# Let $f:(0,1)\to \mathbb{R}$ be a differentiable function.

Let $f:(0,1)\to \mathbb{R}$ be a differentiable function. Prove that if $f^{-1}({0})$ is uncountable, then there exists $c\in (0,1)$ such that $f(c)=f'(c)=0$.

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What have you tried? Also, have you considered Robert's response and tried to do anything with it? Since this is HW, we'd like to see that you have considered something. Regards –  Amzoti Jan 7 '13 at 18:51

Method 1

Claim Every uncountable subset of $(0,1)$ contains a limit point.

Proof: Let $S$ be an uncountable subset of $(0,1)$. If $S$ does not contain a limit point, then every $x \in S$ is isolated from the other elements of $S$, i.e., we can find an $\epsilon_{x} > 0$ such that $(x - \epsilon_{x},x + \epsilon_{x}) \cap (S \setminus \{ x \}) = \varnothing$. We can also choose $\epsilon_{x}$ sufficiently small so that $(x - \epsilon_{x},x + \epsilon_{x}) \subseteq (0,1)$. Then $\left\{ \left( x - \dfrac{1}{2} \epsilon_{x},x + \dfrac{1}{2} \epsilon_{x} \right) ~ \Bigg| ~ x \in S \right\}$ is an uncountable collection of disjoint non-empty open subintervals of $(0,1)$. We thus obtain a contradiction, because by picking a unique rational number from each subinterval, we end up with an uncountable collection of rational numbers --- an impossibility. Therefore, $S$ must contain a limit point. $\quad \spadesuit$

As ${f^{\leftarrow}}[\{ 0 \}]$ is assumed to be uncountable, it contains a limit point $c$, according to the claim. Hence, there is a sequence $(x_{n})_{n \in \mathbb{N}}$ in ${f^{\leftarrow}}[\{ 0 \}] \setminus \{ c \}$ that converges to $c$. As $f(x_{n}) = 0$ for each $n \in \mathbb{N}$, it follows that $$f'(c) = \lim_{n \to \infty} \frac{f(x_{n}) - f(c)}{x_{n} - c} = \lim_{n \to \infty} \frac{0 - 0}{x_{n} - c} = 0.$$

Conclusion: The limit point $c$ is the number that you want.

Method 2

This is based on the comment made above by copper.hat. Let $a \in (0,1)$, and suppose that $f(a) = 0$ and $f'(a) \neq 0$. For all $x$ sufficiently close to $a$ (but not equal to $a$), we have $$\left| \frac{f(x) - f(a)}{x - a} - f'(a) \right| < \frac{1}{2} \cdot f'(a).$$ It follows that $f(x) \neq f(a) = 0$ for all such $x$, so we can find an $\epsilon_{x} > 0$ such that $(x - \epsilon_{x},x + \epsilon_{x}) \cap ({f^{\leftarrow}}[\{ 0 \}] \setminus \{ x \}) = \varnothing$. Once again, $\epsilon_{x}$ can be chosen sufficiently small so that $(x - \epsilon_{x},x + \epsilon_{x}) \subseteq (0,1)$. Then $\left\{ \left( x - \dfrac{1}{2} \epsilon_{x},x + \dfrac{1}{2} \epsilon_{x} \right) ~ \Bigg| ~ x \in {f^{\leftarrow}}[\{ 0 \}] \right\}$ is an uncountable collection of disjoint non-empty open subintervals of $(0,1)$. As explained already in Method 1, this leads to a contradiction.

Conclusion: There must be a number $c \in (0,1)$ that satisfies the given properties.

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Hint: an uncountable set contains a point that is not isolated.

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Please explain by details –  aliakbar Jan 7 '13 at 18:35
If $f(x) = 0$ and $f'(x) \neq 0$, then $f(x) \neq 0$ in some punctured neighborhood $(x-\epsilon,x +\epsilon) \setminus \{x\}$ of $x$. –  copper.hat Jan 7 '13 at 19:19