![]() However, the only way this holds for any $\epsilon \gt 0$, is for $f(c) = k$. Walk, College Math Journal, September 2011. For some interesting extra reading check out: The intermediate value theorem is NOT obviousand I am going to prove it to you, S.M. But the intermediate value would not be true in that case. Hence the Intermediate Value Theorem does not apply, and we can make no definitive statements concerning the question above. You can define continuous functions on the rational numbers. $$k - \epsilon \lt f(c) \lt k + \epsilon$$ It requires the property of completeness. In mathematical analysis, the intermediate value theorem states that if is a continuous function whose domain contains the interval a, b, then it takes on any given value between and at some point within the interval. (As above, the case not inferred by the continuity of $f$ is an obvious one.)įocusing on the right side of this string inequality, $f(x_1) \lt f(c) + \epsilon$, we subtract $\epsilon$ from both sides to obtain $f(x_1) - \epsilon \lt f(c)$. However, since $c - \delta \lt x_1 \lt c + \delta$, we also know that $f(c) - \epsilon \lt f(x_1) \lt f(c) + \epsilon$. If there wasn't, then $c$ would not have been the supremum of $S$ - some value to the right of $c$ would have been. There also must exist some $x_1 \in [c, c + \delta)$ where $f(x_1) \ge k$. Note, that in the case not inferred by the continuity of $f$ (i.e., when $x = c$), this just says $f(c) - \epsilon < f(c) < f(c) + \epsilon$, which should be obvious.įocusing on the left side of this string inequality, $f(c) - \epsilon \lt f(c)$, we add $\epsilon$ to both sides to obtain $f(c) \lt f(x_0) + \epsilon$. However, since $c-\delta \lt x_0 \lt c+\delta$, we also know that $f(c) - \epsilon \lt f(x_0) \lt f(c) + \epsilon$. If there wasn't, then $c$ would not have been the supremum of $S$ - some value to the left of $c$ would have been. Now that there must exist some $x_0 \in (c - \delta, c]$ where $f(x_0) \lt k$. Thus, by the epsilon-delta definition, we know that for any $\epsilon > 0$, we can find a $\delta > 0$ so that (with the exception of $x=c$) whenever $c-\delta \lt x \lt c+\delta$, we must have $f(c) - \epsilon \lt f(x) \lt f(c) + \epsilon$.Ĭonsider any such value $\epsilon \gt 0$ and the value of $\delta$ that goes with it. The case were $f(b) \lt k \lt f(a)$ is handled similarly.ĭefine a set $S = \ f(x) = f(c)$. ![]() Without loss of generality, let us assume that $k$ is between $f(a)$ and $f(b)$ in the following way: $f(a) \lt k \lt f(b)$. ![]() If $f(x)$ is continuous on $$ and $k$ is strictly between $f(a)$ and $f(b)$, then there exists some $c$ in $(a,b)$ where $f(c)=k$.
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