# Showing two operations make a vector space

I'm asked to show that { $f:\mathbb R \rightarrow \mathbb R | f$ continuous} with operations

(a) $(f+g)(t)=f(t)+g(t)$

(b) $(rf)(t) = rf(t)$

make a vector space. I'm trying to show (i)-(iii) below, and I work through them, but I'm not sure if I'm understanding how they relate to (b) correctly.

(i) $r(v+w)=rv+rw$

$(r(f+g)(t))=r(f+g)(t)=r(f(t)+g(t))=r*f(t)+r*g(t)$

(ii) $(r+s)v=rv+sv$

$((r+s)f(t))=(r+s)f(t)=r*f(t)+s*f(t)$

(iii) $(rs)v = r(sv)$

$((rs)f(t))=(rs)*f(t)=r*s*f(t)=r(s*f(t))$

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What do you denote by $\ast$, multiplication in $\mathbb{R}$? If so, why do you introduce that symbol? –  k.stm Jan 22 '13 at 18:11
(a) and (b) + the set contains the zero vector is exactly how you show something is a space. Furthermore in your last line, you only need 1 scalar, not 2 –  Bob Jan 22 '13 at 18:12
@K.Stm, yeah it's multiplication in the reals, I just introduced it o make it clearer, but it may have been a bad idea. –  user1157605 Jan 22 '13 at 18:16
@Bob, can you clarify on my last line? I got the (i)-(iii) conditions from the book. Were my (i) and (ii) proofs correct, but my (iii) wrong? –  user1157605 Jan 22 '13 at 18:17
@user1157605 Your prrof of (iii) looks fine andit is necessary to show that (b) defines an operation. Don't you also need to check that $1v=v$? –  Hagen von Eitzen Jan 22 '13 at 18:20