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 broadly the intermediate value theorem.
This image is bounded.
This image interval is closed, so f attains both its bounds.

1 Proof of assertion 1
2 Proof of assertion 2
3 Proof of assertion 3
4 Caveats

[edit] Proof of assertion 1

See: Intermediate value theorem#Proof

[edit] 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 A_n+1 ⊂ A_n to be a closed interval of width δ_n < 1/n on which f is unbounded.

By the nested interval property, the intersection

$B = \bigcap_{n=1}^{\infty} A_n$

is non-empty, so define x₀ to be a point in B. This point is in each A_n, and any other point in any A_n is at most 1/n away from x₀. Letting

$C_n = \left(x_0 - \frac{2}{n}, x_0 + \frac{2}{n}\right),$

we have A_n ⊂ C_n, so f is unbounded on C_n. 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.

[edit] 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 $f (x) < M$ for $x \in [a,b]$

Let $A = \{ x \in \mathbb{R} : f(x) \leq M \}$ Now A will also have least upper bound m, this must be in [a,b], as $A \subseteq [a,b]$ and [a,b] contains all its limit points.

This point will then be the maximum as $f(m) \leq M \mbox{ as } m \in [a,b]$ Further $f(m) \geq M$ 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 $f (m) = M$

Similarly considering $g: [a,b] \to \mathbb{R}, \ g(x) = -f(x)$

and noting that $\ \ \max g(x) = \min f(x)$ 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.

[edit] Caveats

It is important to note that this theorem only applies to continuous real functions. It does not apply to functions with function domain the rational numbers. As the rationals do not satisfy the least upper bound axiom, they are not complete.

To illustrate, this consider

$f: [0,2] \cap \mathbb{Q} \to \mathbb{R}$