# Real Analysis : Closure, Convergence, Differentiability, Integrability and Sequences

Prove or disprove the following statement. Provide detailed answer and justify all steps.

1). There is a nonempty set S in R such that closure of S is equal to R and the closure of its complement closure(R-S) also is equal to R.

My thoughts on this problem: Q ( rationals) and R-Q (irrationals), but how to prove that the closure of Q is R and that the closure of R-Q is R also.

2). There is a collection of closed intervals I_n in R ( with n in N (natural numbers) ) such that the compact interval [0,1] = union from n=1 to infinity ( I_n), but the interval [0,1] is not equal ( nor is a subset of) any finite union of the intervals I_n. ( If any, explicitly exhibit your intervals I_n)

3). Let a < b. If f: [a,b] -> R is continuous on [a,b], and differentiable on (a,b), then f': (a,b) -> R is differentiable.

4). There is a sequence of differentiable functions f_n : (0,1) -> R such that f_n(x) converges ( pointwise) ) to a function f: (0,1) -> R, but the sequence of derivatives f'_n does not converge ( pointwise) at any point of the interval (0,1). ( Hint: You might want to think of a sequence of some sort of polynomials on some interval, but please show me step by step and prove any claims).

5). There is a sequence of functions f_n : [a,b] -> R such that for each n in N, f_n is DRS-integrable over [a,b] with respect to alpha(x) = x, and f_n(x) converges (pointwise) to a function f: [a,b] -> R, but limit n to infinity of integral from a to b ( f_n(x) dx doesn't equal to the integral from a to b of limit n goes to infinity of f_n(x) dx.

© BrainMass Inc. brainmass.com October 9, 2019, 4:42 pm ad1c9bdddfhttps://brainmass.com/math/real-analysis/closure-convergence-differentiability-integrability-sequence-39396

#### Solution Preview

Mathematics, Calculus

Other

Prove/Disprove convergence.

--------------------------------------------------------------------------------

Advanced Calculus/Real Analysis

Prove or disprove the following statement. Provide detailed answer and justify all steps.

1). There is a nonempty set S in R such that closure of S is equal to R and the closure of its complement closure(R-S) also is equal to R.

My thoughts on this problem: Q ( rationals) and R-Q (irrationals), but how to prove that the closure of Q is R and that the closure of R-Q is R also.

I gave a first try this to you in a previous note -- hope you got it. You are quite right about Q and I = R - Q, excellent startt in between every two rationals there is an irrational and vice versa. Let a, b be any two distinct rational numbers. Take c = average of a, b so a<c<b. These three numbers are not equal so there must be a first decimal place, say the nth place, where the digits of b and c are not equal and the truncations to the nth place a'<c'<b' Then to the truncated c' append 0100100010001. . . to create c" which is clearly irrational. and a<c"<b

Thus the closure of Q must include I. Knowing that R is complete and that R = Q + I (R is defined as all possible decimal numbers, Q all ...

#### Solution Summary

Closure, Convergence, Differentiability, Integrability and Sequences are investigated. The collection of closed intervals are determined.