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[answered] MATH 104 - WEEKLY ASSIGNMENT 13 DUE 28 NOVEMBER 2016, BY 16

Need full proofs to 2,3,4,5,8

Just started Analysis on my own, so I need full proofs that use the definitions of "open" and "closed". Will tip for turning in before deadline

MATH 104 - WEEKLY ASSIGNMENT 13

DUE 28 NOVEMBER 2016, BY 16:00 (1) (i) Let (X, d) be a metric space. Let x 6= y ? X. Show that there exist open sets U1

and U2 , w.r.t. the metric d, such that x ? U1 , y ? U2 , and U1 ? U2 = ?.

(This is known as: the topology induced by a metric separates points.)

(ii) Let X be a set, with at least two elements. Show that the trivial topology on X is

not induced by a metric. (2) Let L? ([0, 1]) := {f : [0, 1] ? R, s.t. f is bounded}. This is a vector space (with scalars

in R). For any f ? L? ([0, 1]), we define

kf k? = sup{|f (x)| : x ? [0, 1]}.

This is known as the infinity norm.

(i) Show that kf k? ? R, for all f ? L? ([0, 1]).

(ii) Let C([0, 1]) := {f : [0, 1] ? R, s.t. f continuous}. This is the vector space of

continuous functions on [0, 1]. Show that C[(0, 1]) ? L? ([0, 1]).

(iii) Show that k ? k? : C([0, 1]) ? R is a norm.

(iv) What is the metric on C([0, 1]) that this norm induces?

(v) Explain why kf ? gk? < r means that the graph of g is within distance r from

the graph of f (i.e., if we vertically translate the graph of f by r upwards and by

r downwards, the graph of g is somewhere within the region we cover).

(vi) Let (fn )n?N be a sequence of functions in C([0, 1]). Suppose that kfn ? f k? ? 0,

for some f : [0, 1] ? R. Show that:

(a) fn (x) ? f (x), for all x ? [0, 1].

(This shouldn?t be a surprise, as kfn ? f k? ? 0 means that the graphs of

fn approach the graph of f in a ?uniform? manner, by (v). This means that

convergence w.r.t. k ? k? , known as uniform convergence, implies pointwise

convergence).

(b) f is continuous.

(This is a theorem, known as: the uniform limit of continuous functions

is continuous.)

(vii) Find a sequence of functions (fn )n?N in C([0, 1]), and an f ? L? ([0, 1]), with

fn (x) ? f (x) for all x ? [0, 1], but with kfn ? f k? = 1 for all n ? N. (So,

pointwise convergence doesn?t imply uniform convergence.)

1 2 DUE 28 NOVEMBER 2016, BY 16:00 (3) Let (X, d) be a metric space, and Y ? X. Note that, if we restrict the metric d on Y ,

? y) = d(x, y) for

then we get a metric on Y (i.e., the function d? : Y ? Y ? R, with d(x,

all x, y ? Y , is a metric on Y . This is trivial.) We call the topology induced on Y by d?

the subspace topology on Y .

? for all U ? X open in X (w.r.t. d).

(i) Show that U ? Y is open in Y (w.r.t. d),

? then UY = U ? Y , for some

(ii) Show that, for any UY ? Y open in Y (w.r.t. d),

U ? X open in X (w.r.t. d).

(Hint: Show first that Bd?(y, r) = Bd (y, r) ? Y , for all y ? Y and r > 0.)

(Note that you have proved by (i) and (ii) that the open sets in Y w.r.t. the

subspace topology are the sets of the form U ? Y , where U is open in X.)

(iii) Show that the closed sets in Y w.r.t. the subspace topology are the sets of the form

F ? Y , where F is closed in X. (4) Let (R, d) be R equipped with the usual metric.

(i) Show that Q is neither open nor closed in R.

(ii) Consider Q as a metric space, with metric the restriction of d on Q (as in (3)). Let

a, b ? R \ Q, with a < b. Show that the set {q ? Q : a < q < b}) is both open and

? (Hint: Use (3).) What is the closure of {q ? Q : a < q < b})

closed in Q, w.r.t. d.

in Q?

(iii) Consider [0, 1] as a metric space, with metric the restriction of d on [0, 1]. Show

that, w.r.t. this metric, [0, 1/2) is open in [0, 1]. (5) (i) Let (xn )n?N be a sequence in R3 . We write xn = (x1,n , x2,n , x3,n ) (? R ? R ? R),

for each n ? N. Let x = (x1 , x2 , x3 ) (? R ? R ? R). Show that xn ? x (w.r.t.

the usual metric on R3 ) if and only if xi,n ? xi for all i = 1, 2, 3 (w.r.t. the usual

metric on R).

(I.e., convergence in R3 is convergence coordinate-wise. Of course the above

generalises to any Rm .)

(ii) Show that the cube [0, 1]3 is closed in R3 (w.r.t. the usual metric). Show the same

for the rectangle [a1 , b1 ] ? [a2 , b2 ] ? [a3 , b3 ]. (6) Show that in Rn with the usual metric d (which, remember, is induced by the 2-norm),

Bd (x, r) = x + rBd (0, 1), for all x ? Rn and r > 0.

(Note that this equality doesn?t make sense unless our space is a vector space. It can

be generalised to any vector space, equipped with any metric induced by a norm.) MATH 104 - WEEKLY ASSIGNMENT 13 3 (7) (Important.) Let (X, k ? k) be a normed space. Let (xn )n?N be a sequence in X. Show

that, if xn ? x (w.r.t. k ? k), then kxn k ? kxk. ed (0, 1).

(8) Show that in Rn , with the usual metric d, Bd (0, 1) = B

(I.e., the closure of the open ball is the closed ball. This generalises to any centre and

radius. There are many ways to prove the above; maybe the fastest is by using (5) and

(7).) (9) (Important). Let (X, d) be a metric space, and F ? X. Show that F? = {x ? X :

d(x, F ) = 0}.

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