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I would like to show the following:

"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 $n$."

After corrections:

Since $f$ is continuous on the open domain $\mathbb{R}$, then we know each $f^{-1}(I_n)$ is open.

Now, for each $k \in K$, there is $\varepsilon_k > 0$ such that $(k - \varepsilon_k, k + \varepsilon_k) \subseteq f^{-1}(I_n)$ for some $n$.

Now, $K \subseteq \bigcup_{k \in K}(k - \varepsilon_k/2, k + \varepsilon_k/2)$. And since $K$ is compact, $K \subseteq \bigcup_{i=1}^{n}(k_i - \varepsilon_i/2, k_i + \varepsilon_i/2)$.

Set $\delta = \min\{\varepsilon_1/2, \dots, \varepsilon_n/2\}$.

Let $x \in K$. Then, $x \in (k_j - \varepsilon_j, k_j + \varepsilon_j)$ for some $j$. Let $t \in (x - \delta, x + \delta)$.

Then, $|t - k_j| \leq |t - x| + |x - k_j| \leq \delta + \varepsilon_j/2 \leq \varepsilon_j/2 + \varepsilon_j/2 = \varepsilon_j$.

Thus, $t \in (k_j - \varepsilon_j, k_j + \varepsilon_j)$.

So, $(x - \delta, x + \delta) \subseteq (k_j - \varepsilon_j, k_j + \varepsilon_j) \subseteq f^{-1}(I_n)$

Therefore, there is a number $\delta > 0$ such that for every $x \in K$, $(x - \delta, x +\delta)$ is contained in $f^{-1}(I_n)$ for some $n$.

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    [Good read.](http://meta.math.stackexchange.com/questions/3589/is-it-ok-to-ask-whether-my-proof-solution-is-correct-if-i-dont-have-any-particu/3590#3590)2012-02-24
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    I tried to do this as suggested. However, I'm a new user so I can't post an answer to my own question for 8 hours.2012-02-24
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    Even I, as a high reputation user cannot immediately post an answer to my own question. If no one answers in the next eight hours, you should do as suggested. Next time, simply indicates that you intend to post your solution as soon as the software allows you to and either ask people to wait or see what do they come up with in the meantime.2012-02-24

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