Topological Closure in Coarser Topology is Larger

Theorem
Let $X$ be a set.

Let $\tau_1$ and $\tau_2$ be topologies on $X$ such that $\tau_1 \subseteq \tau_2$.

That is, $\tau_1$ is coarser than $\tau_2$.

Let $S \subseteq X$.

Then we have:


 * $\map {\cl_2} S \subseteq \map {\cl_1} S$

where $\cl_1$ and $\cl_2$ denote topological closure in $\struct {X, \tau_1}$ and $\struct {X, \tau_2}$ respectively.

Proof
Since $\tau_1 \subseteq \tau_2$ we have:


 * $\set {C \text { closed set in } \struct {X, \tau_1} : S \subseteq C} \subseteq \set {C \text { closed set in } \struct {X, \tau_2} : S \subseteq C}$

from Closed Set in Coarser Topology is Closed in Finer Topology.

From Intersection is Decreasing, we have:


 * $\ds \bigcap \set {C \text { closed set in } \struct {X, \tau_2} : S \subseteq C} \subseteq \bigcap \set {C \text { closed set in } \struct {X, \tau_1} : S \subseteq C}$

From the definition of topological closure, we have:


 * $\map {\cl_2} S \subseteq \map {\cl_1} S$