University:
California State University, Northridge
Course:
MATH 360 | Abstract Algebra I
Academic year:

2024
Views:

201

Pages:

3
Author:

sonalhava9qv

R1 R1 6 g, then 1. (i) d1 (f, f ) = 0 |f (x) − f (x)| dx = 0 0 dx = 0. And, if f = F (x) = |f (x)−g(x)| is not identically zero. Hence ∃x0 so F (x0 ) = 2 > 0. And by continuity, ∃δ Rsuch that ∀x, x0 −δ x < x0 +δ, Rx < R 1F (x) > . So, R1 x0 −δ +δ F (x) dx + x00−δ F (x) dx x0 +δ F (x) dx ≥ d1 (f, g) = 0 F (x) dx = 0 R x0 +δ , dx = 2δ > 0 x0 −δ R1 0 = d1 (f, g) = 0 |f (x) − g(x)| dx, implies |f (x) − g(x)| = 0 for all x, and thus f = g . (ii) d1 (f, g) is symmetric in definition, as |f − g| = |g − f |, and thus equals d1 (g, f ). R1 R1 (iii) d1 (f, h) + d1 (h, g) = 0 |f (x)−h(x)|+ |h(x)−g(x)| dx ≥ 0 |f (x) − g(x)| dx = d1 (f, g). n 2. {fn } : ∀ > 0, ∃N (> 1/) so that ∀n > N, x ∈ R we have 0 < x2 +n 2 ≤ n < . Thus, {f } converges uniformly (and thus also point-wise) to n 2 n 0 on all of R. q 2 {gn } : For any given x, we can pick a sufficiently large N (> x ) so that, n > N implies gn is  close to, 1. Thus, {gn } converges point-wise. But, for any n, if we pick x = n to get gn = 12 . This means that {gn } does not converge uniformly to 1. 3. Let U = {x : d(x, A) < d(x, B )} and V defined anaglgously, with d the distance function defined in problem set 2 problem 1. Clearly, U and V are disjoint and contain A and B respectivly. Now, it remains to show that they are open. Let x ∈ U ,  = 31 (d(x, B) − d(x, A)), and y ∈ B (x). Then, d(y, B) − d(y, A) > d(x, B) − d(x, A) − 2 > 0, and we have y ∈ U and U open. So U and V are as desired. 4. I claim that for U, V open, U ⊂ Ū ⊂ V , there exists W open so that Ū ⊂ W ⊂ W̄ ⊂ V . To see this, consider Ū and X − V , disjoint closed sets. Then there exists W ⊃ Ū , W 0 ⊃ X − V open and disjoint. But then Ū ⊂ W ⊂ X − W 0 ⊂ X − V . Since X − W 0 is closed and contains W , it also contains W̄ . And we have U ⊂ Ū ⊂ W ⊂ W̄ ⊂ V , as desired. 1 Now, letting S1 ⊂ X − B, and S0 ⊃ A as produced by X − B and IntA with the above lemma applied twice. I inductivly create open Sk/2i one level of “i” at a time. At each point if q < r, then S̄q ⊂ Sr . Sk/2i (k odd), is generated by the above lemma between S(k−1)/2i and S(k+1)/2i . So of course, all the sets contain the closure of S0 and are contained in S1 . I define f (x) = inf({1} ∪ {r : x ∈ Sr }). This is set is bounded below and non empty, so the inf exists. It is clear that f (x) = 0 if x ∈ A, f (x) = 1 if x ∈ B, and has range [0, 1]. It now remains to show that f is continuous. I first show f (x) = sup({0} ∪ {r : x ∈ X − S̄r }). In doing this, we only need to consider r in the sets that are terminating binary fractions, as for no other r is Sr defined and thus is it possible for r to satisfy the condition to be in the sets. If, r > f (x) then ∃r0 , r > r0 > f (x) with x ∈ Sr0 ⊂ Sr ⊂ S̄r , making x 6∈ X − S̄r . As 0 ≤ f (x), we can upperbound the sup with f (x). And, ∀r < f (x) ≤ 1, then ∃r0 , r < r0 < f (x) with x 6∈ Sr0 ⊃ S̄r and thus x ∈ X − S̄r . Since all binary fraction 0 < r < f (x) (it is key here that f (x) is bounded above 1 so that all of these r produce valid Sr ) are in the set, and these are dense in the reals, we have f (x) as a lower bound for the sup as well. This fails if f (x) = 0, but then, the additional 0 element saves us. Regardless we have now proved the identity. Next, we show f −1 ([0, r)) is open (r is now any real number). I claim f −1 ([0, r)) = ∪s 0, Br (x)∩S 6= ∅. Let, an be a point in S ∩ B1/n(x). Clearly, an → x 6∈ S, and thus all subsequences converge to x 6∈ S and thus do not converge in S , contradicting the sequential compactness of S . (ii) Suppose S is not totally bounded. Then let r be a radius such that there is no r net of S. Now define an as follows: Let a0 be any point in S and let an ∈ S, such that ∀m < n, d(am , an ) ≥ r. Such an an exists by the lack of an r net of S. But S sequentially compact implies ∃m, n such that d(am , an ) < r which is impossible. So, S must be totally bounded. 7. Clearly if E is totally bounded, then it is totally bounded relative to any metric space containing it. Conversely, if E is totally bounded relative to X, then there is some finite set of point p1 , p2 , . . . pn ∈ X, so that every point in E is within /2 of these points. Without loss of generality, we may assume that there is some qi ∈ E in each of these /2 balls, or else we omit the corresponding pi from the original enumeration. But now, ∪i B (qi ) ⊃ ∪i B/2 (pi ) ⊃ E, by the triangle inequality. And thus for every  we have an epsilon net of E centered around points in E, making E totally bounded. 8. Suppose no such r exists. Then for each r, in particular for each 1/n, there is some xn such that ∀α, B1/n(xn ) 6⊆ Uα . Now, X is compact, so exists ni ∈ Z+ , x ∈ X, so that xni → x. Now, {Uα } covers X, so ∃α, so that x ∈ Uα . As these sets are open, ∃r such that Br (x) ⊂ Uα . And, by convergence, ∃k > 2r , so that d(xk , x) < 2r . But then, B1/k (xk ) ⊂ Br (x) ⊂ Uα , contradicting our construction of {xn }, and implying the existence of such an r. 3

Homological Algebra Solution for Assignment 3

Related Documents

Recommended Documents

Description

Related Documents

Almost There!

Free up your schedule!

Take 5 seconds to unlock