# Variation on Lebesgue Number Lemma. Proof check.

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. – Asaf Karagila Feb 24 '12 at 0:36
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. – Jake G. Feb 24 '12 at 0:56
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. – Asaf Karagila Feb 24 '12 at 1:04

So if $x\in f^{-1}[I_n]$, then for each $n=1,\dots,j$ there is an $\varepsilon_1,\dots,\varepsilon_j$ such that $$(x-\varepsilon_n,x+\varepsilon_n)\subseteq f^{-1}[I_n]\;.$$
Look at what you’ve actually said: if $x$ is in a particular $f^{-1}[I_n]$, then that same $x$ is in an open interval within each $f^{-1}[I_k]$ as $k$ runs from $1$ through $j$. You’re talking about a single fixed $n$ in the hypothesis of the sentence and about all of them at once in the conclusion. Moreover, this is a claim about a single fixed $x$; you very likely need different radii for other $x$’s. Thus, you cannot go on to conclude that these $\varepsilon_n$’s have anything to do with any other point of $K$.
HINT: Don’t try to work directly with the sets $f^{-1}[I_n]$. Instead, for each $x\in K$ find $\varepsilon_x>0$ such that $(x-\varepsilon_x,x+\varepsilon_x)$ is a subset of some $f^{-1}[I_n]$; it doesn’t matter which one. For each $x\in K$ let $$B(x)=\left(x-\frac{\varepsilon_x}2,x+\frac{\varepsilon_x}2\right)\;,$$ take a finite subcover of $\{B(x):x\in K\}$, and try to use that to get your $\delta$.