Topological Closure in Coarser Topology is Larger

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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:

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

$\blacksquare$