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, such that $\tau_1$ 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
Let $\CC_1$ be the set of closed sets $C$ in the topological space $\struct {X, \tau_1}$ such that $S \subseteq C$.

Let $\CC_2$ be the set of closed sets $C$ in the topological space $\struct {X, \tau_2}$ such that $S \subseteq C$.

Let $C \in \CC_1$.

Then from Closed Set in Coarser Topology is Closed in Finer Topology:
 * $C \in \CC_2$

Hence by definition of subset:
 * $\CC_1 \subseteq \CC_2$

From Intersection is Decreasing:


 * $\bigcap \CC_2 \subseteq \bigcap \CC_1$

From the definition of topological closure:
 * $\cl_1 := \bigcap \CC_1$

and:
 * $\cl_2 := \bigcap \CC_2$

Hence the result:
 * $\map {\cl_2} S \subseteq \map {\cl_1} S$