Continuity Property
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Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a .. b}\right]$.
Then the image of $\left[{a .. b}\right]$ under $f$ is also a closed interval.
Proof
Let $I = \left[{a .. b}\right]$.
Let $J = f \left({I}\right)$.
- From Image of Interval by Continuous Function, $J$ is an interval.
- From Image of Closed Real Interval is Bounded, $J$ is bounded.
- From Max and Min of Function on Closed Real Interval‎, $J$ includes its end points.
Hence the result.
$\blacksquare$
Note
Not to be confused with the Continuum Property.
