(3) Finally by the previous points, these sets achieve the infimum over all of ³ ³ proving the existence of an element in Γ³ . Moreover we get the phase transitions in ³ and show non uniqueness for the first phase transition. This is done in Section 4.

2. Finding the sets in F (³ )

Our goal in this section is to find elements of F³ with a smaller double bubble perimeter than given sets A with µ (A) = 1, and B with µ (B) = ³ .

Definition 2.1. (1)

A := [aleft, aright] × [abottom, atop], where aleft = inf{x : (x, y) ∈ A for some y ∈ R} aright = sup{x : (x, y) ∈ A for some y ∈ R} abottom = inf{y : (x, y) ∈ A for some x ∈ R} atop = sup{y : (x, y) ∈ A for some x ∈ R} Lemma 1. Ã (A ) ≤ Ã (A) and µ (A) ≤ µ (A ).

Proof. By definition, A ⊂ A . Therefore, by monotonicity of Lebesgue measure, µ (A) ≤ µ (A ).

Now we show that à (A ) ≤ à (A). Let H1 be the horizontal line passing through atop, and H2 be the horizontal line passing through abottom. Similarly define V1 and V2 to be the vertical lines passing through aleft and aright, respectively. Let DH1,H2be the distance from H1 to H2, and DV1,V2 be the distance between V1 and V2. "A must touch H1 in at least one point, say p1, and similarly must touch H2 in at least one point, say p2. See Figure 3 for an illustration of the notations.

Since A is open and "A is simple, there must be at least two disjoint paths in "A from p1 to p2, abbreviate them » i(t) = (xi(t), yi(t)) : [ 0, 1 ] → R2, for i ∈ {1, 2}. The vertical portion of these paths must be at least DH1,H2 i.e. for any 0 ≤ t1 ≤ t2 ≤ · · · ≤ tn ≤ 1, and i ∈ {1, 2} we have that

DH1,H2

N X j=1

V2 p1

H1

H2

DV1,V2 yi(tj+1) − yi(tj) ≥ DH1,H2 Similarly there must be at least two disjoint paths from a point on V1 to a point on V2, whose horizontal distances must be at least DV1,V2. We have so far found that the boundary of A must measure at least 2 · DV1,V2 + 2 · DH1,H2, which is the length of the boundary of A . That is, Ã (A ) ≤ Ã (A), as claimed.

Lemma 2. If (A, B) ∈ ³ ³ , and B ⊂ A , then there exists (A˜, B˜) ∈ F³ , such that à DB(A˜, B˜) ≤ à DB(A, B), and (A˜, B˜) are either kissing rectangles or embedded rectangles. ˜ B ˜ A ˜ A ˜ B Proof. If B is contained in A there are several options to consider, namely that one, two, three, or no edges of B could touch the same number of edges in A . Let’s take the instance when none of the edges of B touch any of the edges of A , that is, btop < atop, bright < aright, aleft < bleft, and abottom < bbottom. From this case we can easily derive the results for the other cases. Let H1 = R × {atop}, H2 = R × {btop}, H3 = R × {bbottom}, and H4 = R × {abottom}. Now we define the distance between H1 and H2 to be DH1,H2 = atop −btop, the distance between H2 and H3 to be DH2,H3 = btop −bbottom, and the distance between H3 and H4 to be DH3,H4 = bbottom − abottom. Similarly we call the vertical line through aleft V1, the vertical line through bleft V2, the vertical line through bright V3, and the vertical line through aright V4, naming the distances between these lines just as before. See Figure 4 for illustration. V1

V2

V3

V4 DH1,H2 DH2,H3 DH3,H4 DV1,V2

DV2,V3

Now, in "A ∪ "B we need to find three paths with vertical lengths of DH2,H3, and in "A two paths with vertical lengths of DH1,H2 and the same for DH3,H4. Further, we need these paths to be disjoint so we don’t count anything more than once. Then we would do the analogous process for horizontal distances. First, we find points p1 ∈ H1 ∩ "A , and p2 ∈ H4 ∩ "A . There must be two distinct paths in "A between these two points, which we call » i(t) = (xi(t), yi(t)) : [ 0, 1 ] → R2, for i ∈ {1, 2}, both of which must have vertical distance of at least DH1,H2 + DH2,H3 + DH3,H4. That is, for any 0 ≤ t1 ≤ t2 ≤ ... ≤ tN ≤ 1, we have for i ∈ {1, 2},

N X (|yi(tj+1) − yi(tj )|) ≥ DH1,H2 + DH2,H3 + DH3,H4.

j=1

To complete our search for enough vertical length, it remains to find one final path with vertical length DH2,H3 that we have yet to count among these crossings. For a path ³ : [ 0, 1 ] → R2 we say that a subpath ³ [a, b] is a crossing of S := R × (bbottom, btop) if ³ (a) ∈ H2, ³ (b) ∈ H3 and for all a < t < b, ³ (t) ∈/ {H2, H3} (or in the other direction). If "A contains more than two crossings, then each of these crossings must have vertical length at least DH2,H3, and we are done. So we may assume that "A only contains two crossings of S. We denote these two crossings ¿ 1, ¿ 2. Since "A contains exactly two crossings of S, S \ ¿ 1 ∪ ¿ 2 consists of three open sets, exactly two of which are unbounded and have joint boundary with both H2 and H3. We call these two open sets S1 and S2. Since B is open and connected and contained in S, B must be contained in one of these unbounded open sets, say w.l.o.g. S2 (note that if there are more than 2 crossings then B can be contained in a bounded set). Since B is open and "B is rectifiable, there are at least two distinct paths in "B from H2 to H3, which we call » 3 = (x3(t), y3(t)), and » 4 = (x4(t), y4(t)). Both of these paths must be in S2. By planarity only one of these paths might intersect ¿ 2 say w.l.o.g. » 3. This means that we have not counted the vertical part of » 4, and it must have vertical length at least DH2,H3. That is to say, that for any 0 ≤ t1 ≤ t2 ≤ ... ≤ tN ≤ 1, we have

N X (|y4(tj+1) − y4(tj)|) ≥ DH2,H3.

j=1

This is our third such length, and we are done finding vertical lengths. We need only find horizontal lengths now. But this is done in exactly the same manner as in our search for vertical lengths. We could even rotate our figures 90 degrees either left or right, so that vertical lines become horizontal and vice versa, and perform the exact same proof as above.

We now construct our figure. First, we have found a total length of

2 · (DH1,H2 + DH3,H4 + DV1,V2 + DV3,V4) + 3 · (Dh2,H3 + DV2,V3) .

This gives us enough length to construct A and still have left over a total length of DH2,H3 + DV2,V3. With these lengths, we construct a box in the corner of A with dimensions DH2,H3 × DV2,V3. This box, which we call B˜ in the corner has volume the same as B , and therefore volume at least ³ . We can shrink it easily so that it has volume exactly ³ , and by abuse of notation still call this possibly smaller rectangle B˜. On the other hand, A ∪ B ⊂ A , and therefore µ (A ) ≥ 1 + ³ . Therefore, µ (A \ B˜) ≥ 1 + ³ − ³ . So we can easily move the sides of A that don’t share joint boundary with B˜ inwards until the volume is exactly 1. This set we call A˜, and have completed our construction.

Now, suppose that bbottom = abottom, or bleft = aleft, etc. That is one of the sides of B is contiguous with one of the lines of A . The process would be as above, except we would have H3 = H4. This means we wouldn’t have to find two paths with vertical length DH3,H4. The rest of the proof would be the same. Similarly, if two sides of B are contiguous with two sides of A , say abottom = bbottom and aleft = bleft, then we wouldn’t have to find paths with vertical length DH3,H4 and we wouldn’t have to find paths with horizontal length DV1,V2. The rest of the proof would be the same.

The previous lemma took into account all of the cases when all four corners of B are contained in A . The only other two options are if two or one corner of B is contained in A .

Lemma 3. If (A, B) ∈ ³ ³ , and exactly one corner of B is contained in A , then there exists (A˜, B˜) ∈ F³ , such that à DB(A˜, B˜) ≤ à DB(A, B).

Proof. For the one corner case we argue that we can find sets, with a better double bubble perimeter and more joint volume than the original shapes, that looks like the general case of Figure 2:

Here, the rectangle can be either A or B , say A , and the other set is B \ A . Once we create these two sets, we are not necessarily done because the volumes may not be correct. This may cause somewhat more of a problem than in previous cases, but in any case the method remains similar.

In this proof we are assuming that µ (A) = 1, and µ (B) = ³ . Since A and B only intersect in one corner, either atop > btop, or btop > atop. We can suppose without loss of generality that atop > btop. Let H1 = R × {atop} be the horizontal line passing through atop, H2 = R × {btop} be the horizontal line passing through btop, H3 = R × {abottom} be the horizontal line passing through abottom, and H4 = R × {bbottom} be the horizontal line passing through bbottom. Note that the height of A ∩ B is the same as the distance between H2 and H3, which we will call DH2,H3. Furthermore, let the distance between H1 and H2 be DH1,H2, and the distance between H3 and H4 be DH3,H4. Now, let V1 = {bleft} × R be the vertical line passing through bleft, V2 = {aleft} × R be the vertical line passing through aleft, V3 = {bright} × R be the vertical line passing through bright, and V4 = {aright} × R be the vertical line passing through aright. We define the distance between V1 and V2 to be DV1,V2, the distance between V2 and V3 to be DV2,V3, and the distance between V3, and V4 to be DV3,V4. See Figure 5.

V1

V2

V3

V4 DH1,H2 DH2,H3 DH3,H4 H1 H2 H3

H4

Notice that to construct A and B \ A in this way we need two lengths of DH1,H2, DH3,H4, DV1,V2, and DV3,V4, as well as three lengths of DH2,H3, and DV2,V3. In other words, for à DB((A , B \ A )) to be at most à DB((A, B)), we need to find two vertical lengths of DH1,H2 in "A , two vertical lengths of DH3,H4 in "B , and three vertical lengths of DH2,H3 between "A and "B . Similarly for horizontal lengths.

First, there must be a point p1 ∈ H1 ∩ "A , and another point p2 ∈ H3 ∩ "A . Between these two points there must be at least two disjoint paths, which we call » i(t) = (xi(t), yi(t)) : [ 0, 1 ] → R2, i ∈ {1, 2}, in "A , both of which have vertical length at least DH1,H2 + DH2,H3. That is, as before, for any 0 ≤ t1 ≤ t2 ≤ ... ≤ tN ≤ 1, N X(|yi(tj+1) − yi(tj)|) ≥ DH1,H2 + DH2,H3 j=1

Similarly in "B , we can find two disjoint paths of vertical length at least DH2,H3 + DH3,H4. Since there can be no joint boundary below H3, the portions of these paths that measure at least DH3,H4 have not been counted yet. It remains to find a path in "B with vertical length at least DH2,H3 that we have yet to count.

To find this path we define the infinite strip of height DH2,H3, S := R × (abottom, btop). If "A has more than two crossings of S, then each of these crossings has vertical length of at least DH2,H3, as above, and we are done. So we may assume that "A contains exactly two crossings of S (it can’t be less than two as we noted above). Abbreviate these crossings ¿ 1, ¿ 2. In this case, consider S \ ¿ 1 ∪ ¿ 2. This consists of three open sets, but only two are unbounded sets, and in one of these open sets that we find B ∩ S. We call these sets S1 and S2 such that ¿ 1 ⊂ "S 1 and ¿ 2 ⊂ "S 2. Suppose without loss of generality that B ∩ S ⊂ S2. There is at least one point p3 ∈ "B ∩ H2, and at least one point p4 ∈ "B ∩ H3, and there must be two distinct paths in "B from p3 to p4. We call these paths » i(t) = (xi(t), yi(t)) : [ 0, 1 ] → R2, i = 3, 4. By planarity, only one of the paths, either » 3 or » 4 can have joint boundary with ¿ 2, say » 3. This means that we have yet to include the vertical length of » 4, and we have just found our third vertical length of DH2,H3. That is, for any 0 ≤ t1 ≤ t2 ≤ ... ≤ tN ≤ 1,

N X(|y4(tj+1) − y4(tj)|) ≥ DH2,H3.

j=1

Finding the three horizontal lengths of DV2,V3 follows the same argument. The reason we can be sure that we won’t double count anything is because, inside A ∩ B we have only counted vertical distance, and in the argument to find our three horizontal lengths of DV2,V3 we would only count horizontal lengths. So, we have proved that à DB(A , B ) ≤ à DB(A, B). It is clear that µ (A ∪ B) ≤ µ (A ∪ B ). Since A ⊂ A , 1 = µ (A) ≤ µ (A ). However, it is possible that µ (B \ A ) < µ (B) = ³ . This we must correct. For this purpose, let us refer to the notation we established in the introduction for the general case.

a c A b d e

B \ A

f

We look at the two intervals of lengths c, f and assume w.l.o.g that f > c (see Figure 6). Now we slide B \ A down towards the longer edge f . By doing so we do not enlarge the boundary and we can only increase the area. There are two options now: (1) If c + d + e ≤ a we get kissing rectangles where the rectangle on the left has greater area than B \ A . (see Figure 7).

Next we move area from A to the rectangle on the right until reaching the appropriate area ³ . Consider f 2 ≥ f such that µ (B˜) = f 2 · (c + d + e) = ³ , and let b2 = b − (f 2 − f ). Our final sets are of the general case class as can be seen in Figure 8. Since we have only enlarged the total area we are guaranteed that µ (A˜) ≥ 1. Reducing the area is easy and we have dealt with this before. b − (f2 − f) c + d + e ˜ B f2 ˜ B ˜ f d2 f a A˜ ˜ b

c

Now we need only deal with the case when two corners of B are in A . Lemma 4. If (A, B) ∈ ³ ³ , and two corners of B are contained in A , then there exists (A˜, B˜) ∈ F³ , such that à DB(A˜, B˜) ≤ à DB(A, B).

Proof. Here we must have one of the following: btop > atop, bright > aright, bbottom < abottom, or bleft < aleft. Let’s suppose, without loss of generality, that bright > aright. We construct our configuration in a similar way as before. First, since two corners of B are in A , it follows that aleft < bleft ≤ aright. However, if bleft = aright, we can just replace A with A and B with B . This will increase the volume of both A and B, and increase their joint boundary. Then we can reduce the volumes as necessary, which will only decrease the double bubble perimeter. So, we can assume that aleft < bleft < aright. Let H1 be the horizontal line passing through atop, H2 be the horizontal line passing through btop, H3 be the horizontal line passing through bbottom, and H4 the horizontal line passing through abottom. We define DHi,Hi+1 as before, i = 1, 2, 3. Similarly define V1, V2, V3, V4 as the vertical lines passing through aleft, bleft, aright, and bright, respectively. Let DV1,Vi+1 be defined as before, i = 1, 2, 3. Notice that we haven’t eliminated the possibility that H1 = H2, or H3 = H4, or both. See Figure 11.

We wish to find two disjoint paths with vertical lengths at least DH1,H2, two disjoint paths with vertical length as least as long as DH3,H4, three disjoint paths with vertical

V1

V2

V3

V4

DV2,V3 DV3,V4 lengths at least as long as DH2,H3, two disjoint paths with horizontal length at least as long as DV1,V2, three disjoint paths with horizontal lengths at least as long as DV2,V3, and two disjoint paths with horizontal lengths at least as long as DV3,V4 . We achieve this much the same as in the previous Lemma.

Finding the horizontal lengths is nearly identical. We can then proceed to construct our sets. If move B up until H1 = H2 we reduce the double bubble perimeter and we get a shape of the general case type. Now we can fix the volumes in the same way as in the previous Lemma.

à DB(Γ³ ) ≤ à DB(Ç ³ ) ≤ ilim à DB(Ç i) = à DB(Γ³ ),

³> and thus Ç ³ ∈ Γ³ and Γ³ is non empty, establishing part I of Theorem 1. Since we have that Ç ³ ∈ Γ³ , Lemma 8 parts II and III establishes Theorem 1 parts II and III.