# Proof: If $A\subseteq X$ Is Connected Then $A\subseteq U$ or $A\subseteq V$

Let $$X$$ be a topologic space. Let $$U,V\subseteq X$$ be open, disjoint such that $$U\cup V=X$$ and let $$A\subseteq X$$ be connected. Then $$A\subseteq U$$ or $$A\subseteq V$$

Let assume that $$A\not\subseteq U$$ or $$A\not\subseteq V$$, because $$U\cup V=X$$ we get $$A\cap V\neq \emptyset$$ and $$A\cap U\neq \emptyset$$ corresponding. $$A\cap V$$ and $$A\cap U$$ are open, non empty, disjoint sets such that: $$(A\cap U) \cup (A\cap V)=A\cap(U\cup V)=A\cap X = A$$

Therefore $$A$$ is not connected. Contradiction

Why can we assume that $$A\cap V$$ and $$A\cap U$$ are open? I understand that $$U,V$$ are open, but why $$A$$ is open?

• Sets aren't open. They're open in a space. In this case, $A\cap U$ need not be open in $X$, but it is open in $A$ (by definition). – Chrystomath Sep 14 '20 at 14:10

A subspace $$A$$ is connected iff we cannot write it as a disjoint union of relatively open (i.e. open in $$A$$!) subsets that are both non-empty.
$$A \cap U$$ and $$V \cap A$$ are by definition open in $$A$$ and they are disjoint (as $$U$$ and $$V$$ are) and they cover $$A$$ (all we need for that is $$A \subseteq U \cup V$$ and that is certainly the case here). So as $$A$$ is given to be connected one of them must be empty. If $$A \cap U = \emptyset$$ so $$A \cap V = A$$ which implies $$A \subseteq V$$. Likewise if $$A \cap V = \emptyset$$, $$A \subseteq U$$ and we're done.
When you speak about connectedness, you have to consider the subspace topology on $$A$$. Because $$U,V$$ are open in $$X$$, then by definition of this topology, $$A \cap U$$ and $$A \cap V$$ are open in $$A$$.
They are open in $$A$$, by the definition of the subspace topology. $$A$$ is not necessarily open in $$X$$; but it is open in itself.