Continuous Image of Closed Interval is Closed Interval
Jump to navigation
Jump to search
This article is complete as far as it goes, but it could do with expansion. In particular: add topological proof using compactness and connectedness You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Expand}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
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): $\S 9.8$