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Let $f$: $\mathbb{R}\to\mathbb{R}$ be a continuous function. Suppose that $K$ is a compact subset of $\mathbb{R}$, and $f(K)$ $\subseteq$ $\bigcup_{n=1}^{\infty} I_n$, where each $I_n$ is an open interval. Prove that there is a number $\delta > 0$ that for every $x \in K$, $(x - \delta, x + \delta)$ is contained in $f^{-1}(I_n)$ for some $n$.

I will post my suggested proof as an answer. If incorrect, please comment.

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Choose $x\in K$; then $f(x)\in f(K)$ so there is some $n$ for which $f(x)\in I_n$; to wit, $x\in f^{-1}(I_n)$. Because $f$ is continuous, each $f^{-1}(I_n)$ is open. We have an open cover of a compact set; now look at its Lebesgue number.

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Lemma : Lebesgue number
Let $(X,d)$ a compact metric space and $(U_j)_{j\in J}$ an open cover of $X$.
It exists $\lambda>0$ such as $\forall x\in X$, $\exists j_0\in J$, $B(x,\lambda)\subset U_{j_0}$.
We call $\lambda$ the Lebesgue number of the open cover.

Proof : by reductio ad absurdum, suppose that this statement is false.
Then $\forall n\in\mathbb N^*$, it exists $x_n$ such as $B(x_n,\frac{1}{n})$ isn't in any $U_j$.
By compactness, it exists a subsequence $(x_{\varphi(n)})_n$ which converges to $x\in X$.
Let $j_0\in J$ such as $x\in U_{j_0}$. Let $\varepsilon>0$ such as $B(x,\varepsilon)\subset U_{j_0}$, then, after a certain rank, $B(x_{\varphi(n)},\frac{1}{\varphi(n)})\subset B(x,\varepsilon)\subset U_{j_0}$ and so you get a contradiction.
(First, it exists $N_0$ such as $n\ge N_0\Rightarrow\frac{1}{\varphi(n)}<\frac{\varepsilon}{2}$, next, it exists $N_1$ such as $n\ge N_1\Rightarrow\|x-x_{\varphi(n)}\|<\frac{\varepsilon}{2}$. Let $N=\max(N_0,N_1)$ and then for $n\ge N$, let $y\in B(x_{\varphi(n)},\frac{1}{\varphi(n)})$, then $\|x-y\|\le\|x-x_{\varphi(n)}\|+\|x_{\varphi(n)}-y\|\le\varepsilon$). $\blacksquare$

Your result :
By continuity, $f^{-1}(I_n)$ is an open set. Now, $f^{-1}(I_n)$ is an open cover of $K$ (Let $x \in K$, then $f(x)\in f(K)$, so it exists $n$ such as $f(x)\in I_n$ and finally $x\in f^{-1}(I_n)$), and let $\delta$ its Lebesgue number. You have your result : $\forall x\in K$, it exists $n\in\mathbb N$, such as $(x-\delta,x+\delta)\subset f^{-1}(I_n)$.

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