Real Analysis

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Defining metrics on a set
jason carlson
Fichas por jason carlson, actualizado hace más de 1 año
jason carlson
Creado por jason carlson hace más de 6 años
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1. Define a "metric on a set" 2. What is the usual metric on R(or C)? 3. What is the usual metric on R^n? 4. What is the discrete metric on any set X? 5. Taxicab metric on R^n? 6. What is the p-norm metric on R^n? 7. How are the taxicab metric and p-norm related? Bonus (already did it on first try): Prove that the max{|x_i - y_i|} is equivalent to the limit of the p-norm as p approaches infinity on x,y in R^n. Note: We use the sup when n can be infinite. 1. A metric on a set X is a function d: X x X -> R s.t. for x,y,z in X: i) d(x,y) >= 0 (positivity) ii) d(x,y) = 0 iff x=y (definiteness) iii) d(x,y)=d(y,x) (symmetry) iv) d(x,y) <= d(x,z) + d(z,y) (triangle). 2. It is a function d: R x R-> R s.t. d(x,y) = |x-y| for all x,y in R. 4. It is the function d: Rn x Rn-> R s.t. d(x,y) = sqrt( sum(x_i^2)) for x,y in R. 5. It is the function d: Rn x Rn-> R s.t. d(x,y) = sum({|x_i - y_i| : 1 <= i <= n}) for x,y in R. 6. It is the function d_p: Rn x Rn-> R s.t. d(x,y) = ( sum(|x_i - y_i|^p))^(1/p) for x,y in R. 7. the taxicab metric is the result when p = 1. p = infinity gives the max{|x_i - y_i|: 1<= i <= n}. Bonus: Too long to type here.
Prove: d(x,y) <= d(x,z) + d(y,z) iff d(x,y) >=|d(x,z) - d(z,y)| for x,y,z in a metric space. -> d(x,y) <= d(x,z) + d(y,z) for all x,y,z -> d(x,z) + d(y,z) <= d(x,z) + d(y,x) + d(x,z) -> d(y,z) - d(x,z) < = d(y,x). By symmetry we have d(x,z) - d(y,z) <= d(y,x) = d(x,y). Together, this gives the intended result. <- d(x,y) >= d(x,z) - d(z,y) - > d(x,z) <= d(x,y) + d(z,y) -> d(x,y) <= d(x,z) + d(z,y) (by symmetry)
Prove: d(x,y) <= d(x,z) + d(z,y) iff d(x,y) <= d(x,z_1) + d(z1, z2) + ...+ d(zn, y) for x,y,z in a metric space. For the forwards direction, simply reapply the triangle inequality with induction. For the backwards direction, we have it automatically true by setting n=1.
Define a norm on a vector space V (over R or C) A norm on V is a function ||*||: V -> R s.t. for all x,y in V i) ||x|| >= 0 (positivity) ii) ||x|| = 0 iff x = 0 (definiteness) iii) ||c*x|| = |c|*||x|| for c in a field F (homogeneity) iv) ||x + y|| <= ||x|| + ||y|| (triangle)
1. What is the automatic metric that can be associated to any norm ||*|| on a vector space V? 2. Prove that this metric is in fact a metric over R^n. 1. The metric is d(x,y) = ||x - y||. 2. ||x-y||>= 0 by def. If ||x-y||= 0, x-y =0 -> x=y. If x=y, x-y= 0 -> ||x-y|| =0. To prove symmetry, we note (x-y) = - (y-x). By homogeneity, the norms of x-y and y-x should be the same. Lastly, ||x-y+z-z|| <= ||x -z|| + ||z-y|| -> d(x,y) <= d(x,z) + d(z,y).
1. Define a bounded subset of a metric space. 2. What is the set of bounded functions from a set S to R^k? 3. What is the uniform norm on the set of bounded functions (which is a vector space)? 4. Prove that the uniform norm is a norm (mostly prove the triangle inequality). 1. Let (X, d) be a metric space. We say Y \subsetOf X is bounded if there exists a x0 in X and a positive real number M s.t. d(y, x0) <= M for all y in Y. 2. B(S, R^k) = {f: S -> R^k : the range of f is bounded} 3. ||f||_u = sup{||f(x)|| : x in S} where ||*|| is the standard norm in R^k 4. Clearly sup satisfies positivity, definiteness, and homogeneity (since ||*|| does over R^k). Now we show triangle: ||f+g||_u = sup{||f(x)+g(x)|| : x in S}. But ||f(x) + g(x)|| <= ||f(x)|| + ||g(x)|| -> sup{||f(x)+g(X)||: x in S} <= sup{||f(x)|| + ||g(x)|| : x in S} = sup{||f(x)|| : x in S} + sup{||g(x)|| : x in S} = ||f||_u + ||g||_u.
1. How can we think of R^n as a function? 2. How can we think of C^n of a function? 3. How do the above functions relate to the uniform norm? 4. What norm is this the limit of? Why are they different? 1. B(S, R) where S is {1,..., n} (bounded functions from {1,...,n} to R) since we can't have an element as infinity in R^n tuples. 2. B(S, C) 3. The uniform norm ||*||_u is the maximum element in the tuples. 4. It is the limit of the p-norm (and is usually denoted ||*||_infinity). BUT, the p-norm is defined on R^n while the uniform norm is defined on B({1,...,n}, R)
1. Define an inner product on a vector space V. 2. verify that the inner product on the complex numbers satisfies <z, ax + by> = conj(a)<z, x> + conj(b)<z,y>. 3. Verify that in any complex inner product space, <x+y,x+y> = <x,x> + 2*Re<x,y> + <y,y> 4. What is the standard inner product on R^n? On C^n? 5. Is the inner product linear in the second variable in R^n? In C^n? 6. How can we relate the usual norm on R^n to the inner product on R^n? C^n should be the same. 1. Let V be a vector space over R or C. An inner product on V is a function <·, ·> : V ×V → F such that i) <x, x> ≥ 0. ii) <x, x> = 0 only if x = 0. iii) <x, y> = <y, x>. iv) <ax + by, z> = a<x, z> + b<y, z> for all a, b ∈ F (and x, y, z ∈ V 2. <z, ax + by> = conj(<ax + by, z>) = conj(a<x,z> + b<y,z>)= conj(a)*conj(<x,z>) + conj(b)*conj(<y,z>) = conj(a)*<z, x> + conj(b)*<z,y>. 3. <x+y, x+y> = <x, x+y> + <y, x+y> = <x, x> + <x, y> + <y, x> + <y, y> = <x, x> + 2*Re<x,y> + <y,y> (in <x,y>, we have one term with i pos. the other neg., in <y,x> it is the opposite). 4. The standard inner product on R^n is the dot product. On C^n we have the following: <x,y> = x_1*conj(y1)+...+x_n*conj(y_n) 5. In R^n yes, since conj(f) for f in R is f. In C, it is not since conj(a)*<z, x> + conj(b)*<z,y> != <z, x> + <z,y> if a,b are not real. 6. ||x|| = sqrt(<x, x>), ||x+y|| = sqrt(<x-y,x-y>).
1. Define a subspace of a metric space. 2. Is the Torus/ circle a subspace of Euclidean space? Why? 1. Let (X,d) Ybe a metric space and suppose Y \included X. We then say (Y, d|Y x Y) is a subspace of (X, d). 2. Yes. Restrict d_l^2 to {(x,y) : x^2 + y^2 = 1} = Y. Then (Y, d_l^2 | Y x Y) is a subspace of (X, d_l^2) (with X = R^2)
1. Is X= {0,1}x{0,1}x... = {(x_1, x_2, ...) : x_i in {0,1} for all i} countable? If not, what is its cardnality. 2. What is the ultrametric inequality? Is it stronger than the triangle inequality or weaker? 3. define k(x,y) = min{i : x_i != y_i} (why is this possible?). Then define d(x,y) = 1/k(x,y) if x!= y and 0 if x=y. Prove the ultrametric inequality. 1. No. It is uncountable and equivalent in cardnality to P(N) (set of all subsets of N). This is because we can thing of each x_i representing whether or not i is in a specific set. This gives a bijection to P(N). 2. The ultrametric inequality says: d(x,y) <= max{d(x,z), d(z,y)} for any x,y,z. This is clearly stronger than the triangle inequality (it implies the triangle inequality is true) 3. You can take the minimum by the well-ordering property of N (for every subset of N there exists a smallest element). Suppose d(x,y) > d(x,z). Then k(x,y)< k(x,z). We will now show k(x,y)>= k(z,y). We have x_k(x,y) != y_k(x,y) and x_i = y_i for i<k(x,y). Similarly, we have z_k(z,x) != x_k(z,x) and z_i = x_i for i<k(z,x). If k(x,y) < k(z,y), then z_i = y_i for all i <= k(x,y). But z_i = x_i = y_i for i < k(x,y) < k(z,x) and z_i != y_i for i = k(x,y). So k(z,y) <= k(x,y). This means d(x,y) <= d(z,y). To finish, we note that the case d(x,y) > d(y,z) is the same, and we can therefore conclude the hypothesis.
1. Define the open ball in a metric space. 2. Define the closed ball in a metric space. 3. What are the intervals in R for the closed ball and open ball centered at a with radius r? 4. Define a open set in a metric space. 5. Define a closed set in a metric space. 1. Let (X, d) be a metric space. Then for a in X and r> 0 we define the open ball centered at a with radius r as: B_r(a)={x in X : d(x,a) < r}. 2. Same thing as before, but B_r(a) = {x in X : d(x,a) <= r}. 3. B_r(a) = (a-r, a+r) and B_r(a) = [a-r, a+r] for open/closed resp. 4. Let (X,d) be a metric space and let E \subsetOf X. We say E is open in X iff for every x in E there exists B_r(x) \subsetOf E. 5. Let (X,d) be a metric space and let E \subsetOf X. We say E is closed in X iff E^c is open (E^c = X\E).
1. Prove open balls are open sets. 2. Prove closed balls are closed sets. 1. Let B_r(a) with r>0 be an arbitrary open ball centered at a in X with (X,d) a metric space. Choose x in B_r(a). Take r2= r- d(a,x). We will now show B_r2(x) \subsetOf B_r(a). Let y in B_r2(x). Then d(x,y) < r2 = r - d(a,x). But this means d(a,y) < d(a,x) + d(x,y) = d(a,x) +r - d(a,x) = r. So y in B_r(a). This implies B_r2(x) \subsetOf B_r(a).
1. Prove that the union of any collection of open sets in a metric space is open. 2. Prove that any finite intersection of open sets is open (and why finite?). 3. Prove that any finite union of closed sets is closed. 4. Prove that any intersection of a collection of closed sets is closed. 1. Let {A_i : i in I} be an arbitrary family of sets s.t. A_i is open in (X,d) (some metric space). Then let x in \union{A_i}_{i in I}. Then x in A_i for some A_i. Since A_i open, we have there exists B_r(x) \subsetOf A_i. But then B_r(x) is a subset of \union{A_i}_{i in I}. 2. Let {A_i : 1<= i <= n} be a collection of finitely many open sets in an arbitrary metric space (X,d). Suppose x in \intersect{A_i : 1<= i <= n}. For each A_i there is an open ball B_r^i(x) that is contained in A_i. Then clearly \intersect{B_r^i(x) : 1<= i <= n} \subsetOf \intersect{A_i : 1<= i <= n}. Note: If this were infinite, then we could take any element x in X and intersect open balls centered at X s.t. x is the only element of this intersection. This would mean {x} is open. But this is only what the discrete metric should do. 3. By demorgan's laws this is equivalent to showing any finite intersection of open sets is open, which we have proven. 4. By Demorgan's laws, this is equivalent to showing any union of a collection of open sets is open, which we have shown.
1. Prove X and 0 are both open and closed. 2. If X=R^n, are there any other sets that are both open and closed? 3. Prove that a singleton set in a metric space is closed. 4. Prove that a finite subset of a metric space is closed. 5. Prove that (in R^n with the usual metric) open half-spaces ({x : x_i > c}, {x : x_i < c}) are open. 6. Similarly, show that closed half-spaces are closed. 1. Let x in X. Then B_r(x) in X for all r>0 clearly. Moreover, 0 is open (vacuously true). This implies they are both false too (since they are complements of eachother) 2. No. This is a special property of R^n. 3. We will show {x}^c is open to show {x} is closed in (X,d). Let y in {x}^c = X\{x}. Then d(x,y) > 0 (since d(x,y) = 0 iff x=y). This means B_d(x,y)(y) \subsetOf X\{x}. So X\{x} is open and {x} is therefore closed. 4. This is just a finite union of singleton sets -> finite union of closed sets -> closed. 5. Let y in {x : x_i > c}. Then set r = y_i - c > 0. Then suppose z in B_r(y). Then z_i - y_i <= |z_i - y_i| <= d(y,z) < r. So y_i - z_i< y_i - c -> z_i > c. So z in {x : x_i > c} -> B_r(y) \subsetOf {x : x_i > c}. {x : x_i < c} follows similarly. 6. {x : x_i >= c} is closed if {x : x_i < c} is open. But this follows from before.
1. Define a open box in R^n. 2. Define a closed box in R^n. 3. Why are they open/closed respectively? 4. Are most(?) sets open, closed, open and closed, or neither? 1. An open box in R^n is a set of the form (a_1, b_1)x...x(a_n, b_n) for -inf <= a_i <= b_i <= inf for each i. 2. A closed box in R^n is a set of the form [a_1, b_1]x...x[a_n, b_n] for -inf <= a_i <= b_i <= inf for each i. 3. An open box is a finite intersection (of at most 2n) open half-spaces. A closed box is a finite intersection (of at most 2n) closed half-spaces. 4. Neither (think [2,3)). This may only apply to R/R^n
1. Give an example of a subset of R that is neither open nor closed (with proof). 1. [0, 1). [0,1) is not open since any open ball centered at 0 will intersect R\[0,1). Moreover, we know the compliment is not open since [0,1)^c = (-inf, 0) U [1, inf) and any open ball centered at 1 intersects [0,1). Thus [0,1) is not closed. We conclude [0,1) is neither closed nor open.
1. Define the interior of a subset of a metric space. 2. Define the closure of a subset of a metric space. 1. Let (X,d) be metric space and E \subsetOf X. The interior of E is the set int(E) = U {u : u \subsetOf X and u is open}. 2. Let (X,d) be metric space and E \subsetOf X. The closure of E is the set E^{bar} = \intersect {u : X \subsetOf u and u is closed}.
1. How is the closure of a subset E of a metric space X related to whether or not E is closed? 2. How is the interior of a subset E of a metric space X related where or not E is open? 1. E^{bar} = E iff E is closed (trivial). 2. interior(E) = E iff E is open.
Remark 2.12. We observe that i) int(E) is an open set, and is the largest open set contained in E. ii) E^{bar} is a closed set, and is the smallest closed set containing E. iii) E^{bar} = (int(E^c))^c and int(E) = ((E^c)^{bar})^c Prove iii). Proof of iii): (E^{bar})^c = U{K^c : E \subsetOf K and K closed} = U{K^c : K^c \subsetOf E^c and K^c open} = U{K : K \subsetOf X\E and K open} = int(E^c). But this implies E^{bar} = (int(E^c))^c. The second formula is found by replacing E with E^c.
1. Prove x in int(E) iff there exists r > 0 s.t. B_r(x) \subsetOf E. 2. x in E^{bar} iff for every r > 0 B_r(x) \intersect E != 0. 1. Suppose x in int(E). Then x in U{u : u \subsetOf E and u open}. Thus, x in u for some u open and u \subsetOf E. But then x in B_r(x) \subsetOf u \subsetOf E for some r > 0. Suppose there exists r > 0 s.t. B_r(x) \subsetOf E. Since B_r(x) is open and a subset of E, x is in the union of all open subsets of E. 2. We note that x in (E^{bar})^c iff x in int(E^c). But x in int(E^c) iff there exists B_r(x) \subsetOf E^c for some r > 0. But this is true iff B_r(x) \intersect E = 0. Negating the far LHS and the far RHS gives x in E^{bar} iff B_r(x) \intersect E != 0.
In R, 1. What is int((a,b])? 2. What is int(Z)? 3. What is int(Q)? 4. What is (0,1]^{bar}? 5. What is Z^{bar}? 6. What is Q^{bar}? 1. (a,b) 2. 0 since every open ball for any point in Z is not contained in Z. 3. 0 since any open ball will intersect the irrationals. 4. [0,1]. 5. Z (Z is closed in R). 6. R.
1. Define the neighborhood of a point in a metric space. 2. Give an example of a set that is the neighborhood of one point but not another point and explain why. 1. Let (X,d) be a metric space. Moreover, let x in X. We then say N \subsetOf X is a neighborhood of x if x in int(N). 2. [0,1] is a neighborhood of 0.0001 but not 0.
Define the support of a function f: X -> R^k where X is a metric space Let (X,d) be a metric space. The support of a function f: X -> R^k is the set supp(f) = {x in X : f(x) != 0}^{bar}
Define when a subset of a metric space is dense. Let (X,d) be a metric space. E \subsetOf X is dense in X if E^{bar} = X.
1. When is a metric space separable? 2. Is R separable? Why? 3. Does the fact that the R\Q (irrationals) are dense in R imply R is separable? Why? 1. When it contains a countably dense subset. 2. Yes. Q is a countably dense subset of R. Note that N is not a countably dense subset of R, but R is still separable. 3. No. The irrationals are dense, but they are not countable (they are uncountable).
1. Define an accumulation point (limit point/ cluster point) of a subset of a metric space. 2. What denotes the set of all accumulation points for a subset E of a metric space? 1. Let (X,d) be a metric space with E \subsetOf X and a in X. The point a is an accumulation point of the set E if for all r>0 B_r(a) \intersect E is infinite. 2. E'
Assume X =R: 1. What is {1, 1/2, 1/3,...}' ? 2. What is {0, 1, 1/2, 1/3, ...}' ? 3. What is Z' ? 4. What is Q' ? Bonus: Prove the above equalities. 1. {0}. 2. {0}. 3. {}. 4. R. Note that if x in Q, then any open ball that contains x intersects inf. many points in Q. If x not in Q, the same thing happens. Thus, all x in R are in Q'.
1. Prove lemma: a in E' iff for all r>0, (E\{a}) \intersect B_r(a) !=0. 2. Prove E^{bar} = E U E'. TO BE CONTINUED
Define a bounded set in a metric space We say a subset Y in a metric space (X,d) is bounded if there exists a open or closed ball that contains Y. (This is because if there is a point for each point in Y s.t. the distance is less than or equal to M, we can combine these M's and points to a single point and M to obtain a open or closed ball)
CANTOR SET TO BE CONTINUED
1. Define a sequence in any set X 1. A sequence in X is a function x: N -> X.
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