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I assume "closed" and "open" refer to the topology on C([0,1],R) coming from the metric defined by the supremum norm, which I will denote by || ||. So to prove that A is closed in C([0,1],R), it is enough to show that whenever (f_n) is a sequence of functions in A that is convergent, to a function g in C([0,1],R), then g is also in A.

Suppose, for the purpose of finding a contradiction, that g is not in A. By the definition of A, we conclude that there must be some s in [0,1] with g(s) < 0. Since the sequence (f_n) converges to g in the uniform norm, putting the positive number -g(s)/2 in as "epsilon" in the "epsilon-N" definition of this statement, we learn that there is a positive integer N with the property that whenever n >= N, one has ||f_n - g|| < -g(s)/2. In particular, ||f_N - g|| < -g(s)/2. By definition of || || we have that ||f_N - g|| >= |f_N(s) - g(s)|, so we conclude that |f_N(s) - g(s)| < -g(s)/2. This implies that

g(s)/2 < f_N(s) - g(s) < -g(s)/2

and adding g(s) to both sides of this second inequality we learn that f_N(s) < g(s)/2. Since g(s) < 0 we deduce that f_N(s) < 0 also, and hence that f_N is not in A, contradicting the fact that the entire sequence (f_n) was assumed to be in A.

Alternatively, if you know a few theorems, you could do this more clearly without contradiction. Suppose that (f_n) is a sequence in A converging to g in C([0,1],R). Fix any t in [0,1]. Since uniform convergence implies pointwise convergence, the sequence of real numbers f_n(t) converges to the real number g(t). Since f_n is in A for all n, this sequence of real numbers f_n(t) is a sequence of nonnegative real numbers. Since the nonnegative real numbers are a closed subset of the real numbers, the limit g(t) is also nonnegative. Since t was an arbitrary element of [0,1] we deduce that g(t) >= 0 for all t in [0,1] and hence that g is in A.

[The first proof, by contradiction, is the same basic idea, only the necessary facts about uniform convergence implying pointwise convergence, and the closedness of the set of nonnegative real numbers in the real numbers, are sort of built into the proof, and not assumed.]

To see that A is not open in C([0,1],R) consider the function g in C([0,1],R) given by g(x) = 0 for all x. Clearly g is in A. Fix any e > 0, and define the function h by

h(x) = -e/2 for all x in [0,1].

It is clear that h is in C([0,1],R) (constant functions are evidently continuous), and ||g - h|| = e/2 < e, so h is is in the open ball of C([0,1],R) with radius e with center g. But h is not in A (since e.g. h(0) = -e/2 is not nonnegative). This shows that the open ball of C([0,1],R) with radius e and center g is not a subset of A. Since e > 0 was arbitrary, *no* open ball with center g is a subset of C([0,1],R). This shows that g is not in the interior of A and hence that A is not open.

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