Math 2.0

statement theorem

Boundedness theorem

Statement

Let a≤ba \le b and let f ⁣:[a,b]→Rf \colon [a, b] \to \mathbb{R} be continuous. Then ff is bounded. That is, there is KK with ∣f(x)∣≤K|f(x)| \le K for all x∈[a,b]x \in [a, b].

Remarks

Suppose not. Then there are points xn∈[a,b]x_n \in [a, b] with ∣f(xn)∣>n|f(x_n)| > n. By the Bolzano–Weierstrass theorem, a subsequence converges to some c∈[a,b]c \in [a, b]. Continuity at cc gives a contradiction. The interval must be closed and bounded. The function 1/x1/x on (0,1](0, 1] is continuous but not bounded. Mathlib states a form for compact sets, as IsCompact.bddAbove_image.

Explanations

L1 human

On a closed bounded interval, a continuous function cannot shoot off to infinity. Its graph stays between two horizontal lines.

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  1. Limit point of a set definition
  2. Sequence definition
  3. Limit of a function definition
  4. Real sequence definition
  5. Subsequence definition
  6. Continuity at a point definition
  7. Bounded sequence definition
  8. Limit of a sequence definition
  9. Monotone sequence definition
  10. Continuous function definition
  11. Convergent sequence definition
  12. Monotone convergence theorem for sequences statement
  13. Every sequence has a monotone subsequence statement
  14. Bolzano–Weierstrass theorem statement
  15. Boundedness theorem statement

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Sources

  1. Web https://en.wikipedia.org/wiki/Extreme_value_theorem, Proof of the boundedness theorem