The new space, call it U, consists of the points of S, and the points of T without T0. (Since T0 is already represented by S0, we need not count it twice.) A set is open in U if it is the union of two open sets A in S and B in T such that A∩S0 and B∩T0 correspond under h. In other words, A and B agree at the common border. Note that A and B could miss the border entirely; whence their intersections with the border (empty) still correspond under h. Set B to the empty set, for example, and A could be any open set in S that misses the border. This remains open in U, along with open sets in T-T0, and open sets A+B that coincide at the border.
Take a moment to verify this is a topology. The union of open sets A1+B1, A2+B2, A3+B3 … is built from two open sets A1∪A2∪A3∪… + B1∪B2∪B3∪ …, which have a common border under h. Thus the arbitrary union of open sets is open. In the same way, the intersection of two open sets is open.
Remember that S0 and T0 are homeomorphic under h. Within the topology of S0, inherited from S, A∩S0 is open, because A is open. Apply h and find an open set in T0. This has to come from an open set B in T. Therefore A+B will do the trick.
The technical definition is (practically) the same, but there is only one space S, and S0 and T0 both come from S. Open sets have the form A+B where h equates A∩S0 with B∩T0, and h also equates A∩T0 with B∩S0. An argument similar to the above shows this is a topology.
Continuing the earlier example, sew the left and right sides of a square, then sew the top and bottom together. The first step creates the paper tube, and if the paper was rubber, you could bend the tube around until the top circle meets the bottom circle, forming a torus. (Try this with a vacuum cleaner tube.) Often the square is a simpler representation of the torus. As you move off the top edge, you reappear at the bottom, and as you move off the right side, you reappear at the left.
A three dimensional torus, denoted T3, can be constructed by sewing opposite faces of a cube together. Top = bottom, front = back, left = right. The space has a finite volume, yet you can travel in one direction forever. Our universe could be such a space??
Going backwards, the one dimensional torus T1 is a line segment with its ends joined together. This is simply a circle. In general, Tn is the topological product of n circles, one for each dimension of travel.
A more interesting example sews the circle to itself with a 180 degree phase shift. In other words, h carries each point on the circle to its opposite point. Let an ant crawl along this bowl, up to the edge. As he passes through the edge, he appears on the other side crawling down.
The same space can be created in another way. Let S be the unit sphere, and let S0 = T0 = S. The key is the homeomorphism; h equates each point with its opposite. This is projective space, denoted P2. This space, and the aforementioned bowl with its projective rim, are both equivalent to the set of lines passing through the origin, where euclidean distance on the unit sphere determines the open sets.
Let's use a square to represent a mobius strip. Let h merge the top and bottom edges of the square, but this time h implements a reflection. The top left corner is mapped onto the bottom right corner, and so on. If a fish swims up through the top, his head reappears at the bottom, but he is reflected. His dorsel fin is on the other side. If he swims the circuit again he will be reflected back into his original orientation.
Draw a vertical line up the center of the square. Place the fish to the right of this centerline and let him swim up through the top. He appears at the bottom, reflected, to the left of the centerline. Let him swim up through the top again and he is back on the right. The space remains connected, even if it is cut down the middle. Try this with your paper mobius strip. Poke a hole with your scissors and cut it down the middle, all the way around. The result is a band with a double (360 degree) twist. (Limeric)
Cut a mobius strip into thirds and find two separate loops that are interlocked. One has a 180 degree twist and the other has a 360 degree twist.
Take a tall sheet of paper and curl it up until its left and right edges coincide. This creates a tall tube. Paint a red line where the left and right edges meet. This is the seam, if you will. Orient the tube so that the seam runs up the right side.
Assume the paper can twist and stretch, like rubber. In an earlier example we bent the tube around and joined the two ends to make a torus. The red seam met itself to create a circle that ran along the inside of the torus. A blue seam joins the two circles, where top and bottom meet. However, to make a klein bottle, you must reflect one of the circles before sewing the two ends together. How can we do that?
Return to the tall tube with its red seem running up the right side. Twist the top 180 degrees, so that the red seam spirals around the front and winds up at the top left. Now expand the bottom and shrink the top, so the tube has the shape of a bottle with a tall thin neck. Pretend like you are a glass blower, and bend the neck over to the right. This glass can do more than twist and stretch, it can pass through itself. After you have pulled the tall narrow neck to the right, and around, and down, push it through the side of the wider base. The top has now reentered the bottle, pointing down. Fan out the top circle, like a funnel, and sew it to the bottom circle. Notice that the red seam meets itself. This time the blue seam joins the top and bottom circles through a reflection.
No, it can't be done. Here is an outline of the proof.
When viewed as a square, the klein bottle is compact, without a boundary. It's continuous image in 3 space is also compact, without a boundary. (A point p is on the boundary if it can be removed, and the surface is still locally simply connected.)
Can the klein bottle cut 3 space into pieces? Is there an inside and an outside? Suppose there is an inside, and start filling it up with air, like a tire. Place a fish on the surface with its distinguishing dorsel fin. The right side of the fish has air under pressure (the inside), while the left side of the fish, facing you, is on the outside of the surface. As the fish swims along the surface of the bottle, suppose it suddenly loses pressure on its right side. It has passed through a wall, from one chamber into another. The surface passes through itself, the way we pushed the neck through the base in the earlier depiction. We're trying to avoid this; therefore the right side of our fish is always under pressure. As the fish swims the mobius circuit, and back to start, it is reflected. By continuity, there is still high presssure on its right side, which is now facing us. All points on both sides of the surface are under pressure. There is no inside and outside. In other words, the complement of our klein bottle is one connected component.
This introduces the notion of an orientable surface, which generalizes to higher dimensions. A surface is nonorientable if you can return to start and find yourself reflected. Such a surface cannot separate space into two pieces, an inside and an outside, unless it creates a small chamber by passing through itself, which is usually not what we want. The klein bottle is nonorientable, and it cannot have a well defined interior.
Well - a mobius strip (nonorientable) has no interior either, yet it exists in 3 space.
True, it exists in 3 space, but it has edges; the klein bottle does not. That makes all the difference. The klein bottle has to wrap up onto itself, without having an inside.
Assume a surface is compact, without boundary, in 3 space, and it has no inside (like the klein bottle). Place a fish on the surface and draw a path through 3 space that connects its left side to its right side. If you think about the mobius strip, the path arcs around the edge of the strip and touches the other side. Assuming the surface is not infinitely crinkled, pull the path in until it hugs the surface. In fact, the path projects onto the surface. The fish can swim along, as we follow the path. By continuity, the path is always on the left side of the fish, but at some point the path switches over to the right side of the fish. This happens at an edge. In other words, the surface has to have a boundary. If there is no boundary, then there is no path through space from one side to the other.
In summary, a nonorientable, compact surface with no boundary, such as the klein bottle, cannot have an inside, yet it must. Therefore it cannot be represented in 3 space.
Naturally this is a rough outline. A rigorous proof would consume dozens of pages, and is beyond the scope of this article.
Note that the klein bottle is easily embedded in four space. The neck need not punch through the wall at the base; it can use the fourth dimension to sneak in the back door. Suddenly it is just there, inside the walls of the base, whence the top fans out to meet the bottom.
even a space pasted to itself can exhibit the same problem. Sew the left and right edges of a square together to make a tube. Let the base consist of rational open squares at the left, irrational open squares at the right, and arbitrary open squares down the middle. Once again base sets do not combine across the border.
If p q and r are in S, the triangular inequality is inherited from S. So let p and q lie in S while r lies in T. Let x be the point in S0 that minimally connects p and r. (If there is no such x, let x be a point that is within ε of our lower bound.) Let y ∈ S0 join q and r. Suppose the distance from p to r is too long, relative to the other two sides. Traveling from p to y to r, instead of passing through x, is even longer. this takes |h(y),r| out of the picture, and |p,y| > |p,q| + |q,y|, which is a contradiction.
If |p,q| is too long then it is longer than |p,x| + |x,y| + |y,q|. Once again this is impossible, hence U is a metric space, and the open balls form a base for a topology.
Is this the topology of U? Let O be an open set in S, missing S0, and select p in O. Enclose p in an open ball B, which is contained in O. Turning towards the balls of U, and using the same radius, B is still an open ball. (Nothing from T is dragged in.) Therefore O is covered in open balls and is open. The only tricky part is when O is a set a+B, with a common border W in S0 and T0. If p is in O, and outside of the border, use the above reasoning to place p inside an open ball inside O. (We need S0 and T0 to be closed for this step.) Finally, let p lie in S0. Find an open ball in S containing p and an open ball in B containing p. (These balls are cut off by the border.) Choose the lesser radius and build a ball about p in U. This lies within the union of the two open balls, and within A+B. Every open set in U has been covered with open balls.
Conversely, let O be an open ball in U. Cover the points of O with open sets of the form A, B, or A+B. This is similar to the above; I'll leave the details to you. Therefore the joint metric establishes the topology of U.
You can also sew a metric space to itself, and the result is another metric space. (We're still assuming S0 and T0 are closed, and h preserves distance.) The distance from p to q is the original distance in S; or the lower bound of |p,x| + |h(x),q|, or |p,h(x) + |x,q|.
Suppose p q and r violate the triangular inequality. If you need them, select x y and z as intermediates, to bring pairs of points closer together. You need at least one intermediate, or p q and r violate the triangular inequality within S.
Assume z, passing from p to q, is the only intermediate, and suppose |p,q| is too long, relative to the other two sides. The distance from p to q within S is even longer, and that is a contradiction. If |p,r| is the (excessivley) long side, then it is longer than |p,z| + |h(z),r|, and we should have used z to bring p and r closer together. (This is true even if x stands between p and r.) If |q,r| is the long side, one can use z to bring q and r closer together. (This is true even if y stands between q and r.)
If two sides have intermediates and the third side does not, and one of the first two sides is too long, let q → y → r be the long side, and apply the previous paragraph.
Let x and y be the two intermediates, while p,q is the long side. If x and y both point towards r, so to speak, then |p,q| > |p,x| + |x,y| + |y,q|, which is impossible. The same holds if x and y point away from r; just use h(x) and h(y) in the above inequality.
Continuing the above, let x point towards r and let y point away. We know that p,q is too long, but vectoring through x/h(x) is even longer. Thus |h(x),q| > |h(x),r| + |r,y| + |h(y),q|. This is a triangle with one intermediate that violates the triangular inequality. This has already been dealt with.
Finally suppose all three intermediates x y and z are in play. If p → z → q is the long side, then p,q direct is even longer. If x and y lead towards, or away from r, this is ruled out by an earlier paragraph. So let x point towards r and let y point away. Passing through z, or perhaps h(z), is too long, but passing through x is even longer. This builds the same contradictory triangle we saw in the previous paragraph. Therefore the triangular inequality holds, and U is a metric space.
A proof similar to the above shows the topology of the metric space is indeed the topology of U.