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Visualizing Aleph_1
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2007-12-01, 02:42
Post: #1
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Visualizing Aleph_1
The Basic Idea The idea is simple: instead of restricting ourselves to 1 dimension, where we get a picture that is very difficult to visualize much less understand, we use 2 dimensions instead, by 'wrapping' each stage around to fill 2D, and iterate that into a fractal-like structure. The Infinite Image Here's our first crack at it: first, we begin with a unit square. This represents 0, the first finite ordinal. Then we apply the successor function to get 1, which we represent by a rectangle to the right of the unit square, with the same height but half the width. Next, we append to the right of these two a third rectangle, still the same height, but with 1/4 the original width. We take the limit of the process by repeating this, ad infinitum, halving the width of each subsequent rectangle. This gives us w, the first infinite limit ordinal. It fits in the 2x1 rectangular area (since the half-widths has a limit at 2). This idea, of course, is not new. Many have used analogous schemes to visualize the transfinite ordinals. The difference comes in the next step. Instead of merely applying the successor function, we make the next step by copying the rectangles corresponding to w, halving their heights, and placing the copies above the originals. Now we have a 2x1.5 rectangular area representing the structure of 2*w. Then we take another copy of w, make its height 1/4 the original, and stack it on 2*w, to get a 2x1.75 rectangular area representing the structure of 3*w. Now we take the limit, and end up with a 2x2 rectangular area, representing w*w. Then, we take this 2x2 area, scale it by (1,1/2) so that its height remains the same, but with half its original width, and place this squished copy to the right of the original. So we get a 3x2 area representing Then we take the 4x2 area, scale it to half its height, and stack it on itself, and repeat this, halving the height each time, and take the limit, and we get a 4x4 area representing The pattern is now clear. At each step, we take the limit of pasting copies of the square area into halved areas, switching between the X and Y axes. Every two steps yields a square area, and so we can repeat this process. Now, take the limit of this process, and we get an infinitely large square representing The Finite Image The picture above has the disadvantage of being infinitely large, so it still doesn't give us a full view of the completed set. Is it possible to create a completed view of the set? We shall try. First, notice that every two steps of the above process yields a square 4 times larger than the original (twice the original's dimensions). This suggests that we can scale the picture after every two steps, so that it will still fit in a unit square area. We can repeat this process any even number of steps, to get a picture of the ordinal corresponding to an ever larger initial segment of Finally, we can take the limit as the number of steps go to infinity: we get a fractal-like image, which happens to be the same image we get if we take the limit of repeatedly substituting each rectangle in the 2x2 area for w*w with a suitably scaled version of itself. This final image gives us a view of the completed set Perhaps. Although this interpretation is very tempting, since it finally gives our mind a pictorial way to grasp the full structure of
So, sadly to say, although this fractal picture is very compelling, in that it offers us a finite view of the completed set |
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2007-12-01, 02:58
Post: #2
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RE: Visualizing Aleph_1
As a followup, I discuss the implications of the above attempts to visualize an uncountable ordinal.
Firstly, the two images tell us that the structure of Secondly, we must ask, why does the fractal picture have fatal flaws? The limit process we have used seems like it ought to accurately represent the completed Thirdly, consider the implications this may have for the Continuum Hypothesis: In every finite step in the generation of the fractal picture, we have rectangles that correspond to individual elements in the set. Our construction ensures that these rectangles completely tile the unit square. As we take the limit of this process, each of these rectangles are reduced to points. Since the unit square is fully covered in the resulting fractal, that means the cardinality of the set of collapsed squares must be (at least) the cardinality of the continuum. This seems to imply the Continuum Hypothesis... if it were not for the fact that the resulting fractal can't possibly represent We can take this even further. Suppose we take the fractal picture of The problem is that the picture thus arrived at is identical to the fractal picture for Or perhaps this is an argument for equating The disturbing thing about all of this, is that the limit that yielded this fractal involves only a countable number of steps. Even more disturbing is the question, if this fractal does not represent the actual structure of the uncountable ordinal, then what image would? If taking the limit of the finitary steps, which do correspond with the initial segments of the ordinal, gives us a picture which has gone "too far", then what exactly sits between the finitary steps and the limiting fractal which may correspond with the actual ordinal? How is it possible for there to be something between a limit and all possible finite steps leading up to the limit? Surely the limit must sit at most a countable number of steps beyond the finite steps; does this mean there are multiple distinct countable infinities (how can that be?!), with a lesser countable corresponding to an "intermediate" image that represents the actual ordinal? Perhaps this is a reason to reconsider the Axiom of Regularity, and allow sets to contain themselves. The so-called New Foundations show that it is possible to have a consistent system that contains Quine Atoms, but not Russell's Paradox. In any case, these considerations give us a little glimpse into the pathological properties that infinite sets may have. |
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2008-02-01, 11:59
Post: #3
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RE: Visualizing Aleph_1
Actually, I discovered that the picture isn't quite this simple after all. I was under the impression that the equivalent of the Ackermann function on
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