[증명] f is continuous iff the graph is closed

Theorem

Let f : X → Y; let Y be compact Hausdorff. f is continuous if and only if the graph of f, Gf = { x × f(x) | x ∈ X } is closed in X × Y.

 

Proof

 

Start with f is continuous. (the if only part)

Fix a point x ∈ X. Let y be a point in Y such that y₀ ≠ f(x) (so that x × y ∉ Gf)

Let Nf be a neighborhood of f(x) and Ny be a neighborhood of y in Y such that Nf ∩ Ny = Ø. (They exist because Y is Hausdorff)

f is continuous, so f⁻¹(Nf) is open.

⇒ f⁻¹(Nf) × Ny is open.

 

For all x₀ × y₀ ∈ f⁻¹(Nf) × Ny, f(x₀) ∈ Nf, so f(x₀) ∉ Ny ⇒ f(x₀) ≠ y₀

⇒ f⁻¹(Nf) × Ny ∩ Gf = Ø ⇒ f⁻¹(Nf) × Ny ⊂ X × Y - Gf

 

⇒ for all x × y ∈ X × Y - Gf, there exists open set f⁻¹(Nf) × Ny such that x × y ∈ f⁻¹(Nf) × Ny ⊂ X × Y - Gf

⇒ X × Y - Gf is open

⇒ Gf is closed

 

 

Now, start with Gf is closed (the if part)

Let A be closed in Y

⇒ X × A is closed in X × Y (the product of two closed spaces is closed)

⇒ Gf ∩ X × A is closed in X × Y

⇒ π₁(Gf ∩ X × A) is closed in X (because when Y is compact, the projection is a closed map)

 

π₁(Gf ∩ X × A) = f⁻¹(A), so f⁻¹(A) is closed.

 

⇒ f is continuous.

 

End of Proof.

 

*Note that the only if part is possible when Y is Hausdorff, and the if part is possible when Y is compact.

For iff to hold, Y needs to be both compact and Hausdorff.

This theorem is also known as the CLOSED GRAPH THEOREM.