Path-Connected Set in Subspace

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

Let $A \subseteq B \subseteq S$.

Let $T_B = \struct {B, \tau_B}$ be the topological space where $\tau_B$ is the subspace topology on $B$.

Then
 * $A$ is path-connected in $T_B$ $A$ is path-connected in $T$.

Proof
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.

Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$.

By the definition of a path-connected set, $A$ is path-connected in $T$ every two points in $A$ are path-connected in $\struct {A, \tau_A}$.

Similarly, $A$ is path-connected in $T_B$ every two points in $A$ are path-connected in $\struct {A, \tau'_A}$.

From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$.

Hence every two points in $A$ are path-connected in $\struct {A, \tau_A}$ every two points in $A$ are path-connected in $\struct {A, \tau'_A}$.