Bijection is Open iff Closed

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Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: T_1 \to T_2$ be a bijection.


Then $f$ is open if and only if $f$ is closed.


Proof

Let $f$ be a bijection.

Suppose $f$ is an open mapping.

From the definition of open mapping:

$\forall H \in \tau_1: f \sqbrk H \in \tau_2$

As $f$ is a bijection:

$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$

By definition of closed set:

$S_1 \setminus H$ is closed in $T_1$

and as $f$ is an open mapping:

$f \sqbrk {S_1 \setminus H} = S_2 \setminus f \sqbrk H$

is closed in $T_2$.

Hence by definition $f$ is a closed mapping.


A similar argument demonstrates that if $f$ is closed then it is open.

$\blacksquare$


Also see


Sources