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Lecture 1: August 25 Introduction. Topology Grew out of Certain Questions in Geometry and Analysis About 100 Years Ago

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Lecture 1: August 25 Introduction. Topology Grew out of Certain Questions in Geometry and Analysis About 100 Years Ago Lecture 1: August 25 Introduction. Topology grew out of certain questions in geometry and analysis about 100 years ago. As Wikipedia puts it, \the motivating insight behind topology is that some geometric problems depend not on the exact shape of the objects involved, but rather on the way they are put together. For example, the square and the circle have many properties in common: they are both one dimensional objects (from a topological point of view) and both separate the plane into two parts, the part inside and the part outside." In other words, topology is concerned with the qualitative rather than quantitative aspects of geometric objects. The fundamental objects in topology are \topological spaces" and \continuous functions"; both were defined in more or less their current form by Felix Hausdorff in 1914. Like the concept of a group in algebra, topological spaces are very useful for unifying different parts of mathematics: they show up naturally in analysis, geometry, algebra, etc. Most mathematicians therefore end up using both ideas and results from topology in their research. On the other hand, topologist nowadays do not study all possible topological spaces { instead, they focus on specific classes such as 3-dimensional manifolds. In the course, we will look at the most important definitions and results from basic point set topology and elementary algebraic topology. Our textbook will be the second edition of Topology by James Munkres, but I will not present things in exactly the same order. Most of the homework questions, however, will be from the textbook. Metric spaces. The goal of today's class is to define topological spaces. Since it took people some time to find a good definition, let us try to retrace at least a small portion of this process. One concern of 19th century mathematics was to create rigorous foundations for analysis. This lead to the study of continuous and differentiable functions on subsets of the real line R and of Euclidean space Rn. Here the notion of \distance" between points plays an important role: for example, a function f : R ! R is continuous if, for every x 2 R and every real number " > 0, one can find another real number δ > 0 such that jf(y) − f(x)j < " for every y 2 R with jy − xj < δ. By abstracting from the properties of distance in Euclidean space, people arrived at the idea of a \metric space". Definition 1.1. Let X be a set. A metric on X is a function d: X × X ! R with the following three properties: (a) One has d(x; y) ≥ 0 for every x; y 2 X, with equality if and only if x = y. (b) d is symmetric, meaning that d(x; y) = d(y; x). (c) The triangle inequality: d(x; z) ≤ d(x; y) + d(y; z) for every x; y; z 2 X. The pair (X; d) is then called a metric space. The name of the triangle inequality comes from the fact that, in a triangle in Euclidean space, the length of each side is smaller than the sum of the lengths of the two other sides. Drawing pictures in the plane can be useful to visualize what is going on { but keep in mind that things like \straight line" or \triangle" do not make actually make sense in a general metric space. The only notions that make sense are those that can expressed in terms of distances between points. One such 1 2 notion, which is also useful in Euclidean space, is that of an \open ball": if x0 2 X is a point, and r > 0 a positive real number, the open ball of radius r and center x0 is the set Br(x0) = x 2 X d(x0; x) < r : We can get some idea of what a given metric space looks like by visualizing open balls of different radii. Here are some examples of metric spaces: Example 1.2. Euclidean space Rn with the usual notion of distance. Denote the n points of R in coordinates by x = (x1; x2; : : : ; xn), and define the length È 2 2 kxk = x1 + ··· + xn: Then the distance between two points x; y 2 Rn is given by È 2 2 d(x; y) = kx − yk = (x1 − y1) + ··· + (xn − yn) : It may seem pretty obvious that this satisfies the axioms for a metric, but let us make sure. In the first condition, d(x; y) ≥ 0 is clear from the definition; moreover, d(x; y) = 0 if and only if xi − yi = 0 for every i = 1; : : : ; n if and only if x = y. The 2 2 second condition is also clear since (xi − yi) = (yi − xi) . Checking that the third condition holds requires a little bit more work. We first prove the following inequality for lengths: (1.3) kx + yk ≤ kxk + kyk: After taking the square and expanding, we get kx + yk2 = (x + y) · (x + y) = kxk2 + 2x · y + kyk2; where x · y = x1y1 + ··· + xnyn is the dot product. From the Cauchy-Schwarz inequality in analysis, we obtain x · y ≤ kxkkyk; and therefore 2 kx + yk2 ≤ kxk2 + 2kxkkyk + kyk2 = kxk + kyk ; which proves (1.3). Returning to the third condition, we now have d(x; z) = kx − zk = (x − y) + (y − z) ≤ kx − yk + ky − zk = d(x; y) + d(y; z); and so the triangle inequality holds and d is a metric. Example 1.4. Another metric on Rn is given by setting d(x; y) = max jxi − yij: 1≤i≤n Unlike before, it takes almost no effort to verify all three axioms. The \open balls" in this metric are now actually open cubes. 2 Example 1.5. Let X ⊆ R be the union of all the vertical lines x1 = n and all the horizontal lines x2 = n, for n 2 Z. The \taxicab metric" on X is defined by setting d(x; y) = jx1 − y1j + jx2 − y2j: Here is the proof of the triangle inequality: d(x; z) = jx1 − z1j + jx2 − z2j ≤ jx1 − y1j + jy1 − z1j + jx2 − y2j + jy2 − z2j = d(x; y) + d(y; z): 3 Example 1.6. Another interesting example is the \railroad metric" on R2. Choose a point P 2 R2 in the plane, and define ¨kx − yk if the three points x; y; P are collinear; d(x; y) = kx − P k + ky − P k otherwise: To see the analogy with railroads, think of P as being the capital city of a country, in which all railroad lines go through the capital. I will leave it as an exercise to show that this defines a metric. Example 1.7. On an arbitrary set X, one can define the trivial metric ¨0 if x = y, d(x; y) = 1 if x 6= y. In this case, open balls of radius 0 < r < 1 consist of only one point. The usual "-δ-definition of continuity carries over to the setting of metric spaces. Suppose that (X; dX ) and (Y; dY ) are two metric spaces. Definition 1.8. A function f : X ! Y is said to be continuous if, for every point x 2 X and every real number " > 0, one can find a real number δ > 0 such that 0 0 0 dY f(x); f(x ) < " for every x 2 X with dX (x; x ) < δ. More graphically, the condition says that f should map the entire open ball Bδ(x) into the open ball B" f(x) . Equivalently, we can look at the preimage −1 0 0 f B" f(x) = x 2 X f(x ) 2 B" f(x) ; and the condition is that it should contain an open ball of some radius δ > 0 around the point x. Sets that contain an open ball around any of their points are called \open"; this is the same use of the word open as in \open intervals". The precise definition is the following. Definition 1.9. Let X be a metric space. A subset U ⊆ X is called open if, for every point x 2 U, there is some r > 0 such that Br(x) ⊆ U. Note that the empty set is considered to be open: the condition in the definition is vacuous in that case. It is also obvious that X itself is always open. The following lemma shows that open balls { as defined above { are indeed open. Lemma 1.10. In a metric space X, every open ball Br(x0) is an open set. Proof. By picture. If x 2 Br(x0) is any point, then d(x; x0) < r, and so the quantity δ = r − d(x; x0) is positive. Intuitively, δ is the distance from the point x0 to the boundary of the ball. Now Bδ(x) ⊆ Br(x0); indeed, if y 2 Bδ(x), then we have d(y; x0) ≤ d(y; x) + d(x; x0) < δ + d(x; x0) = r by virtue of the triangle inequality. It is clear from the definition that if fUigi2I is any family of open subsets of X, indexed by some set I, then the union [ Ui = x 2 X x 2 Ui for some i 2 I i2I 4 is again open. Similarly, given finitely many open subsets U1;:::;Un ⊆ X, the intersection U1 \···\ Un = x 2 X x 2 Ui for every i = 1; : : : ; n is also open. In fact, if x 2 U1 \···\ Un, then x 2 Ui; but Ui is open, and so Bri (x) ⊆ Ui for some ri > 0. Now if we set r = min(r1; : : : ; rn), then Br(x) ⊆ U1 \···\ Un; proving that the intersection is again an open set.
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