Set Closure Preserves Set Inclusion

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $B \subseteq A \subseteq S$.

Then:

$B^- \subseteq A^-$

where $A^-$ and $B^-$ are the set closures in $T$ of $A$ and $B$ respectively.

Proof

By definition 1 of set closure:

$B^- = B \cup B'$

where $B'$ is the derived set of $B$ in $T$.

Similarly:

$A^- = A \cup A'$

where $A'$ is the derived set of $A$ in $T$.

From Derived Set Preserves Set Inclusion:

$B' \subseteq A'$

So, by Set Union Preserves Subsets:

$B \cup B' \subseteq A \cup A'$

That is:

$B^- \subseteq A^-$

$\blacksquare$