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We are assuming that only two lines cross per crossing, right?

In that case, the simplest proof that I can think of is a two-coloring of the areas bounded by the loop. For example,

b|w|b
w|b|w
b|w|b

We know that such a coloring exists because the loop is closed and all of the crossings only involve two parts of the rope. I can't think of a neat proof for this fact off the top of my head, but one could probably induct on the number of crossings+the Jordan curve theorem.

Anyway, now we have this coloring. Now consider any black region and any crossing on the boundary of that black region. If we walk along one of the lines in the crossing toward the intersection, the black region will be on our left; on the other line, the region will be on our right.
So let us choose to put the line for which the black region is on our right on top at the crossing.

b |
--|-

If we do this for all regions, then we have our Celtic knot.

So why does this work? Consider any line. Given two adjacent crossings on that line, the segment of the line between those crossings borders exactly one black region. If we start at the middle of the segment and move toward one crossing, the black region is on the right, and thus the line goes on top at that crossing. If we start at the middle of the segment and move toward the other crossing, the black region is on the left, and thus the line goes on the bottom at that crossing.
Thus the line must alternate being on top and on the bottom.

Let's try that again (yeah, diagrams would really help.)

Firstly, we can make the line thingy into one long loop by joining the ends that are sticking out. Well, it doesn't even need to be one long loop, as long as there aren't any ends sticking out. If we do it properly, by joining ends that are next to each other, then we don't create any more crossings.

For example, suppose that we've closed everything and end up with this (simple case for illustrative purposes):



Now we can just split off loops from the outside as so:



Thus we end up with a bunch of closed loops, like so:



Now we fill in all of the closed loops, which we can do by the Jordan Curve theorem:



And stick everything back together. Now we have our coloring.
Now let's look at one of the filled regions, and at one portion of the boundary of that region (lines made red for clarity):



Let's call the three lines the higher line, the middle line, and the lower line (names self-explanatory).
Let's look at the lower intersection. We take each of the lines in that intersection, and draw arrows going along next to the filled region toward that intersection:



Now, for the middle line, the filled region is on the right, and for the lower line, the filled region is on the left. So we put the middle line on top at the crossing:



Now we do that for all of the crossings. Note that when going toward the lower crossing, the middle line has the filled region on the right, but when going toward the higher crossing, it has the filled region on the left. Thus we put it above at the lower crossing, and below at the higher crossing:



Thus we get an alternating over-under pattern.
Now we can just undo the coloring and the extra joinings, and the crossings will still fit the over-under criterion.

(Pictures done really quickly in Photoshop)

btw, does anyone know of a good graphics software program thats free and runs on ubuntu? i dont have photoshop for ubuntu and i dont think gimp has one either. crying

it would be nice if i can post some diagrams.

ok redface
now i see what you mean.

but i dont understand your explanation of how to divide the design into distinct loops with the same outside.
what I'm getting is that you take out one unintersected loop at a time (from the outside edge of the design) but i'm not convinced that the remaining shape will alwyas be a loop. why not a line?
or maybe i should get some sleep and read this again.

gotcha. biggrin

Ha, I'm actually kinda glad this topic was revived. One of my friends gave a talk on alternating diagrams for our junior seminar in knot theory. If no one else covers it, I'm gonna end up having to do the seminar on constructing the fundamental group; if not, I get to do one of the optional topics, either Dehn surgery or Kirby calculus.