1549901369-Elements_of_Real_Analysis__Denlinger_

628 Appendix B • Sets and Functions

(11) VJ,g E F(S, JR), Jg E F(S, JR);

(12) VJ, g, h E F(S, JR), f(gh) = (fg)h;

(13) VJ, g E F(S, JR), Jg= gf;

(14) VJ, g, h E F(S, JR), f(g + h) =Jg + fh;

(15) Vf,g E F(S, JR), andVr E JR, r(fg) = (rf)g = J(rg);

Proof. (1) Let f,g E F(S, JR). By Definition B.3.1 (a), f + g E F(S, JR).

(2) Let f, g, h E F(S, JR). Then, Vx ES,

[f + (g + h)](x) = f(x) + (g + h)(x) by definition off+ (g + h)

= f(x) + [g(x) + h(x)] by definition of g + h

= [f(x) + g(x)] + h(x)] by axiom (A2) of JR

= (f + g)(x) + h(x) by definition off+ g

= [(! + g) + h](x) by definition of (f + g) + h.

Thus, by Definition B.2.2, f + (g + h) = (f + g) + h.

(3) and (13) Exercise 5.

(4) Define the function 0 E F(S, JR) by the rule: Vx ES, O(x) = 0 ER

Let f E F(S, JR). Then , Vx E S ,

(0 + J)(x) = O(x) + f(x) by definition of 0 + f

= 0 + f(x) by definition of the 0 function

= f(x) by axiom (A3) of R

Thus, by Definition B.2.2, 0 + f = f. Also, f + 0 = f, by Part (c) above.

(5) Exercise 6.

(6) See proof of (1) above.

(7) Exercise 7.

(8) Let f , E F(S, JR) and r, s ER Then, Vx ES,

[(r + s)J](x) = (r + s) · J(x) by definition of (r + s)f

= r · f(x) + s · f(x) by axiom (D) of JR

= (r f + sf)(x) by definition of r f +sf.

Thus, by Definition B.2.2, (r + s)f =fr+ sf.

( 9) Exercise 8.

(10) Exercise 9.

( 11) Exercise 10.