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[parent] least and greatest zero (Theorem)

Theorem. If a real function $ f$ is continuous on the interval $ [a,\,b]$ and has zeroes on this interval, then $ f$ has a least zero and a greatest zero.

Proof. If $ f(a) = 0$ then the assertion concerning the least zero is true. Let's assume therefore, that $ f(a) \neq 0$.

The set $ A = \{x\in [a,\,b]\vdots\,\, f(x) = 0\}$ is bounded from below since all numbers of $ A$ are greater than $ a$. Let the infimum of $ A$ be $ \xi$. Let us make the antithesis, that $ f(\xi) \neq 0$. Then, by the continuity of $ f$, there is a positive number $ \delta$ such that

$\displaystyle f(x) \neq 0\quad \mathrm{always\,when}\,\,\vert x-\xi\vert < \delta.$
Chose a number $ x_1$ between $ \xi$ and $ \xi\!+\!\delta$; then $ f(x_1) \neq 0$, but this number $ x_1$ is not a lower bound of $ A$. Therefore there exists a member $ a_1$ of $ A$ which is less than $ x_1$ ( $ \xi < a_1 < x_1$). Now $ \vert a_1-\xi\vert < \vert x_1-\xi\vert < \delta$, whence this member of $ A$ ought to satisfy that $ f(a_1) = 0$. This contains a contradiction. Thus the antithesis is wrong, and $ f(\xi) = 0$.

This means that $ \xi\in A$ and $ \xi$ is the least number of $ A$.

Analogically one shows that the supremum of $ A$ is the greatest zero of $ f$ on the interval.



"least and greatest zero" is owned by pahio.
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See Also: zeroes of analytic functions are isolated

Other names:  zeroes of continuous function
Keywords:  continuous real function

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Cross-references: supremum, least number, contradiction, lower bound, positive, numbers, bounded from below, interval, continuous, real function

This is version 3 of least and greatest zero, born on 2007-01-12, modified 2007-03-04.
Object id is 8742, canonical name is LeastAndGreatestZero.
Accessed 893 times total.

Classification:
AMS MSC: 26A15 (Real functions :: Functions of one variable :: Continuity and related questions )
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