Continuity

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Theorem 1 (continuity). Let $𝐴 \subset ℝ$ , a function $𝑓 : 𝐴 \to ℝ$ is continuous if and only if $\forall o p e n s e t 𝑈 \subset ℝ, \exists s o m e o p e n s e t 𝑉 \subset ℝ, s u c h t h a t 𝑓 - 1 (𝑈) = 𝑉 \cap 𝐴$ .

Proof. Let

𝑥 0 ​ \in 𝐴

𝜀 ​ > 0

. Setting

𝑈 ​ \subset ℝ ​ = (𝑓 (𝑥 0) ​ - 𝜀, 𝑓 (𝑥 0) ​ + 𝜀)

and

𝑓 - 1 (𝑈) ​ = 𝑉 ​ \cap 𝐴

, where

𝑉 ​ \subset ℝ

. Then

𝑥 0 ​ \in 𝑉

. Since

𝑉

𝑉

is an open set, there exists

(𝑥 0 ​ - 𝛿, 𝑥 0 ​ + 𝛿) ​ \subset 𝑉

. Because

𝑥 0 ​ \in 𝑉 ​ \cap 𝐴

𝑉 ​ \cap 𝐴

is non-empty. Hence,

\forall 𝑥 ​ \in ℝ ​ w h i c h m a k e s ​ ​ | 𝑥 ​ - 𝑥 0 ​ | ​ < 𝛿, ​ | 𝑓 (𝑥) ​ - 𝑓 (𝑥 0) ​ | ​ < 𝜀

. Thus,

\forall 𝜀 ​ > 0, \exists 𝛿 ​ > 0, ​ | 𝑥 ​ - 𝑥 0 ​ | ​ < 𝛿 ​ \Rightarrow ​ | 𝑓 (𝑥) ​ - 𝑓 (𝑥 0) ​ | ​ < 𝜀

.
The converse is similar and is left to the reader.

□

View Typst source

#import "@preview/noteworthy:0.4.0": noteworthy
#import "/themes/site/notes.typ": *
#show: web-equations
#import "../shared.typ": *

#show: tylenotes.with(
  title: "Continuity",
  date: "2026-08-11",
  tags: ("topology", "lean"),
  summary: "An equivalent characterization of continuity for functions on A ⊆ ℝ: the preimage of every open set is open in the subspace topology.",
)

#theorem(title: "continuity")[
  Let $A subset RR$,
  a function $f:A -> RR$ is continuous if and only if
  $ forall "open set" U subset RR, exists "some open set" V subset RR, "such that" f^(-1)(U)=V inter A$.
]

#proof[
  Let $x_0 in A$, $epsilon > 0$.
  Setting $U subset RR = (f(x_0)-epsilon, f(x_0)+epsilon)$ and $f^(-1)(U) = V inter A$,
  where $V subset RR$.
  Then $x_0 in V$.
  Since $V$ is an open set, there exists $(x_0-delta, x_0+delta) subset V$.
  Because $x_0 in V inter A$, $V inter A$ is non-empty. 
  Hence, $forall x in RR "which makes" |x - x_0| < delta, |f(x)-f(x_0)|<epsilon$.
  Thus, $forall epsilon > 0, exists delta > 0, |x - x_0| < delta => |f(x)-f(x_0)| < epsilon$. \
  The converse is similar and is left to the reader.
]