# Let $X$ be a topological space and $A,B \subseteq X$ such that $A \cup B = X$ and $A \cap B = \emptyset$ then…

Let $X$ be a topological space and $A,B \subseteq X$ such that $A \cup B = X$ and $A \cap B = \emptyset$ then show or give a counterexample of:

• $A$ no compact, $B$ compact then $X$ compact
• $A$ no compact, $B$ compact then $X$ no compact
• $A$ no compact, $B$ no compact then $X$ no compact.

My attemp:

• Counterexample: $B = [0,1]$ , $A= (1,2)$
• Counterexample: $B = [0,1]$ , $A = (1,2]$
• Proof:

Let $\mathcal{U} = \{ U_i | i \in I\}$ open cover of $A$, then $\exists Q_i$ open sets of $X$, such that $U_i = A \cap Q_i$

Let $\mathcal{V} = \{ V_j | j \in J \}$ open cover of $B$, then $\exists Q_j$ open sets of $X$, such that $V_j = B \cap Q_j$

$A\cup B = X \implies \{ Q_a | a \in I \cup J\}$ is an open cover with no finite subcover $\implies$ X is not compact. (If $\{ Q_a | a \in I \cup J\}$ has a finite subcover $\{ Q_1, ... , Q_n\}$ then we can find finite subsets for $A$ and $b$, (namely $\{ Q_1 \cap A, ... Q_n \cap A\}$ and $\{ Q_1 \cap A, ... Q_n \cap A\}$)

I know this proof is wrong, because it allows to show that "$A$ no compact, $B$ compact then $X$ no compact" and I have a counterexample for that. What is wrong in my proof? Can you give a correct proof?

• Your second countereample is a counterexample to which statement ? – Amr Jan 20 '14 at 16:41
• A counter exemple for the last one could be $A=[0,1)\cup \{2\},\ B=[1,2)$ – aaaaa Jan 20 '14 at 16:43

If $\{Q_a\mid a\in I\cup J\}$ has a finite subcover $Q_1,...,Q_n$ for $A\cup B$, then some of the indices $1,...,n$ belong to $I$ and some belong to $J$. Therefore it would be wise to replace them by $i_1,...,i_m,j_1,...,j_n$ where $i_k\in I$ and $j_l\in J$. Then although the $\{Q_{i_k}\cap A,\ Q_{j_l}\cap A\ \mid k=1,...,m,\ l=1,...,n\}$ form a finite cover for $A$, the sets $Q_{i_k}\cap A$ may not. The sets $Q_{j}\cap A, j\in J$ were not among the $U_i, i\in I$.
The last one is also false. For a counterexample, consider $X = [0,1]$, $A = X\cap \mathbb{Q}$ and $B = X\setminus \mathbb{Q}$. Neither $A$ nor $B$ is compact, but $X$ is.