Set is Closed in Metric Space iff Closed in Induced Topological Space

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $\tau$ be the topology induced by the metric $d$.

Let $F$ be a subset of $M$.

Then:

$F$ is closed in $M$ if and only if $F$ is closed in $\struct {A, \tau}$

Proof

By definition of a closed set in $M$:

$F$ is closed set in $M$ if and only if $A \setminus F$ is open in $M$

By definition of the topology $\tau$ induced by the metric $d$:

$A \setminus F$ is open in $M$ if and only if $A \setminus F$ is open in $\struct {A, \tau}$

By definition of a closed set in $\struct{A, \tau}$:

$A \setminus F$ is open in $\struct {A, \tau}$ if and only if $F$ is closed set in $\struct {A, \tau}$

The result follows.

$\blacksquare$