Finding two compact sets.

Let $K$ be a compact set in a metric space $X$. Show that, if $K\subset U_1 \cup U_2$ for some open set $U_1$ and $U_2$, then there exist compact set $K_1$ and $K_2$ such that $K_1\subset U_1$, $K_2\subset U_2$ and $K=K_1\cup K_2.$

I think that, in general, it's not possible to find disjoint compact sets $K_1$ and $K_2$, and I think this is a counterexample: Question about compact set contained in an union of open sets.

I define $K_1$ and $K_2$ as follow.

$$K_1=K\cap ((U_1-\partial(U_2))\cup \partial(U_1))$$

$$K_2=K\cap ((U_1-\partial(U_1))\cup \partial(U_2))$$

By $\partial(U_1)$ I mean the boundary of $U_1$. To show $K_1$ is compact, I will show it's sequentially compact.

Let $(x_n)\in K\cap (U_1-\partial(U_2)$, as $K$ is compact, $(x_n)$ has a convergent sequence $(x_{n_k})\to \bar{x}\in{K}$.

If $\bar{x}\notin U_1$, then it's in $\partial({U_1})$. (it's a limit point of $U_1)$.

If $(x_n)$ be seqeuence in $K\cap\partial(U_1) \subset K$, as $K\cap\partial(U_1)$ and it's close set of compact set $K$, it's compact. Hence $K_1$ is compact.

Is my argument valid? Thank you.

• Is your definition of $K_2$ correct? – Lord Shark the Unknown Aug 11 '17 at 4:40
• Also, should it be specified that $U_1\cap U_2 = \emptyset$? As an aside, we use \partial to specify the boundary of a set, e.g. $\partial U_1$. – Michael Lee Aug 11 '17 at 4:47
• Your $K_1$ and $K_2$ are not subsets of $U_1$ and $U_2$. $K_1$ may contain part of the boundary of $U_1$, which is not in $U_1$. – Robert Israel Aug 11 '17 at 4:47
• @RobertIsrael Yes. I just added the boundaries to make them closed, but they are not subsets of $U_1$ and $U_2$ anymore. – Parisina Aug 11 '17 at 5:20
• @MichaelLee No they are not disjoint here. – Parisina Aug 11 '17 at 5:22

If $Y$ is a normal space and $\{O_1 ,O_2\}$ is an open cover of $Y$, then there is a closed cover $\{F_1, F_2\}$ of $Y$ such that $F_i \subseteq O_i$ for $i=1,2$.
Proof of the lemma: $C_1 =Y\setminus O_1$, and $C_2 = Y\setminus O_2$. Then $C_1$ and $C_2$ are disjoint closed sets in $Y$ so they have disjoint open neighbourhoods $N_1$ resp. $N_2$. Then for $i=1,2$, $F_i = Y \setminus N_i$ are as required.
Now your question follows almost immediately: if $X$ is metric and $K \subseteq U_2 \cup U_2$ is compact, then $Y = U_1 \cup U_2$ is a normal space (all I need of a metric space is that it is hereditarily normal), and so we find $F_i \subseteq Y$ as required. Then $K_i = K \cap F_i$ are compact (this is certainly true in $Y$ but compactness is absolute) and as the $F_i$ cover $Y$, these sets are as required.
• Very nice! I was working on a more (unnecessarily) complicated proof using the normality of $X$. I like yours much better. – Michael Lee Aug 11 '17 at 6:13