HOMEWORK #4 - MA 504

HOMEWORK #4 - MA 504 PAULINHO TCHATCHATCHA Chapter 2, problem 19. (a) If A ad B are disjoit closed sets i some metric space X, prove that they are separated. (b) Prove the same for disjoit ope set. (c) Fix p X, δ > 0, defie A to be the set of all q X for which d(p, q) < δ, defie B similarly, with > i place of <. Prove that A ad B are separated. (d) Prove that every coected metric space with at least two poits is ucoutable. Hit: Use (c). (a) Let A ad B are disjoit closed sets i some metric space X. We wat to prove that they are separated. Sice A ad B are disjoit ad closed, we have A B = A B =, A B = A B =. Hece by defiitio 2.45, A ad B are separated. (b) Let A ad B are disjoit ope sets i some metric space X. We wat to prove that they are separated. Sice A ad B are ope, the complemets of A ad B, say A c ad B c are closed. Suppose that A ad B are ot separated, ie, either A B or A B is oempty. Assume without loss of geerality that A B. Let x A B. The, sice B is ope, there exists a eighborhood N of x such that N B. But sice x A, N A. Therefore, sice A N A B, A B, a cotradictio. Hece A ad B are separated. (c) Fix p X, δ > 0, defie A = {q X d(p, q) < δ}, B = {q X d(p, q) > δ}. First it ot hard to see that A ad B are ope. Ideed, if q A, the cosider the eighborhood N r (q) = {z X d(q, z) < r}, 0 < r < δ d(p, q). The if z N r (q), we have d(p, z) d(p, q) + d(z, q) < d(p, q) + δ d(p, q) = δ. So z A, ad sice z N r (q) is arbitrary, N r (q) A ad A is ope. Similarly oe ca show that for ay q B, N r (q) B for ay 0 < r < d(p, q) δ, usig that for ay z N r (q) d(p, z) d(p, q) d(q, z) > d(p, q) d(p, q) + δ = δ. Clearly A ad B are disjoit ope sets i X, so by (b) they are separated. (d) Let X be a coected metric space with at least two poits. We wat to show that X ca ot be coutable. 1

2 PAULINHO TCHATCHATCHA For each t (0, 1), let r t = td(x 1, x 2 ). We have that sice X has at least two poits, say x 1, x 2 X, the d(x 1, x 2 ) > 0, ad r t > 0. For each t (0, 1) cosider A t = {q X d(x 1, q) < r t }, B t = {q X d(x 2, q) > r t }. We have that x 1 A ad x 2 B, so A ad B are oempty ope sets. So either X = A t B t or there exists x t X such that d(x 1, x t ) = r t. Sice X is coected, X A t B t, ad for each t (0, 1), there exists x t X s.t. d(x 1, x t ) = r t, but this gives a ijective correspodece f : (0, 1) X, f(t) = x t. Sice (0, 1) is ucoutable, sice we have this ijective correspodece f, X is also ucoutable. Q.E.D. Chapter 2, problem 22. A metric space is called separable if it cotais a coutable dese subset. Show that R k is separable. Let Q k be the set of poits of R k which have oly ratioal coordiates, ie, Q k = Q Q k times = {(q 1,..., q k ) R k q j Q, j = 1,..., k}. Clearly Q k is coutable sice it is a fiite product of coutable sets. We ow wat to show that Q k is dese i R k, ie, for ay x = (x 1,..., x k ) R k ad ay ɛ > 0, there exists q Q k such that q x < ɛ. Sice Q is dese i R, for each j = 1,..., k, there exists q j Q such that Let q = (q 1, q 2,..., q k ). The q j x j < ɛ k, j = 1,..., k. q x 2 = (q 1 x 1 ) 2 + + (q k x k ) 2 < ɛ2 k Hece Q k is dese i R k ad R k is separable. +... + ɛ2 k = ɛ2. Chapter 2, problem 29. Prove that every ope set i R is the uio of a at most coutable collectio of disjoit segmets. Let O R be ope. Assume that O is oempty. For each q O Q, let R q = {r > 0 (q r, q + r) O}. Sice O is ope, by what we showed above R q ad if r 0 R q, the r R q for every 0 < r r 0. Note that if r Rq (q r, q + r) = R, the O = R, otherwise r q = sup R q <.

HOMEWORK #4 - MA 504 3 So assume that sup R q <, ad cosider r q = sup R q. We see that I q = (q r q, q + r q ) = r Rq (q r, q + r) O. We claim that O = q O Q I q. Sice O Q Q, the O Q is coutable, so the uio above is also coutable. Clearly, q O Q I q O. Now if x O, there exists ɛ > 0, such that (x ɛ, x + ɛ) O. By the desity of Q i R, there exists q OQ such that 0 < x q < ɛ/2. We see that (q ɛ/2, q + ɛ/2) O. Ideed, if z (q ɛ/2, q + ɛ/2), the z x z q + q x < ɛ/2 + ɛ/2 = ɛ, ie, z (x ɛ, x + ɛ) O. So ɛ/2 r q by the defiitio of r q, ad we have that Sice x O is arbitrary, we have that x (q ɛ/2, q + ɛ/2) (q r q, q + r q ) = I q. O q O Q I q, ad hece O = q O Q I q. Sice O Q is coutable, say O Q = {q 1, q 2,..., q,...} ad I qj = I j. The O = j=1i j. Let E = I \ 1 j=1 I j, = 2, 3,..., E 1 = I 1. We have that O = j=1i j = =1E. Note that by costructio, each E is either a segmet, a fiite disjoit uio of segmets or empty, ad E E m = if m. Therefore the equility above proves that O is the uio of a at most coutable collectio of disjoit segmets. Chapter 3, problem 1. Prove that covergece of {s } implies covergece of { s }. Is the coverse true? Suppose that s s. We have s s s s. Sice s s, give ɛ > 0, there exists N such that s s < ɛ for all N. By the iequality above we see that s s < ɛ for all N. Sice ɛ > 0 is arbitrary, we see that s s. The coverse is ot true. Ideed, cosider s = ( 1). The {s } does NOT coverge, but s = 1 coverge to 1. Chapter 3, problem 2. Calculate lim ( 2 + ).

4 PAULINHO TCHATCHATCHA We have 2 + = 2 + 2 + + ( 2 + + ) = 2 + 2 2 + + = Now ote that Therefore 2 + + = 1 1 1 + 1 + 1 2 lim ( 2 + ) = 1 2. Chapter 3, problem 3. If s 1 = 2, ad s +1 = 2 + s, ( = 2, 3,...), prove that {s } coverges, ad that s < 2 for = 1, 2, 3,... as., > 0. 2 + + Let s R be such that s = 2 + s. We see that such s exists sice the fuctio f(s) = s 2 + s is cotiuous ad f(4) > 0, f( 2) < 0, so there must be a s, 2 < s < 4, such that f(s) = 0. We have that s +1 s = 2 + s = 2 + s = 2 + s 2 + s 2 2 + s + 2 + s ( + s + 2 + s) = s s 2 + s + 2 + s s s 2, 2 The iequality follows form the fact that s, s > 0, so 2 + s, 2 + s > 2, so 2 + s + 2 + s > 2 1 2 ad 2 + s + 2 + s < 1 2 2. We have ow that s s 2 2 = s s 2 s + s 2 s + s = s s 2 2( s + s). Now ote that, by defiitio, s 2 for all, ad s > 2, so Therefore we showed s s 2 2( s + s) s s 2 2(2 1/4 + 2 1/4 ) s s. 4 s +1 s s s. 4

HOMEWORK #4 - MA 504 5 If we cotiue this process we have s +1 s s s s 1 s s +1 k s, k. 4 2 4 2 2k So for k =, we have s +1 s s 1 s = s 2 2 +1 s 2 s 4 2 2 2 = 1, 2. 2 22 2 1 Sice 2 0, as, we have that s 2 2 s. Sice x x for ay x 1, ad x 1/2 x x, x 1. Clearly oe has that s 2 = 2 + 2 1/2 < 2 + 2 = 2. Now usig iductio, assumig s < 2, s +1 = 2 + s < 2 + 2 = 2. Therefore s < 2 for = 1, 2, 3,.... Problem A. Show that a sequece {p } is covergig to a poit p if, ad oly if, every subsequece of {p} coverges to p. Suppose that p p. Let {p j } be a subsequece of {p }. Sice p p, for every ɛ > 0, there exists N such that p p < ɛ, N. I particular p j p < ɛ, j N. Hece {p j } coverges to p. Now assume that {p } does ot coverges to p. The give ɛ > 0, for every k N there exists k k such that p k p ɛ. The {p k } does ot coverge to p. Therefore if every subsequece of {p} coverges to p, the {p } is covergig to a poit p. Problem B. Show that a sequece {p } is Cauchy if, ad oly if, diam(e N ) 0 as N (here, E N = {p N, p N+1,...}). Let E N = {p N, p N+1,...}. We have diam(e N ) = sup{ p p m :, m N}. So if {p } is Cauchy, the for every ɛ > 0 there exists M such that p p m < ɛ,, m M. This implies that diam(e N ) ɛ, for all N M. Sice ɛ > 0 is arbitrary, we have that diam(e N ) 0 as N. Now if diam(e N ) 0 as N, the we have for every ɛ > 0, there exists M such that diam(e N ) = sup{ p p m :, m N} < ɛ, N M, i particular p p m < ɛ,, m M, ie, sice ɛ > 0 is arbitrary {p } is Cauchy.