- Continuity property
In

mathematics , the**continuity property**may be presented as follows.:Suppose that "f" : ["a", "b"] →

**R**is acontinuous function . Then the image "f"( ["a", "b"] ) is a closed bounded interval.The theorem is the combination of the

intermediate value theorem and theextreme value theorem , comprising the three assertions:

# The image "f"( ["a", "b"] ) is an interval.

# This image is bounded.

# This image interval is closed, so "f" attains both its bounds.The first assertion is broadly the intermediate value theorem; the latter two are equivalent to the extreme value theorem.

**Proof of assertion 1**"See:"

Intermediate value theorem#Proof **Proof of assertion 2**"See:"

Extreme value theorem#Proof of the boundedness theorem **Proof of assertion 3**"See:"

Extreme value theorem#Proof of the extreme value theorem **Caveats**It is important to note that this theorem only applies to

**continuous real functions**. It does not apply to functions withfunction domain the rational numbers. As the rationals do not satisfy the least upper bound axiom, they are not complete.To illustrate, this consider

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

$x\; mapsto\; e^\{(\; -\; (x\; -\; sqrt\{2\})^2\; )\}$

f would obtain its maximum value at $sqrt\{2\}$but this is not in the set.

If f is not continuous consider as a counterexample

$f:\; [0,1]\; o\; mathbb\{R\}$

$x\; mapsto\; egin\{cases\}frac\{1\}\{x\}\; frac\{1\}\{x\}\; in\; mathbb\{Z\}\; \backslash 0\; mbox\{otherwise\}end\{cases\}$

This is unbounded, but [0,1] is bounded.

Further, one should carefully note that the set must be closed, otherwise the maximum and minimum values might not be obtained.

**References*** Binmore, K. G. "Mathematical Analysis: A Straightforward Approach", Cambridge University Press, (1982). ISBN 0521288827.

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