# 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 Intervalâ€Ž, $J$ 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): $\S 9.8$