surface area of sierpinski tetrahedron

[Karl131058]

Hi people,

I have got a dumb question for you - or maybe not?! The argument is quite lengthy, sorry for that.

The sierpinski tetrahedron is, as you all know, defined as THE (one and only) thing that consists of 4 copies of itself with half the size, moved towards the corners of a tetrahedron. (now that was an IFS without any formulae )

The standard construction starts with a tetrahedron; next step consists of 4 smaller tetrahedrons touching at corners (alternatively described as cutting out the central octahedron); repeat ad infinitum.

Some websites claim (e.g. wikipedia, bourke, ...) with all authority they can muster that the endresult has zero volume but finite surface area equal to that of the first tetrahedron. They even hint at a connection between this finite area and the fact that the box-count-dimension of that thing is exactly 2: four copies of half size , dim=ln(4)/ln(2)=2!

I beg to disagree - not on the dimension being 2 or volume being zero, but about the finite area.

Their argument is that with every step of iteration 1 tetrahedron gets replaced by 4 half-sized copies, which have a quarter of the original surface area each, so the area remains constant for each step of iteration. So the final result after infinitely many steps must have the same surface area.

Now let's construct the thingie differently:
Start with only the edges of a tetrahedron; infinitely thin. Then iterate as described: 4 copies positioned correctly reproduce the original 6 edges, plus 12 additional edges of halved length; repeat infinitely.
The final result of this process IS the Sierpinski Tetrahedron, too, because it clearly consists of 4 copies of itself with half the size, moved towards the corners of a tetrahedron.
Since at every step there is only a finite number of infinitely thin intervals, the surface area is ZERO all the time, and again the final result after infinitely many steps must have the same surface area.

Now which is it: zero or the original tetrahedrons area (which is NOT zero)??

Notice that the constructions are quite different: the first method removes half of the volume at every step, while the other doubles the total length of all edges at every step. But either the finite surface area disappears when going to infinitely many steps, OR a non-zero surface suddenly appears when going to infinitely many steps. Both seems unlikely, or equally likely, if you will.

It's just a question of different geometry to start with; you know you can choose ANY set because the final result only depends on the IFS!

I suppose that the correct answer goes something like "you can't give sets of this kind a meaningful surface area expressed in square meters" - hopefully accompanied by a hint "read THIS book for more info".

I would really look forward to th ISBN of "THIS book" 

Greetings
Karl

[cKleinhuis]

hmm, i am thinking about the 2d menger sponge, a 2d mnger sponge is constructed like:
-begin with a square e.g. sidelength 1 -> surface 4*1
- take out the center square of size 1/3 -> surface 4*1 ( the border of the original ) + 4*1/3
...

as you can see the surface hase really become larger, you are right that when talking about the sierpinski gasket, the new surfaces exhibit no larger surface, but hence it is going to infinity, it has to be inifinitely large surface area ... at least i think it is the explanation ... has anyone a better explanation of this at hand ??!

 

[Karl131058]

Well, I would TEND to agree on this one, but it seems difficult ...
It would be the same about the total edge length of the sierpinski tetrahedron, it gets DOUBLED everytime, no matter if you take a solid tetrahedron or a wireframe tetrahedron as start geometry, so this seems(!) to go to infinity.

Concerning the surface area of the sierpinski tetrahedron, the middle triangles of the sides are removed BUT new(!) surfaces appear to close the 4 subtetrahedra. An analogous situation in lower dimension might be the following variant on the standard cantor set:
start with the unit interval on the x axis, remove the middle third but DON'T throw it away - instead move it 1/3 into positive y-direction. You get something looking like _-_ (imagine the "minus" being of equal length to the underscore).
The total length of these three disjoint pieces is still one. Now continue as before: with all these three intervals, take out the middle third and move it its own length into the positive y-direction. With every final step you have a final number of intervals with constant total length, BUT shouldn't it be converging into countable many copies of the original cantor dust, all "well" separated from another - so I SUPPOSE this should be a dust too, though a lot more dense than the original (hausdorff-dimension would be 1, 3 copies of size 1/3 each step)
If I weren't already on my second glass of wine I might start an IFS right now to look at it, but my wife is waiting for me  and this has to wait until tomorrow morning.

Greetings
Karl

[Karl131058]

Hi people,

I am really sorry. I should have remembered one of the most import rules: Think at least twice before posting!

After some thinking I realised that my argument doesn't hold, so the sierpinski tetrahedron probably has finite surface area after all.

In the first message I claimed, that the set of all iterated edges IS the sierpinski tetrahedron, because it "clearly" consists of four copies of itself with half the size, moved towards the corners of a tetrahedron.
That conclusion is WRONG, it only proves that it is a SUBset of the full sierpinsky tetrahedron. One might as well start with only the cornerpoints of the first tetrahedron and push them through all the iterations; the resulting set also is a set which consists of 4 copies of itself scaled and moved as above, but that does NOT make it the WHOLE sierpinsky tetrahedron, only a subset of it. This set of iterated corners contains only countably many points, and even the "edges only" thingie contains (infinitely many) intervals, i.e. MORE points.

The relevant keyword here is "dense" 
I have been dense; and those two sets ("corners only" and "edges only") are (mathematically) dense in the complete sierpinsky tetrahedron, but they both are not the complete thing - just like all rational number between 0 and 1 are dense in the unit interval, but are NOT the whole unit interval.

There is more to it, but I will spare you 
(hint: "nested intervals"; the mathematically inclined will either groan or shake their heads at my stupidity, the others probably don't care)

If anybody should be interested in pointing out points in the full sierpinski tetrahedron that are not part of the "iterated edges" thingie I will gladly explain my errors to them - but somehow I don't think many will show up.

Have a nice day
Karl