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Let $X$ be a Hausdorff space, and $K\subset X$ compact, and $U_1,U_2\subset X$ open sets, and suppose that $K\subset U_1\cup U_2$, show that exist $K_1\subset U_1$ and $K_2\subset U_2$ compact in $X$ such that $K=K_1\cup K_2$ and $K_1\cap K_2=\emptyset$.

I was able to almost prove it, if $U_1\cap U_2=\emptyset$, then I can use that K is $T_4$, and use the propertie that $K\setminus U_1$ and $K\setminus U_2$ are disjoint compact sets because they are closed subsets of $K$, and then define $K\setminus U_2=K_1$, and $K\setminus U_1=K_2$.

Then for the case that $U_1$ and $U_2$ are not disjoint I've tried to follow a tip from the exercise, I've found open sets $V_1$ and $V_2$, that $K\setminus U_1\subset V_1$ and $K\setminus U_2\subset V_2$, and then set $K_1=K\setminus V_1$ and $K_2=K\setminus V_2$. But then things become a mess, and I could not prove the result from that. So I would be glad if anyone could help to finish this proof.

Thanks in Advance.

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    $\begingroup$ Are you sure that $K_1\cap K_2$ is demanded to be empty? I would rather expect the demand that $K_i\subseteq U_i$ for $i=1,2$. In your question the covering by the $U_i$ seems to be irrelevant. You can just take $K_1=K$ and $K_2=\varnothing$. $\endgroup$ – drhab Jul 25 '17 at 13:37
  • $\begingroup$ Sorry, I will fix, one of the conditions is to $K_i \subset U_i$, but the textbook says that $K_1\cap K_2=\emptyset$, I've tried a lot, but this part is the one causing problems, I think it may be wrong, but I couldnt prove also $\endgroup$ – e.turatti Jul 25 '17 at 13:43
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    $\begingroup$ But if e.g. $X=K=[0,3]$ and $U_1=[0,2)$, $U_2=(1,3]$. Then the existence of compact (hence closed) $K_i$ with $K_1\cup K_2=[0,3]$ and $K_1\cap K_2=\varnothing$ and $K_i\subseteq U_i$ tells us that $[0,3]$ is not connected... $\endgroup$ – drhab Jul 25 '17 at 13:51
  • $\begingroup$ Perfect, Thanks! Do you want to answer this as a counter example? Then I give the right question to it. $\endgroup$ – e.turatti Jul 25 '17 at 13:53
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The statement cannot be true.

E.g. let $X=K=[0,3]$ and $U_1=[0,2)$, $U_2=(1,3]$.

Then the existence of compact (hence closed) $K_i$ with $K_1\cup K_2=[0,3]$ and $K_1\cap K_2=\varnothing$ and $K_i\subseteq U_i$ (leading to $K_i\neq K$) for $i=1,2$ tells us that $[0,3]$ is not connected.

However, it is a well known fact that $[0,3]$ is connected.

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  • $\begingroup$ Just about what I would have said. I would have let $X=K=[0,1]$ with $U_1=[0,1)$ and $U_2=(0,1].$.......................+1 $\endgroup$ – DanielWainfleet Jul 25 '17 at 14:36

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