Continuity property
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In mathematics, the continuity property may be presented as follows.
- Suppose that f : [a, b] → R is a continuous function. Then the image f([a, b]) is a closed bounded interval.
The theorem is a stronger form of the intermediate value theorem, comprising the three assertions:
- The image f([a, b]) is an interval. This is the intermediate value theorem.
- This image is bounded.
- This image interval is closed, so f attains both its bounds.
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Proof of assertion 1
See: Intermediate value theorem#Proof
Proof of assertion 2
We proceed by contradiction. Suppose f is unbounded on some interval [a′, b′]. Then if a′ < c′ < b′, then f is unbounded on either [a′, c′] or [c′, b′]. This allows us to find an interval [y, y + δ] on which f is unbounded for arbitrarily small δ.
However, this contradicts the continuity of f. Let A1 be a closed interval of length δ1 < 1 on which f is unbounded. We recursively define An+1 ⊂ An to be a closed interval of width δn < 1/n on which f is unbounded.
By the nested interval property, the intersection
- <math>B = \bigcap_{n=1}^{\infty} A_n<math>
is non-empty, so define x0 to be a point in B. This point is in each An, and any other point in any An is at most 1/n away from x0. Letting
- <math>C_n = \left(x_0 - \frac{2}{n}, x_0 + \frac{2}{n}\right),<math>
we have An ⊂ Cn, so f is unbounded on Cn. However, 2/n can be made arbitrarily small, so for all ε and for all δ, there exists an x ∈ (x0 − δ, x0 + δ) such that |f(x) − f(x0)| > ε. Thus f is discontinuous at x0, a contradiction. Thus f([a, b]) must be bounded.
Proof of assertion 3
This comes from the least upper bound property of the real line
By the least upper bound axiom, we know that there is a minimum M such that <math> f(x) < M <math> for <math> x \in [a,b] <math>
Let <math> A = \{ x \in \mathbb{R} : f(x) \leq M \} <math> Now A will also have least upper bound m, this must be in [a,b], as <math> A \subseteq [a,b] <math> and [a,b] contains all its limit points.
This point will then be the maximum as <math> f(m) \leq M \mbox{ as } m \in [a,b] <math> Further <math> f(m) \geq M <math> as we can find a member of A, b such that f(b) is arbitrarily close to M otherwise M is not minimal, so if f(m)< M,M cannot be an upper bound.
Thus <math> f(m) = M <math>
Similarly considering <math>
g: [a,b] \to \mathbb{R}, \ g(x) = -f(x)
<math>
and noting that <math> \ \ \max g(x) = \min f(x) <math> ,we see f obtains its minimum
Thus f obtains its minimum and maximum at at least one point so, by the intermediate value theorem, it obtains all values in between
Caveats
It is important to note that this theorem only applies to continuous real functions. It does not apply to the rationals, as these do not satisfy the least upper bound axiom; they are not complete.
To illustrate this consider
<math> f: [0,2] \cap \mathbb{Q} \to \mathbb{R} <math>
<math>
x \mapsto e^{( - (x - \sqrt{2})^2 )}
<math>
f would obtain its maximum value at <math> \sqrt{2} <math> but this is not in the set.
If f is not continuous consider as a counterexample
<math> f: [0,1] \to \mathbb{R} <math>
<math> x \mapsto \begin{cases} \frac{1}{x} & \frac{1}{x} \in \mathbb{Z} \\ 0 & \mbox{otherwise} \end{cases} <math>
This is unbounded, but [0,1] is bounded.
Further one should reiterate that the set must be closed, otherwise the maximum and minimum values maybe not be obtained.