Continuity Defined by Closure

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: T_1 \to T_2$ be a mapping.

Then $f$ is continuous :
 * $\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$

where $H^-$ denotes the closure of $H$ in $T_1$.

That is, the image of the closure is a subset of the closure of the image.

Also see

 * Closure of Image under Continuous Mapping is not necessarily Image of Closure demonstrating that it is not necessarily the case that $f \sqbrk {H^-} = \paren {f \sqbrk H}^-$


 * Closed Image of Closure of Set under Continuous Mapping equals Closure of Image demonstrating that if $f \sqbrk {H^-}$ s closed then it does follow that $f \sqbrk {H^-} = \paren {f \sqbrk H}^-$