Vtyjkyuk

Let $K subset mathbbR $ be a compact set, and let $U_1cdots U_n ,nin mathbbN $ be finitely many open sets with $K subset U_1 cupcdots cup U_n$. Show that there exists $deltagt0 $ , for any $xin K$ the interval $(x-delta,x+delta)$ lies entirely in one of the sets $U_j$.

I was wondering if I can do this by contradiction.

If for any $deltagt0$, there exists $x_0in K$ such that the interval $(x-delta,x+delta)$ do not lie in any of the sets $U_j$ hence also do not lie in $K$. So $x_0$ is an isolated point. But $E$ is compact (i.e bounded and closed). Hence $x_0$ contradicts that $E$ is closed.

Any wrongness in the prove?

It seems the proof is not correct. I'll consider other ways to figure out. Every hints would be helpful. Thanks and regards.

Some comments on your answer

Your sentence: If for any $deltagt0$, there exists $x_0in K$ such that the interval $(x-delta,x+delta)$ do not lie in any of the sets $U_j$ hence also do not lie in $K$. So $x_0$ is an isolated point. But $K$ is compact (i.e bounded and closed). Hence $x_0$ contradicts that $K$ is closed.

1. The $x_0$ should not be denoted like that. It depends on $delta$. So better to denote it by $x_delta$.


2. 
   So $x_0$ is an isolated point. Why? You should provide a convincing argument.


3. 
   Hence $x_0$ contradicts that $K$ is closed. Why? A closed subset may have isolated points.

Proof of the requested result

For each $xin K$, there exists $delta_x >0$ such that $(x-delta_x,x+delta_x)$ is included in one of the $U_j$. The set of open intervals $S=(x-delta_x/2,x+delta_x/2) ; x in K$ is an open cover of $K$. As $K$ is compact, one can extract from $S$ a finite open subcover $overline S = (x_1-delta_x_1/2,x_1+delta_x_1/2), dots,(x_m-delta_x_m/2,x_m+delta_x_m/2)$.

$delta = dfrac12minlimits_1 le i le m delta_x_i$ is satisfying what we’re requested. Indeed for $x in K$, it exists $i in 1, dots m$ and $j in 1, dots n$such that $x in (x_i-delta_x_i,x_i+delta_x_i) subseteq U_j$.

Then
$$(x-delta , x+delta) subseteq (x_i-delta_x_i,x_i+delta_x_i) subseteq U_j.
$$

Hints: each $U_i$ is a union of open intervals. A finite number of these cover $K$. So the proof actually reduces to the case when each $U_i$ is an open interval. Arrange these intervals so that the left end points are in increasing order. Look at the intersection of adjoining intervals. Take $delta$ smaller than the lengths of these intersections.