Continuity theorem for the closure of a set

Given a continuous function $ f : X \to Y $ and a set $ A \subset X $, if a point $ x \in X $ belongs to the closure of the set $ A $ (i.e., $ x \in Cl(A) $), then the image of the point $ f(x) $ belongs to the closure of the image of the set $ f(A) $, i.e., $ f(x) \in Cl(f(A)) $.

In simpler terms, the theorem tells us that the continuity of $ f $ preserves the relationship of a point being part of the closure of a set.

If a point $ x $ is "close" (belongs to the closure) to a set $ A $, then its image $ f(x) $ will also be "close" (in the closure) to the image of $ A $.

A practical example
Proof

A practical example

Consider the continuous function $ f : \mathbb{R} \to \mathbb{R} $ defined as $ f(x) = x^2 $ in the topological space $ X =\mathbb{R} $ and the set $ A = (0, 2) \subseteq \mathbb{R} $.

$$ A = (0,2) $$

The closure of $ A $ is the set $ Cl(A) = [0, 2] $, since we include the boundary points $ x = 0 $ and $ x = 2 $, which don’t belong to $ A $ but are its limit points.

$$ Cl(A) = [0,2] $$

The image of the set $ A $ under the function $ f(x) = x^2 $ is $ f(A) = (0, 4) $, as $ f(x) = x^2 $ for $ x \in (0, 2) $ takes values from $ 0 $ to $ 4 $, excluding the endpoints.

$$ f(A) = (0,4) $$

The closure of the image $ f(A) $ is the closed interval $ Cl(f(A)) = [0, 4] $, as we add the points $ 0 $ and $ 4 $, which are the boundary values of $ f(x) = x^2 $ when $ x \to 0 $ and $ x \to 2 $, respectively.

$$ Cl(f(A)) = [0,4] $$

According to the theorem, if $ x \in Cl(A) $, then $ f(x) \in Cl(f(A)) $.

For $ x = 0 \in Cl(A) $, the image $ f(0) = 0 \in Cl(f(A)) $
For $ x = 2 \in Cl(A) $, the image $ f(2) = 4 \in Cl(f(A)) $
For any $ 0 < x < 2 \in Cl(A) $, the image $ f(x) \in Cl(f(A)) $

This confirms the theorem: for every point $ x $ in the closure of the set $ A $, the image $ f(x) $ belongs to the closure of the image $ Cl(f(A)) $.

Proof

Let $ f : X \to Y $ be a continuous function, with $ x \in X $ and $ A \subset X $.

We are assuming that the image of $ x $ does not lie in the closure of $ f(A) $ in $ Y $.

$$ f(x) \notin Cl(f(A)) $$

Since $ f(x) \notin Cl(f(A)) $, there exists an open neighborhood $ B \subseteq Y $ that contains $ f(x) $ and satisfies $ B \cap f(A) = \emptyset $.

This follows directly from the definition of closure: if $ f(x) \notin Cl(f(A)) $, there must be an open neighborhood around $ f(x) $ that does not intersect $ f(A) $.

Because $ f $ is continuous, the preimage $ f^{-1}(B) $ of this open set $ B $ is an open neighborhood of $ x $ in $ X $.

This is guaranteed by the continuity of $ f $, which ensures that open sets are mapped to open sets in the preimage.

Since $ B \cap f(A) = \emptyset $, the preimage neighborhood $ f^{-1}(B) $ does not intersect $ A $, meaning $ f^{-1}(B) \cap A = \emptyset $.

In other words, the open neighborhood $ f^{-1}(B) $ of $ x $ in $ X $ contains no points from $ A $.

Therefore, because there is an open neighborhood of $ x $ that does not intersect $ A $, by the definition of closure, we conclude that $ x $ does not belong to the closure of $ A $, i.e., $ x \notin Cl(A) $.

Thus, we have shown that if $ f(x) \notin Cl(f(A)) $, then $ x \notin Cl(A) $.

Note: The proof relies on the fact that if the image of $ x $ under $ f $ is not "close" to $ f(A) $ (meaning it does not lie in the closure of $ f(A) $), then $ x $ itself cannot be "close" to $ A $ (meaning it does not lie in the closure of $ A $).

And so on.