Continuous Image of Closed Interval is Closed Interval


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): $\S 9.8$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.