Let $K, L$ compacts subsets of $\mathbb{R}$ with usual topology, and $ĸ\cap L = \emptyset$ prove that, exists $G, H$ open subsets Let $K, L$ compacts subsets of $\mathbb{R}$ with usual topology, and $ĸ\cap L = \emptyset$ prove that, exists $G, H$ open subsets of R that $K  \subset G,  L  \subset H$ then $cl(G)\cap cl(H) = \emptyset$
Hi, i need help with that excersice, to find the mistake in my solution, i just did it yesterday, but my teacher said that is'nt right, he told me that i can not assume the position and form of G and H, but i don't have any better idea...
i will let my proof.
Giben $K$ and $L$ compacts on $\mathbb{R}$ with usual topolgy, then by Heine-Borel-Lebesgue theorem, $K, L$ are closed, for that reason because they are compact at $\mathbb{R}$, they must be intervals, and we have two cases:
$K$ is left to $L$, and, $L$ is left to  $K$, because are intervals.
I'll prove the first case:
$K$ is left to $L$, then we can say that max($K$)<min($L$), now we'll calculate $d(\max($K$), \min($L$))=\varepsilon$, now just to be secure, let's take $\varepsilon/3$ and create these subsets:
$G= (\min(L) - \varepsilon/3, \max(K) + \varepsilon/3)$,
$H= (\min(K) - \varepsilon/3, \max(L) + \varepsilon/3)$
and is obviusly that $K  \subset G$, $L\subset H$, the we have that
$\overline{G}= [\min(L) - \varepsilon/3, \max(K) + \varepsilon/3]$,
$\overline{H}= [\min(K) - \varepsilon/3, \max(L) + \varepsilon/3]$
then because $\max(K) =\min(L) - d(\max(K), \min(L))$, and $\max(K) =\min(L) - \varepsilon$, we have
$\max(K) =\min(L) - d(\max(K), \min(L))$
$\max(K) < \min(L)) -2\varepsilon/3$
$\max(K) + \varepsilon/3 < \min(L)) -\varepsilon/3$
then $\overline{G}\cap \overline{H} = \emptyset$
 A: You get into trouble right away, when you say that because $K$ and $L$ are compact, they must be intervals. This is very far from true. For instance, every finite subset of $\Bbb R$ is compact, as is the set $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$.
HINT: What you should do first is show that for each $x\in K$ there are open sets $U_x$ and $V_x$ such that $x\in U_x$, $L\subseteq V_x$, and $U_x\cap V_x=\varnothing$. To do this, use the fact that $\Bbb R$ is Hausdorff: for each $y\in L$ there are open sets $G_y$ and $H_y$ such that $x\in G_y$, $y\in H_y$, and $G_y\cap H_y=\varnothing$. $L$ is compact, and $\{H_y:y\in L\}$ is an open cover of $L$, so there is a finite $F\subseteq L$ such that $\{H_y:y\in F\}$ covers $L$. Let $U_x=\bigcap_{y\in F}G_y$ and $V_x=\bigcup_{y\in F}H_y$; I’ll leave it to you to verify that $U_x$ and $V_x$ are open, $x\in U_x$, $L\subseteq V_x$, and $U_x\cap V_x=\varnothing$.
Now try to use a very similar idea to show that there are open sets $U$ and $V$ such that $K\subseteq U$, $L\subseteq V$, and $U\cap V=\varnothing$.
