the periods of the disc-like components increasing according to the Fibonacci sequence:
There is also approximate self-similarity about Feigenbaum points at the tip of period-doubling cascades:
The Mandelbrot set is asymptotically self-similar around Misiurewicz points (pre-periodic points in the boundary of the Mandelbrot set filaments, typically spiral centers and tips of filaments):
Iterative renormalization: a "minibrot" in the filaments of a parent minibrot, repeatedly zooming to the same location relative to the next child minibrot, pruning off the additional decorations by choice of colouring algorithm. This is a generalization of Feigenbaum points.
Ultimately, "interesting" zoom paths require aesthetic choices of where to zoom.
answered Mar 26 at 3:05
ClaudeClaude
2,585523
answered Mar 26 at 3:05
ClaudeClaude
2,585523
answered Mar 26 at 3:05
ClaudeClaude
2,585523
answered Mar 26 at 3:05
ClaudeClaude
2,585523
2,585523
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2181175%2fvalue-to-use-as-center-of-mandelbrot-set-zoom%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You are asking if there are points where any neighborhood of that point contains points which are in the Mandelbrot set and points that are not? My guess is that there are infinitely many. I believe one would be at -2, though I'm not sure it would make such an interesting zoom.
$endgroup$
– Josh B.
Mar 10 '17 at 23:33
$begingroup$
Different points of the Mandelbrot set have different small scale behaviors. No single close-up of one point will give a meaningful view of the whole Mandelbrot set. In fact, the more you zoom in to a single point, the more special the view will be, losing more and more information about the global structure of the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:25
$begingroup$
Also, since the Mandelbrot set is closed, its complement is open. Each point of the complement is therefore the center of an open ball disjoint from the Mandelbrot set, and so zooming sufficiently closely into each such point you see only a vacuum entirely disjoint from the Mandelbrot set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:27
1
$begingroup$
One last thing, for each point $p$ in the topological frontier of the Mandelbrot set, every open ball around $p$ contains points both of the Mandelbrot set and of its complement, and therefore no matter how much you zoom into $p$ you will never see a solid region of points that are all within the set.
$endgroup$
– Lee Mosher
Mar 14 '17 at 18:30