Continuity defined using closed sets

Given two topological spaces $ X $ and $ Y $, a function $ f: X \to Y $ is continuous if and only if the preimage of every closed set $ C \subseteq Y $ is a closed set in $ X $.

This theorem offers an alternative way to define the continuity of a function between two topological spaces.

Continuity is usually defined by stating that the preimage of every open set in $ Y $ must be open in $ X $.

However, this theorem shows that closed sets can be used as well: a function $ f: X \to Y $ is continuous if the preimage $ f^{-1}(C) $ of any closed set in $ Y $ is closed in $ X $.

Note: This highlights the duality between open and closed sets in the definition of continuity, as any closed set can be seen as the complement of an open set, and vice versa.

A practical example
The proof

A practical example

Let's take a function $ f: \mathbb{R} \to \mathbb{R} $ like $ f(x) = x^2 $, with the standard topology, where open sets are open intervals and their unions.

$$ f(x) = x^2 $$

We need to check that the preimage of every closed set in $ Y $ is closed in $ X $.

Consider a closed set in $ Y $, for example, $ C = [1, +\infty) \subseteq Y $, which is closed in $ Y $ because it includes its lower bound.

The preimage of $ C $ under $ f $ is:

$$ f^{-1}(C) = \{ x \in \mathbb{R} : f(x) = x^2 \in [1, +\infty) \} = (-\infty, -1] \cup [1, +\infty) $$

The set $ f^{-1}(C) = (-\infty, -1] \cup [1, +\infty) $ is closed in $ \mathbb{R} $ since intervals of the form $[-a, +\infty)$ are closed in the standard topology on $ \mathbb{R} $.

Since the preimage of $ [1, +\infty) $, which is a closed set in $ Y $, is closed in $ X $, the condition for continuity is satisfied.

By applying the same reasoning to any closed set in $ Y $, we can conclude that the function $ f(x) = x^2 $ is continuous.

The proof

The proof proceeds in two parts: first, we show that if $ f $ is continuous, then the preimage of every closed set is closed; second, we show that if the preimage of every closed set is closed, then $ f $ is continuous.

1] (⇒) If $ f $ is continuous, then $ f^{-1}(C) $ is closed for every closed set $ C \subseteq Y $:

By assumption, $ f $ is continuous. According to the standard definition, this means that the preimage of every open set in $ Y $ is open in $ X $.

Now, take a closed set $ C \subseteq Y $. Since $ C $ is closed, its complement $ Y \setminus C $ is open in $ Y $.

Because $ f $ is continuous, the preimage of $ Y \setminus C $, that is, $ f^{-1}(Y \setminus C) $, is open in $ X $.

But the preimage of a complement is the complement of the preimage, so $ f^{-1}(Y \setminus C) = X \setminus f^{-1}(C) $.

Therefore, $ X \setminus f^{-1}(C) $ is open in $ X $, which means that $ f^{-1}(C) $ is closed in $ X $.

This shows that if $ f $ is continuous, the preimage of every closed set is closed.

2] (⇐) If the preimage of every closed set is closed, then $ f $ is continuous:

Assume that the preimage of every closed set in $ Y $ is closed in $ X $.

Now, consider an open set $ U \subseteq Y $. We need to prove that $ f^{-1}(U) $ is open in $ X $.

Since $ U $ is open, its complement $ Y \setminus U $ is closed in $ Y $.

By assumption, the preimage of $ Y \setminus U $, that is, $ f^{-1}(Y \setminus U) $, is closed in $ X $.

But $ f^{-1}(Y \setminus U) = X \setminus f^{-1}(U) $, so $ X \setminus f^{-1}(U) $ is closed in $ X $.

This implies that $ f^{-1}(U) $ is open in $ X $, since the complement of a closed set is open.

Therefore, the function $ f $ is continuous.

3] Conclusion

We have proven both directions. Thus, a function $ f: X \to Y $ is continuous if and only if the preimage of every closed set $ C \subseteq Y $ is closed in $ X $.

And so forth.