Example III:
Consider the set C[a,b] of all continuous functions f defined on the closed interval [a,b] > R. Define vector addition:
2. Since R is a field, if r,s,t in R, then r+(s+t)=(r+s)+t.
Then for f,g,h, in C[a,b] and x in [a,b], f(x)+(g(x)+h(x))=((f(x)+g(x))+h(x) and therefore (f+g)+h = f+(g+h).
3. Consider the function 0, where for x in [a,b], 0(x)=0, 0 being the neutral element from R.
0 is in C[a,b], and for f in C[a,b] and x in [a,b],
0(x)+f(x)=0+f(x)=f(x) and hence 0+f=f.
4. For f in C[a,b] consider the function f,
defined by (f)(c)=(f(c)). f is in C[a,b] since it is defined from [a,b] to R and continuous.
5. Since R is a field, for r,s in R, r+s=s+r.
Then for f,g in C[a,b] and x in [a,b], f(x)+g(x)=g(x)+f(x) and hence f+g=g+f.
6. If r in R and f in C[a,b], then r*f is again a continuous function with values in R and hence an element of C[a,b].
7. Since R is a field, if r,s,t in R, r*(s*t)=(r*s)*t.
Then if r,s in R and f in C[a,b], for x in [a,b], (r*s*f(x))=r*(s*f(x)) and hence (r*s)*f = r*(s*f).
8. Since R is a field, 1*r=r for all r in R.
If f is in C[a,b], it follows for x in [a,b]: (1*f)(x)= 1*f(x)=f(x) and hence 1*f=f.
9. Since R is a field, if r,s,t in R then r*(s+t)=(r*s)+r*t.
Then for r in R, f,g in C[a,b], and x in [a,b], r*(f(x)+g(x))= (r*f(x)+r*g(x) and hence r*(f+g)=r*f+r*g.
10. Since R is a field, if r,s,t in R, then (r+s)*t=r*t+s*t.
Then for r,s in R, f in C[a,b] and x in [a,b], we have (r+s)f(x)=r*f(x)+s*f(x) and hence (r+s)*f=r*f+s*f.
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