Connected Subspace Lie in One Component of Separation

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $A_1 \mid A_2$ be a separation of $T$.

Let $X$ be a connected set of $T$.

Then there exist $i, j \in \set { 1, 2 }$ with $i \ne j$ such that $X \subseteq A_i$, and $X \cap A_j = \O$.

Proof
By definition of separation, $A_1$ and $A_2$ are disjoint.

Define $B_i = X \cap A_i$ for $i \in \set {1, 2}$.

From Intersection is Subset, $B_i \subseteq A_i$ for $i \in \set {1, 2}$.

From Subsets of Disjoint Sets are Disjoint, $B_1$ and $B_2$ are disjoint.

The union of $B_1$ and $B_2$ is:

By definition of separation, $A_1$ and $A_2$ are open in $T$.

Let $S_X = \struct {X, \tau_X}$ be the topological subspace of $T$.

Then $B_1$ and $B_2$ are open in $S_X$.

If both $B_1$ and $B_2$ are non-empty, they would form a separation of $S_X$, which contradicts the assumption that $X$ is connected.

Hence either $B_1 = \O$, or $B_2 = \O$.