Samuelb88
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Theorem. Suppose that f(x) is continuous on [a,b], a, b \in \mathbb{R}. Then \exists c, a<c<b such that f(c)=0.
Proof. Let c=\sup\{ x : f(x) < 0 \} and observe that c<0. Now let x_n \rightarrow c as n \rightarrow \infty and observe that x_n < 0. Get by transmission of continuity that f(x_n) \rightarrow f(c) as n \rightarrow \infty. Thus we have f(c) = \lim_{n \rightarrow \infty} f(x_n) \leq 0. Now suppose that f(c) < 0, we will derive a contradiction. Now we know that \exists \epsilon such that \forall y \in (c-\epsilon, c+\epsilon), f(y) < 0. Then either y<c<0 or c<y<0. Suppose that c<y<0, but this would contradict our definition of c. Therefore we conclude that our initial supposition was wrong and therefore f(c)=0, as required.
Here's my question: I think I'm missing something about the definition of c, that is, how can we define c to be the least upper bound of the negative values of a continuous function since we can get as close as we want to zero without ever reaching zero?
Proof. Let c=\sup\{ x : f(x) < 0 \} and observe that c<0. Now let x_n \rightarrow c as n \rightarrow \infty and observe that x_n < 0. Get by transmission of continuity that f(x_n) \rightarrow f(c) as n \rightarrow \infty. Thus we have f(c) = \lim_{n \rightarrow \infty} f(x_n) \leq 0. Now suppose that f(c) < 0, we will derive a contradiction. Now we know that \exists \epsilon such that \forall y \in (c-\epsilon, c+\epsilon), f(y) < 0. Then either y<c<0 or c<y<0. Suppose that c<y<0, but this would contradict our definition of c. Therefore we conclude that our initial supposition was wrong and therefore f(c)=0, as required.
Here's my question: I think I'm missing something about the definition of c, that is, how can we define c to be the least upper bound of the negative values of a continuous function since we can get as close as we want to zero without ever reaching zero?