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:
- $\map {\cl_2} S \subseteq \map {\cl_1} S$
$\blacksquare$