Continuity Defined from Closed Sets

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Theorem

Let $T_1$ and $T_2$ be topological spaces.

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


Then $f$ is continuous if and only if for all $V$ closed in $T_2$, $f^{-1} \sqbrk V$ is closed in $T_1$.


Proof

First we show the following.

Let $W \in T_2$.

We note that $f^{-1} \sqbrk T_2 = T_1$.

Hence, from Preimage of Set Difference under Mapping, we have:

$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$


Necessary Condition

Suppose the condition on closed sets holds.

Let $U$ be open in $T_2$.

By Relative Complement of Relative Complement, $T_2 \setminus \paren { T_2 \setminus U } = U$.

Therefore $T_2 \setminus \paren { T_2 \setminus U }$ is open in $T_2$.

Then $T_2 \setminus U$ is closed in $T_2$.

By hypothesis, $f^{-1} \sqbrk {T_2 \setminus U} = T_1 \setminus f^{-1} \sqbrk U$ is closed in $T_1$.

So $f^{-1} \sqbrk U$ is open in $T_1$.

This is true for any $U \in T_2$.

Hence $f$ is continuous.

$\Box$


Sufficient Condition

Now let $f$ be continuous.

Let $V$ be closed in $T_2$.

Then $T_2 \setminus V$ is open in $T_2$.

As $f$ is continuous, $f^{-1} \sqbrk {T_2 \setminus V} = T_1 \setminus f^{-1} \sqbrk V$ is open in $T_1$.

So $f^{-1} \sqbrk V$ is closed in $T_1$.

$\blacksquare$


Sources

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