Continuous Image of Closed Interval is Closed Interval
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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 is Interval, $J$ 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.
$\blacksquare$
Also known as
Some sources refer to this as the continuity property, but this is not standard, and is too easily confused with the Continuum Property.
Also see
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 9.8$