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**friday616** Let f:R --> R be a function that satisfies f(x+y)=f(x)+f(y) for all x, y as an element in R.

(a) Suppose that f is continuous at some point c. Prove that f is continuous on R.

f is cont. at x = 0: let {x_n} be any sequence converging to zero ==> the sequence {x_n + c} converges to c ==> by cont. of f at c we get:

f(x_n + c) --> f(c) ; but f(x_n + c) = f(x_n) + f(c), so in fact:

f(x_n) + f(c) --> f(c) ==> by arithmetic of limits we get f(x_n) --> 0.

Since f(0) = f(0 + 0) = f(0) + f(0) ==> f(0) = 0 , and then we proved

f(x_n) --> f(0) = 0 for any seq. converging to zero.

Let now w be any real number and let {y_n} be any seq. s.t. y_n --> w

==> y_n - w --> 0 ==> by cont. of f at zero we get:

f(y_n - w) --> f(0) = 0 and as before f(y_n - w) = f(y_n) - f(w) --> 0

==> f(y_n) --> f(w) and we're done.

(b) Suppose that f is continuous on R and that f(1)=k. Prove that f(x) = kx for all x as an element in R.